Question 15 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
AnswerA wave travelling along the positive x-direction is given as: $\text{y}_1=\text{a}\sin(\omega\text{t}-\text{kx})$The wave travelling along the negative x-direction is given as:
$\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})$ The superposition of these two waves yields: $\text{y}=\text{y}_1+\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})-\text{a}\sin(\omega\text{t}-\text{kx})$
$=\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})-\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})$
$=-2\text{a}\sin(\text{kx})\cos(\omega\text{t})$
$=-2\text{a}\sin\Big(\frac{2\pi}{\lambda}\text{x}\Big)\cos(2\pi\text{ vt})\ \dots(\text{i})$ The transverse displacement of the string is given as: $\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})\ \dots(\text{ii})$ Comparing equations (i) and (ii), we have: $\frac{2\pi}{\lambda}=\frac{2\pi}{3}$
$\therefore$ Wavelength, $\lambda=3\text{m}$ It is given that: $120\pi=2\pi\text{v}$ Frequency, $\text{v} = 60\text{Hz}$ Wave speed, $\text{v}=\text{v}\lambda$
$=60\times3=180\text{m/s}$
View full question & answer→Question 25 Marks
A pipe $20cm$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430Hz$ source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340m s^{–1}$).
AnswerFirst (Fundamental); No
Length of the pipe, l = 20cm = 0.2m
Source frequency = $n^{th}$ normal mode of frequency, $ν_n= 430Hz$
Speed of sound, $v = 340m/ s$
In a closed pipe, the $n^{th}$ normal mode of frequency is given by the relation:
$\text{v}_\text{n}=(2\text{n}-1)\frac{\text{v}}{4\text{l}}$ n is an interger = 0, 1, 2, 3
$430=(2\text{n}-1)\frac{340}{4\times0.2}$
$2\text{n}-1=\frac{430\times4\times0.2}{340}=1.01$
$2\text{n}=2.01$
$\text{n}\sim1$
Hence, the first mode of vibration frequency is resonantly excited by the given source. In a pipe open at both ends, the $n^{th}$ mode of vibration frequency is given by the relation:
$\text{v}_\text{n}=\frac{\text{nv}}{2\text{l}}$
$\text{n}=\frac{2\text{lv} _n}{\text{v}}$
$=\frac{2\times0.2\times430}{340}=0.5$
View full question & answer→Question 35 Marks
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45Hz$. The mass of the wire is $3.5 \times 10^{–2}kg$ and its linear mass density is $4.0 \times 10^{–2}kg m^{–1}$. What is
- The speed of a transverse wave on the string,
- The tension in the string?
Answer
- Mass of the wire, $m = 3.5 \times 10^{–2}kg$
Linear mass density, $\mu=\frac{\text{m}}{\text{l}}=4.0\times10^2\text{kg m}^{-1}$
Frequency of vibration, v = 45Hz
$\therefore$ length of the wire, $\text{l}=\frac{\text{m}}{\mu}=\frac{3.5\times10^{-2}}{4. 0\times10^{-2}}=0.875\text{m}$
The wavelength of the stationary wave $(\lambda)$ is related to the length of the wire by the relation:
$\lambda=\frac{2\text{l}}{\text{m}}$
where,
n = Number of nodes in the wire
For fundamental node, n = 1:
$\lambda=2\text{l}$
$\lambda=2\times0.875=1.75\text{m}$
The speed of the transverse wave in the string is given as:
$\text{v}=\text{v}\lambda=45\times1.75=78.75\text{m/s}$
- The tension produced in the string is given by the relation:
$\text{T}=\text{v}^2\mu$
$=(78.75)^2\times4.0\times10^{-2}=248.06\text{N}$ View full question & answer→Question 45 Marks
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
AnswerAll the waves have different phases. The given transverse harmonic wave is: $\text{y}(\text{x, t})=3.0\sin\Big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\Big)\ \dots(\text{i})$ For x = 0, the equation reduces to: $\text{y}(\text{x, t})3.0\sin\Big(36\text{t}+\frac{\pi}{4}\Big)$ Also, $\omega=\frac{2\pi}{\text{t}}=36\text{ rad/s}^{-1}$ $\therefore\ \text{t}=\frac{\pi}{18}\text{s}$ Now, plotting y vs. t graphs using the different values of t, as listed in the given table
| t (s) |
0 |
T/8 |
2T/7 |
3T/8 |
4T/8 |
5T/8 |
6T/8 |
7T/8 |
| y (cm) |
$\frac{3}{\sqrt{2}}$ |
3 |
$\frac{3}{\sqrt{2}}$ |
0 |
$-\frac{3}{\sqrt{2}}$ |
–3 |
$-\frac{3}{\sqrt{2}}$ |
0 |
View full question & answer→Question 55 Marks
One end of a long string of linear mass density $8.0 \times 10^{–3}kg m^{–1}$ is connected to an electrically driven tuning fork of frequency $256Hz$. The other end passes over a pulley and is tied to a pan containing a mass of $90kg$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At $t = 0$, the left end (fork end) of the string $x = 0$ has zero transverse displacement $(y = 0)$ and is moving along positive y-direction. The amplitude of the wave is $5.0cm$. Write down the transverse displacement y as function of x and t that describes the wave on the string.
AnswerThe equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: $\text{y}(\text{x, t})=\text{a}\sin(\text{wt}-\text{kx})\ \dots(\text{i})$ Linear mass density, $\mu=8.0\times10^{-3}\text{kg m}^{-1}$ Frequency of the tuning fork, ν = 256Hz Amplitude of the wave, a = 5.0cm = 0.05m …(ii) Mass of the pan, m = 90kg Tension in the string, T = mg = 90 \times 9.8 = 882N The velocity of the transverse wave v, is given by the relation: $\text{v}=\sqrt{\frac{\text{t}}{\mu}}$
$=\frac{882}{8.0\times10^{-3}}=332\text{m/s}$ Angular Frequency, $\omega=2\pi\text{v}$
$=2\times3.14\times256$
$=1608.5=1.6\times10^3\text{ rad/s}\ \dots(\text{iii})$ Wanelength, $\lambda=\frac{\text{v}}{\text{v}}=\frac{332}{256}\text{m}$
$\therefore$ Propagation constant, $\text{k}=\frac{2\pi}{\lambda}$
$=\frac{2\times3.14}{\frac{332}{256}}=4.84\text{m}^{-1}\ \dots(\text{iv})$Substituting the values from equations (ii), (iii) and (iv) in equatio (i) we get the displacement equation:
$\text{y}(\text{x, t})=0.05\sin(1.6\times10^3\text{t}-4.84\text{x})\text{m}.$
View full question & answer→Question 65 Marks
A transverse harmonic wave on a string is described by
$\text{y}(\text{x, t})=3.0\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
AnswerYes; Speed = 20 m/s, Direction = Right to left
Explanation:
Given,
$\text{y}(\text{x, t})=3\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
We know, the equation of a progressive wave travelling from right to left is:
$\text{y}(\text{x, t})=\text{a}\sin(\omega\text{t}+\text{kx}+\phi)\ \dots(2)$
Comparing equation (1) to equation (2), we see that it represents a wave travelling from right to left and also we get:
$\text{a}=3\text{cm},\omega=36\text{rad/s},\text{k}=0.081\text{cm and }\phi=\frac{\pi}{4}$
Therefore the speed of propagation, $\text{v}=\frac{\omega}{\text{k}}=\frac{36}{0.018}=20\text{m/s}$
View full question & answer→Question 75 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Increases with humidity.
AnswerLet $v_m$ and $v_d$ be the speed of sound in moist air and dry air respectively.
Let $\rho_\text{m}$ and $\rho_\text{d}$ be the densities of the moist air and dry air respectively.
Take the relation:
$\text{v}=\sqrt{\frac{\gamma\rho}{\rho}}$
Hence, the speed of sound in moist air is:
$\text{v}_\text{m}=\sqrt{\frac{\gamma\rho}{\rho_\text{m}}}\ \dots(\text{i})$
And the speed of sound in dry air is:
$\text{v}_\text{d}=\sqrt{\frac{\gamma\rho}{\rho_\text{d}}}\ \dots(\text{ii})$
On dividing equations (i) and (ii), we get:
$\frac{\text{v}_\text{m}}{\text{v}_\text{d}}=\sqrt{\frac{\gamma\rho}{\rho_\text{m}}\times\frac{\rho_\text{d}}{\gamma\rho}}=\frac{\rho_\text{d}}{\rho_\text{m}}$
However, the presence of water vapour reduces the density of air, i.e.,
$\rho_\text{d}<\rho_\text{m}$
$\therefore\ \text{v}_\text{m}>\text{v}_\text{d}$
Hence, the speed of sound in mois air is greater than it is in dry air.
Thus, in gaseous medium, the speed of sound increases with humidity.
View full question & answer→Question 85 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Increases with temperature,
AnswerTake the relation:
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
or one mole of any ideal gas, the equation can be written as:
PV = RT
$\text{P}=\frac{\text{RT}}{\text{V}}\ \dots(\text{ii})$
Substituting equation (ii) in equation (i), we get:
$\text{v}=\sqrt{\frac{\gamma\text{RT}}{\text{VP}}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}\ \dots(\text{iii})$
where,
mass, $\text{M}=\rho\text{V}$ is a constant
γ and R are also constants
We conclude from equation (iii) that $\text{v}\propto\sqrt{\text{T}}$
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
View full question & answer→Question 95 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Is independent of pressure,
AnswerTake the relation:
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
where,
Density, $\rho=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{M}}{\text{V}}$
M = Molecular weight of the gas
V = Volume of the gas
Hence, equation (i) reduces to:
$\text{v}=\sqrt{\frac{\gamma\text{PV}}{\text{M}}}\ \dots(\text{ii})$
Now from the ideal gas equation for n = 1:
PV = RT
For constant T, PV = Constant
Since both M and $\gamma$ are constants, v = Constant
Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.
View full question & answer→Question 105 Marks
A steel rod $100cm$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 kHz$. What is the speed of sound in steel?
AnswerLength of the steel rod, l = 100cm = 1m Fundamental frequency of vibration, $ν = 2.53 kHz = 2.53 \times 10^3Hz$ When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.
The distance between two successive nodes is $\frac{\lambda}{2} .$
$\therefore\ \text{I}=\frac{\lambda}{2}$ $\lambda=2\text{l}=2\times1=2\text{m}$
The speed of sound in steel is given by the relation: $\text{v}=\text{v}\lambda$ $=5.06\times10^3\text{m/s}$ $=2.53\times10^3\times2$ $=5.06\text{km/s}$ View full question & answer→Question 115 Marks
Two sitar strings $A$ and $B$ playing the note ‘Ga’ are slightly out of tune and produce beats of frequency $6Hz$. The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to $3Hz$. If the original frequency of A is $324Hz$, what is the frequency of $B$?
AnswerFrequency of string $A, f_A=324 H z$
Frequency of string $B=f_B$
Beat's frequency, $\mathrm{n}=6 \mathrm{~Hz}$
Beat's frequency is given as:
$\mathrm{n}=\left|\mathrm{f}_{\mathrm{A}}+-\mathrm{f}_{\mathrm{B}}\right|$
$6=324+-\mathrm{f}_{\mathrm{B}}$
$\mathrm{f}_{\mathrm{B}}=330 \mathrm{~Hz} \text { or } 318 \mathrm{~Hz}$
Frequency decreases with a decrease in the tension in a string. This is because frequency is irectly proportional to the square root of tension. It is given as:
$\mathrm{v} \propto \text { Underroot } \mathrm{T}$
Hence, the beat frequency cannot be 330 Hz
$\therefore \mathrm{f}_{\mathrm{B}}=318 \mathrm{~Hz}$
View full question & answer→Question 125 Marks
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340Hz) when the tube length is 25.5cm or 79.3cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
AnswerFrequency of the turning fork, ν = 340Hz Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.
Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation: $\text{l}_1=\frac{\pi}{4}$ where, length of pipe, $\text{l}_1=25.5\text{cm}=0.255\text{m}$ $\therefore\ \lambda=4\text{l}_1=4\times0.255=1.02\text{m}$ The speed of the sound is given by the relation: $\text{v}=\text{v}\lambda=340\times1.02=346.8\text{m/s}$ View full question & answer→Question 135 Marks
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
(a) Does the pulse have a definite,
- frequency,
- wavelength,
- speed of propagation?
(b) If the pulse rate is 1 after every 20s, (that is the whistle is blown for a split of second after every 20s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05Hz?
Answer(a)
- No
- No
- Yes
(b) No
Explanation:
- The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.
- The short pip produced after every 20s does not mean that the frequency of the whistle is 1/20 or 0.05Hz. It means that 0.05Hz is the frequency of the repetition of the pip of the whistle.
View full question & answer→Question 145 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is $1.5m$ and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Determine the tension in the string.
AnswerThe velocity of a transverse wave travelling in a string is given by the relation:
$\text{v}=\sqrt{\frac{\text{T}}{\mu}}\ \dots(\text{i})$
Where,
Velocity of the transverse wave, v = 180m/ s
Mass of the string, $m = 3.0 \times 10^{-2}kg$
Length of the string, l = 1.5m
Mass per unit length of the string, $\mu=\frac{\text{m}}{\text{l}}$
$=\frac{3.0}{1.5}\times10^{-2}$
$=2\times10^{-2}\text{kg m}^{-1}$
Tension in the string = T
From equation (i), tension can be obtained as:
$\text{T}=\text{v}^2\mu$
$=(180)^2\times2\times10^{-2}$
$=648\text{N}$
View full question & answer→Question 155 Marks
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is $40 kHz$. During one fast swoop directly toward a flat wall surface, the bat is moving at $0.03$ times the speed of sound in air. What frequency does the bat hear reflected off the wall?
AnswerUltrasonic beep frequency emitted by the bat, $ν = 40 kHz$
Velocity of the bat, $v_b = 0.03v$
where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as:
$\text{v}'=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{b}}\Big)\text{v}$
$=\Big(\frac{\text{v}}{\text{v}-0.03\text{v}}\Big)\times40$
$=\frac{40}{0.97}\text{kHz}$
This frequency is refiected by the stationary wall (vs) toward the bat.
The frequency (v'') of the received sound is given by the relation:
$\text{v}''=\Big(\frac{\text{v}+\text{v}_0}{\text{v}}\Big)\text{v}'$
$=\Big(\frac{\text{v}+0.03\text{v}}{\text{v}}\Big)\times\frac{40}{0.97}$
$=\frac{1.03\times40}{0.97}=42.47\text{kHz}$
View full question & answer→Question 165 Marks
For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same,
- Frequency,
- Phase,
- Amplitude?
AnswerFor the wave on the string described in questions we have seen that l = 1.5m and $\lambda=3\text{m}.$ So, it is clear that $\lambda=\frac{\lambda}{2}$ and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.
- Yes, all the sring particles, except nodes, vibrate with the same frequency v = 60Hz.
- As all string particles lie in one segment, all of them are in same phase.
- Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06m. It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.
View full question & answer→Question 175 Marks
A train, standing in a station-yard, blows a whistle of frequency $400Hz$ in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10m s^{–1}$. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of $10m s^{–1}$? The speed of sound in still air can be taken as $340m s^{–1}$.
AnswerFor the stationary observer: 400Hz; 0.875m; 350m/ s
For the running observer: Not exactly identical
For the stationary observer:
Frequency of the sound produced by the whistle, ν = 400Hz
Speed of sound = 340m/ s
Velocity of the wind, v = 10m/ s
As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, $v_e = 340 + 10 = 350m/ s$
The wavelength $(\lambda)$ of the sound heard by the observer is given by the relation:
$\lambda=\Big(\frac{\text{v}_\text{e}}{\text{V}}\Big)-=\frac{350}{400}=0.875\text{m}$
For the running observer:
Velocity of the observer, $v_0 = 10m/ s$
The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (V').
This is given by the relation:
$\text{v}'=\Big(\frac{\text{v}+\text{v}_0}{\text{v}}\Big)\text{v}$
$=\Big(\frac{340+10}{340}\Big)\times400=411.76\text{Hz}$
Since the air is still, the effective speed of sound = 340 + 0 = 340m/s
The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875m.
Hence, the given two situations are not exactly identical.
View full question & answer→Question 185 Marks
A travelling harmonic wave on a string is described by
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
What are the displacement and velocity of oscillation of a point at x = 1cm, and t = 1s? Is this velocity equal to the velocity of wave propagation?
AnswerThe given harmonic wave is:
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
For x = 1cm and t = 1s,
$\text{y}=(1,1)=7.5\sin\Big[0.0050+12+\frac{\pi}{4}\Big]$
$=7.5\sin\Big[12.0050+\frac{\pi}{4}\Big]$
$=7.5\sin\theta$
Where, $\theta=12.0050+\frac{\pi}{4}=12.0050+\frac{3.14}{4}=12.79\text{ rad}$
$=\frac{180}{3.14\times12.79}=732.81^\circ$
$\therefore\ \text{y}=(1,1)=7.5\sin[732.81^\circ]$
$=7.5\sin(90\times8+12.81^\circ)$
$=7.5\sin(12.81^\circ)$
$=7.5\times0.2217$
$=1.6629\approx1.663\text {cm}$
The velocity of the oscillation at a given point and time is given as:
$\text{v}=\frac{\text{d}}{\text{dt}}\text{y}(\text{x, t})=\frac{\text{d}}{\text{dt}}\Big[7.5\sin\big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\big)\Big]$
$=7.5\times12\cos\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
At x = 1cm and t = 1s:
$\text{v}=\text{y}(1, 1)=90\cos\Big(12.005+\frac{\pi}{4}\Big)$
$=90\cos(732.81^\circ)=90\cos(90\times8+12.81^\circ)$
$=90\cos(12.81^\circ)$
$=90\times0.975=87.75\text{cm/s}$
Now, the equation of a propagating wave is given by:
$\text{y}(\text{x, t})=\text{a}\sin(\text{kx}+\text{wt}+\phi)$
Where,
$\text{k}=\frac{2\pi}{\lambda}$
$\therefore\ \lambda=\frac{2\pi}{\text{k}}$
And $\omega=2\pi\text{ v}$
$\therefore\ \text{v}=\frac{\omega}{2\pi}$
Speed $=\text{v}=\text{v}\lambda=\frac{\omega}{\text{k}}$
Where
$\omega=12\text{ rad/s}$
$\text{k}=0.0050\text{m} ^{-1}$
$\therefore\ \text{v}=\frac{12}{0.0050}=2400\text{cm/s}$
$\therefore$ Hence, the velocity of the wave oscillation at x = 1cm and t = 1s is not equal to the velocity of the wave propagation.
View full question & answer→Question 195 Marks
A train, standing at the outer signal of a railway station blows a whistle of frequency $400Hz$ in still air.
- What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of $10m s^{–1}$, (b) recedes from the platform with a speed of $10m s^{–1}$?
- What is the speed of sound in each case? The speed of sound in still air can be taken as $340m s^{–1}$.
Answer
- (a) Frequency of the whistle, ν = 400Hz
Speed of the train, $v_T= 10m/ s$
Speed of sound, $v = 340m/ s$
The apparent frequency (v') of the whistle as the train approaches the platform is given by the
relation:
$\text{v}'=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{r}}\Big)\text{v}$
$=\Big(\frac{340}{340-10}\Big)\times400=412.12\text{Hz}$
(b) The apparent frequency (v') of the whistle as the train recedes from the platform is given by the relation:
$\text{v}''=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{r}}\Big)\text{v}$
$=\Big(\frac{340}{340+10}\Big)\times400=388.57\text{Hz}$
- The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340m/ s.
View full question & answer→Question 205 Marks
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse $( S )$ and Iongitudinal ( $P$ ) sound waves. Typically the speed of $S$ wave is about $4.0 km s ^{-1}$, and that of $P$ wave is $8.0 km s ^{-1}$. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
AnswerLet $v_S$ and $v_P$ be the velocities of S and P waves respectively.
Let L be the distance between the epicentre and the seismograph.
We have:
$L = v_St_S …(i)$
$L = v_Pt_P …(ii)$
Where,
$t_S $and $t_P$ are the respective times taken by the S and P waves to reach the seismograph from the epicentre
It is given that:
$v_P = 8km/ s$
$v_S = 4km/ s$
From equations (i) and (ii), we have:
$v_S t_S = v_P t_P$
$4t_S = 8t_P$
$t_S = 2 t_P …(iii)$
It is also given that:
$t_S – t_P = 4 min = 240s$
$2t_P – t_P = 240$
$t_P = 240$
And $t_S = 2 \times 240 = 480s$
From equation (ii), we get:
$L = 8 \times 240$
$= 1920km$
Hence, the earthquake occurs at a distance of 1920km from the seismograph.
View full question & answer→Question 215 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is $1.5m$ and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
AnswerA wave travelling along the positive x-direction is given as: $\text{y}_1=\text{a}\sin(\omega\text{t}-\text{kx})$The wave travelling along the negative x-direction is given as:
$\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})$ The superposition of these two waves yields: $\text{y}=\text{y}_1+\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})-\text{a}\sin(\omega\text{t}-\text{kx})$ $=\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})-\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})$ $=-2\text{a}\sin(\text{kx})\cos(\omega\text{t})$ $=-2\text{a}\sin\Big(\frac{2\pi}{\lambda}\text{x}\Big)\cos(2\pi\text{ vt})\ \dots(\text{i})$ The transverse displacement of the string is given as: $\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})\ \dots(\text{ii})$ Comparing equations (i) and (ii), we have: $\frac{2\pi}{\lambda}=\frac{2\pi}{3}$
$\therefore$ Wavelength, $\lambda=3\text{m}$ It is given that: $120\pi=2\pi\text{v}$ Frequency, $\text{v} = 60\text{Hz}$ Wave speed, $\text{v}=\text{v}\lambda$ $=60\times3=180\text{m/s}$
View full question & answer→Question 225 Marks
A pipe $20cm$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430Hz$ source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340m s^{–1}$).
AnswerFirst (Fundamental); No
Length of the pipe, l = 20cm = 0.2m
Source frequency = $n^{th}$ normal mode of frequency, $ν_n= 430Hz$
Speed of sound, v = 340m/ s
In a closed pipe, the $n^{th}$ normal mode of frequency is given by the relation:
$\text{v}_\text{n}=(2\text{n}-1)\frac{\text{v}}{4\text{l}}$ n is an interger = 0, 1, 2, 3
$430=(2\text{n}-1)\frac{340}{4\times0.2}$
$2\text{n}-1=\frac{430\times4\times0.2}{340}=1.01$
$2\text{n}=2.01$
$\text{n}\sim1$
Hence, the first mode of vibration frequency is resonantly excited by the given source. In a pipe open at both ends, the $n^{th}$ mode of vibration frequency is given by the relation:
$\text{v}_\text{n}=\frac{\text{nv}}{2\text{l}}$
$\text{n}=\frac{2\text{lv} _n}{\text{v}}$
$=\frac{2\times0.2\times430}{340}=0.5$
View full question & answer→Question 235 Marks
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45Hz$. The mass of the wire is $3.5 \times 10^{–2}kg$ and its linear mass density is $4.0 \times 10^{–2}kg m^{–1}$. What is
- The speed of a transverse wave on the string,
- The tension in the string?
Answer
- Mass of the wire, $m = 3.5 \times 10^{–2}kg$
Linear mass density, $\mu=\frac{\text{m}}{\text{l}}=4.0\times10^2\text{kg m}^{-1}$
Frequency of vibration, v = 45Hz
$\therefore$ length of the wire, $\text{l}=\frac{\text{m}}{\mu}=\frac{3.5\times10^{-2}}{4. 0\times10^{-2}}=0.875\text{m}$
The wavelength of the stationary wave $(\lambda)$ is related to the length of the wire by the relation:
$\lambda=\frac{2\text{l}}{\text{m}}$
where,
n = Number of nodes in the wire
For fundamental node, n = 1:
$\lambda=2\text{l}$
$\lambda=2\times0.875=1.75\text{m}$
The speed of the transverse wave in the string is given as:
$\text{v}=\text{v}\lambda=45\times1.75=78.75\text{m/s}$
- The tension produced in the string is given by the relation:
$\text{T}=\text{v}^2\mu$
$=(78.75)^2\times4.0\times10^{-2}=248.06\text{N}$ View full question & answer→Question 245 Marks
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
AnswerAll the waves have different phases. The given transverse harmonic wave is: $\text{y}(\text{x, t})=3.0\sin\Big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\Big)\ \dots(\text{i})$ For x = 0, the equation reduces to: $\text{y}(\text{x, t})3.0\sin\Big(36\text{t}+\frac{\pi}{4}\Big)$ Also, $\omega=\frac{2\pi}{\text{t}}=36\text{ rad/s}^{-1}$ $\therefore\ \text{t}=\frac{\pi}{18}\text{s}$ Now, plotting y vs. t graphs using the different values of t, as listed in the given table
| t (s) |
0 |
T/8 |
2T/7 |
3T/8 |
4T/8 |
5T/8 |
6T/8 |
7T/8 |
| y (cm) |
$\frac{3}{\sqrt{2}}$ |
3 |
$\frac{3}{\sqrt{2}}$ |
0 |
$-\frac{3}{\sqrt{2}}$ |
–3 |
$-\frac{3}{\sqrt{2}}$ |
0 |
View full question & answer→Question 255 Marks
One end of a long string of linear mass density $8.0 \times 10^{–3}kg m^{–1}$ is connected to an electrically driven tuning fork of frequency $256Hz$. The other end passes over a pulley and is tied to a pan containing a mass of $90kg$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At $t = 0$, the left end (fork end) of the string $x = 0$ has zero transverse displacement $(y = 0)$ and is moving along positive y-direction. The amplitude of the wave is $5.0cm$. Write down the transverse displacement y as function of x and t that describes the wave on the string.
AnswerThe equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: $\text{y}(\text{x, t})=\text{a}\sin(\text{wt}-\text{kx})\ \dots(\text{i})$ Linear mass density, $\mu=8.0\times10^{-3}\text{kg m}^{-1}$ Frequency of the tuning fork, ν = 256Hz Amplitude of the wave, a = 5.0cm = 0.05m …(ii) Mass of the pan, m = 90kg Tension in the string, T = mg = 90 \times 9.8 = 882N The velocity of the transverse wave v, is given by the relation: $\text{v}=\sqrt{\frac{\text{t}}{\mu}}$
$=\frac{882}{8.0\times10^{-3}}=332\text{m/s}$ Angular Frequency, $\omega=2\pi\text{v}$
$=2\times3.14\times256$
$=1608.5=1.6\times10^3\text{ rad/s}\ \dots(\text{iii})$ Wanelength, $\lambda=\frac{\text{v}}{\text{v}}=\frac{332}{256}\text{m}$
$\therefore$ Propagation constant, $\text{k}=\frac{2\pi}{\lambda}$
$=\frac{2\times3.14}{\frac{332}{256}}=4.84\text{m}^{-1}\ \dots(\text{iv})$Substituting the values from equations (ii), (iii) and (iv) in equatio (i) we get the displacement equation:
$\text{y}(\text{x, t})=0.05\sin(1.6\times10^3\text{t}-4.84\text{x})\text{m}.$
View full question & answer→Question 265 Marks
Find the velocity of source, when the frequency appears to be (a) double (b) half, the original frequency to a stationary listener.
AnswerHere, $\nu_\text{S}=?$
$\nu_\text{L}=0,$
Case (a) v' = 2v. It is possible if source is approaching the stationary listener, i.e., $\nu_\text{S}$ is +.
As $\text{V}'=\frac{\nu-\nu_\text{L}}{\nu-\nu_\text{S}}\text{V}$
$\therefore 2\text{V}=\frac{\nu}{\nu-\nu_\text{S}}\text{V}$
$\nu_\text{S}=\frac{\nu}{2}$
Thus, source should approach the listener with half the velocity of sound propagation.
Case (b) $\text{V}'=\frac{\text{V}}{2}.$ It is possible if source is receding away from the stationary listener, i.e., $\nu_\text{S}$ is negative and $\nu_\text{L}=0.$
$\therefore \text{V}'=\Big(\frac{\nu}{\nu+\nu_\text{s}}\Big)\text{V}$
$\frac{\text{V}}{2}=\Big(\frac{\nu}{\nu+\nu_\text{s}}\Big)\text{V}$
$\nu_\text{s}=\nu$
Thus, the source should recede away from the listener with the velocity of sound propagation.
View full question & answer→Question 275 Marks
A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20° C, calculate
- Speed of sound in air at room temperature
- Speed of sound in air at 0° C
- If the water in the tube is replaced with mercury, will there be any difference in your observations?
Answer
- If a pipe partially filled with water whose upper surface of the water acts as a reflecting surface of a closed organ pipe. If the length of the air column is varied until its natural frequency equals the frequency of the fork, then the column resonates and emits a loud note.
The frequency of tuning fork,f = 512Hz.
For observation of first maxima of intensity,
- We know that $\text{v}\propto\sqrt{\text{T}}$
where tempreature (T) is in kwlcin.
$\frac{\text{v}_{20}}{\text{v}_0}=\sqrt{\frac{273+20}{273+0}}=\sqrt{\frac{293}{273}}$
$\frac{\text{v}_{20}}{\text{v}_0}=\sqrt{1.073}=1.03$
$\text{v}_0=\frac{\text{v}_{20}}{1.03}=\frac{348.16}{1.03}=338\text{m/ s}$
- The resonance will still be observed for 17cm length of air column above mercury. However, due to more complete reflection of sound waves at mercury surface, the intensity of reflected sound increases.
View full question & answer→Question 285 Marks
The length of a sonometer wire between two fixed ends is $110cm$. Where should the two bridges be placed so as to divide the wire into three segments, whose fundamental frequencies are in the ratio $1 : 2 : 3$?
AnswerHere, total length of the wire
$1 = 110cm$
$V_1: V_2: V_3 = 1 : 2 : 3$
If $l_1, l_2, l_3$ are the length of these three parts as $\text{V}\propto\frac{1}{\text{l}}$
$\therefore \text{l}_2:\text{l}_2:\text{l}_3=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}$
$=6:3:2$
Sum of the ratios = 11
$\therefore \text{l}_1=\frac{\text{L}}{1\text{l}}\times6=\frac{110}{11}\times6=60\text{cm}$
$\text{l}_2=30\text{cm}$
$\text{l}_3=20\text{cm}$
Hence, the bridges should be placed at 60cm and 90cm from the zero end of the wire.
View full question & answer→Question 295 Marks
Bullets are fired at regular intervals of 10 seconds from an armoured car. A is moving with a speed of 30m/s towards B. At what interval will an officer seated in car B and dashing with a speed of 60m/s towards A hear the report? Velocity of sound = 330m/s.
AnswerLet T' and T represent the apparent and true periods respectively.
$\therefore \text{V}'=\frac{1}{\text{T}'}$ and $\text{V}=\frac{1}{\text{T}}$
We know, $\text{V}'=\Big(\frac{\nu-\omega+\text{u}_0}{\nu-\omega+\text{u}_\text{s}}\Big)\text{V}$
$\text{T}=\text{T}\Big(\frac{\nu-\omega+\text{u}_\text{s}}{\nu-\omega+\text{u}_0}\Big)$
Given: $\text{T}=10\text{s},\text{u}_0=30\text{ms}^{-1}$
$\nu=330\text{m/s}$ and
$\text{u}_\text{s}=60\text{ms}^{-1},\omega=10\text{m/s}$
Putting the respective values we have,
$\text{T}'=10\times\Big(\frac{330-10-30}{330-10+60}\Big)$
$=10\times\Big(\frac{290}{380}\Big)=\frac{290}{38}=7.6\text{s}$
View full question & answer→Question 305 Marks
- Describe various modes of vibration in an open ends organ pipe.
- Show that the ratio of frequencies of different harmonics with the first harmonics in open pipe is $1 : 2 : 3$
- In an open organ pipe, the fundamental frequency is $30Hz$. If the organ pipe is closed at one end then what will be the fundamental frequency?
Answer
- An open organ pipe is open at both ends. Therefore, an antinode is formed at each end.
$\lambda=\frac{2\text{L}}{\text{n}}$
Where n = 1, 2, 3,.....
- First normal node of vibration
Let $\lambda_1$ be the wavelength of stationary waves set up in the open organ pipe corresponding to n = 1
$\lambda_1=\frac{2\text{L}}{1}\Rightarrow\text{L}=\frac{\lambda_1}{2}$
$\text{v}_1=\frac{\text{v}}{\lambda_1}=\frac{\text{v}}{2\text{L}}$
This is the lowest frequency of vibration and is called fundamentak frequency.
- Second normal mode of vibration $\lambda_2=\frac{2\text{L}}{2}=\text{L}$
$\text{v}_2=\frac{\text{v}}{\lambda_2}=\frac{\text{v}}{\text{L}}$
$\Rightarrow\text{v}_2=\frac{2\text{v}}{\text{L}}=2\text{v}_1\text{ i.e. }\text{v}_2=2\text{v}_1$
- Third normal mode of vibration $\lambda_3=\frac{2\text{L}}{3}=\text{L}\Rightarrow \text{L}=\frac{3\lambda_3}{2}$
$\text{v}_3=\frac{\text{v}}{\lambda_3}=\frac{3\text{v}}{2\text{L}}$
$\Rightarrow\text{v}_3=3\text{v}_1$
- From (i), (ii) and (iii), we have $v_1 : v_2 : v_3 = 1 : 2 : 3$
In an open organ pipe, fundamentak frequency $v_1 = 30Hz$
Let for closed organ pipe, fundamental frequency is $V'_1$.
We know that the fundamental frequency for open organ pipe, $\text{v}_1=\frac{\nu}{2\text{L}}$ and for closed organ pipr, $\text{v}'_1=\frac{\nu}{4\text{L}}$
$\therefore\frac{\text{v}_1}{\text{v}'_1}=\frac{\nu}{2\text{L}}\times\frac{4\text{L}}{\nu}$
$\Rightarrow \text{v}'_1=\frac{\text{v}_1}{2}=\frac{30\text{Hz}}{2}=15\text{Hz}$ View full question & answer→Question 315 Marks
Explain why?
- Inspite of formula $\nu=\sqrt{\frac{\gamma\text{P}}{\rho}},$ the speed of sound in air independent of pressure.
Where $\gamma=$ ratio of specific heats i.e., $\gamma =\frac{\text{C}_\text{p}}{\text{C}_\nu},$ $\rho=$ density, P = pressure.
- Bats can ascertain distances, directions, nature and the size of the obstacles without any "eyes”.
Answer
- $\nu=\sqrt{\frac{\gamma \text{P}}{\rho}}$
Speed of sound in air $\nu=\sqrt{\frac{\gamma \text{RT}}{\text{mV}}.\text{V}}$
where $\rho=$ Density of gas $=\frac{\text{m}}{\text{V}}$
$\Rightarrow\nu=\sqrt{\frac{\gamma\text{RT}}{\text{m}}}=\text{Constant}$
PV = RT at constant temprerature, PV = constant, $\rho =\frac{\text{m}}{\text{V}}$
- Bats can heard ultrasonic vibrations, i.e. whose frequency is greater than 20KHz that cannot be heard by human ear that is why bat can ascertain distances, directions, nature and the size of the obstacles without any "eyes”.
View full question & answer→Question 325 Marks
Two tuning forks A and B give 5 beats/sec. A resounds with a closed column of air 15cm long and B with an open column 30.5cm long. Calculate their frequencies (neglect end correction).
Answer$\text{V}_1=\frac{\nu}{4\times15},\text{V}_2=\frac{\nu}{2\times30.5}=\frac{\nu}{61}$
$m = \text{V}_1-\text{V}_2=\frac{\nu}{60}-\frac{\nu}{61}=\nu\times\frac{1}{61\times60}$
$5=\frac{\nu}{61\times60}$
$=\nu=5\times16\times60\text{cm/s}$
$\therefore \text{V}_1=\frac{\nu}{60}=\frac{5\times61\times60}{60}=305\text{Hz}$
$\text{V}_2=\frac{\nu}{61}=\frac{5\times60\times61}{61}=300\text{Hz}$
View full question & answer→Question 335 Marks
The amplitude of a wave disturbed propagating in the positive x direction is given by
$\text{y}=\frac{1}{1+\text{x}^2}$ at t = 0 and $\text{y}=\frac{1}{1+(\text{x}-1)^2}$ at t = 2s
where x and y in metre. The shape of disturbance does not change during the propagation. What is the velocity of the wave?
AnswerAt t = 0, $\text{y}=\frac{1}{1+\text{x}^2}$
$\therefore 1+\text{x}^2=\frac{1}{\text{y}}$
$\text{x}^2=\frac{1}{\text{y}}-1=\frac{1-\text{y}}{\text{y}}\text{x}=\Big(\frac{1-\text{y}}{\text{y}}\Big)^{\frac{1}{2}}$
At t = 2s, $\text{y}=\frac{1}{[1+(\text{x}-1)^2]}$
$1+(\text{x}-1)^2=\frac{1}{\text{y}}$
$(\text{x}-1)^2=\frac{1}{\text{y}}-1=\frac{1-\text{y}}{\text{y}}$
$(\text{x}-1)=\Big(\frac{1-\text{y}}{\text{y}}\Big)^{\frac{1}{2}}$
$\Rightarrow \text{x}=1+\Big(\frac{1-\text{y}}{\text{y}}\Big)^{\frac{1}{2}}$
Since, $v=\frac{\text{x}_2-\text{x}_1}{\text{t}_2-\text{t}_1}$
$\therefore v=\frac{1}{20}=0.5\text{ms}^{-1}$
View full question & answer→Question 345 Marks
The intensities due to two sources of sound are $I_0$ and $4I_0$. What is the intensity at a point where the phase difference between two waves is:
- $0^\circ$
- $\frac{\pi}{2}$
- $\pi?$
AnswerIf $a_1$ and $a_2$ are the amplitudes of two waves, then the resultant amplitude is given by
$\text{A}=\sqrt{\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi},$
where $\phi$ is the phase difference between two waves
Now, $\text{A}^2=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi$
Expressing this equation in terms of intensity.
$\text{I}=\text{I}_1+4\text{I}_2+2\sqrt{\text{I}_1}\sqrt{\text{I}_2}\cos\phi$
- $\text{I}=\text{I}_0+4\text{I}_0+2\sqrt{\text{I}_0}\sqrt{4\text{I}_0}\cos0^\circ=9\text{I}_0$
- $\text{I}=\text{I}_0+4\text{I}_0+2\sqrt{\text{I}_0}\sqrt{4\text{I}_0}\cos\frac{\pi}{2}=5\text{I}_0$
- $\text{I}=\text{I}_0+4\text{I}_0+2\sqrt{\text{I}_0}\sqrt{4\text{I}_0}\cos\pi=\text{I}_0.$
View full question & answer→Question 355 Marks
A transverse harmonic wave on a string is described by
$\text{y}(\text{x, t})=3.0\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
AnswerYes; Speed = 20 m/s, Direction = Right to left
Explanation:
Given,
$\text{y}(\text{x, t})=3\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
We know, the equation of a progressive wave travelling from right to left is:
$\text{y}(\text{x, t})=\text{a}\sin(\omega\text{t}+\text{kx}+\phi)\ \dots(2)$
Comparing equation (1) to equation (2), we see that it represents a wave travelling from right to left and also we get:
$\text{a}=3\text{cm},\omega=36\text{rad/s},\text{k}=0.081\text{cm and }\phi=\frac{\pi}{4}$
Therefore the speed of propagation, $\text{v}=\frac{\omega}{\text{k}}=\frac{36}{0.018}=20\text{m/s}$
View full question & answer→Question 365 Marks
Differentiate the following:
- Wave velocity and particle velocity.
- Harmonics and overtones.
Answer
-
|
Wave velocity
|
Particle velocity
|
|
The velocity of wave motion through a particular medium is constant. It depends only on the nature of the medium, does not depends upon its frequency or wavelength or intensity.
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Velocity of the particles during their vibration is different at different position.
|
-
|
Harmonics
|
Overtones
|
|
Harmonics are the notes/ seconds of frequency equal to or an integral multiple of fundamental frequency (n). Thus, first, second, third, ... harmonics have frequencies n, 2n, 3n, ... respectively.
|
Overtones are the notes/ seconds of frequency twice/ thrice/ four times ... the fundamental frequency (n). Thus, first, second, third, ... overtones have frequencies 2n, 3n, 4n, ... respectively and so on.
|
View full question & answer→Question 375 Marks
What is meant by beats? Discuss graphical method of formation of beats.
AnswerBeats: The waxing and waning of sound due to interaction between two slightly different frequencies. If $v_1$ and $v_2$ are the two frequencies, $V_b = lv_1- v_2l$. Beats are heard only when $lv_1- v_2l <10$, since the sound persist in our ears for $\frac{1}{10}\text{th}$ of a second. As a result of the super-position of the nearly equal frequencies, a varying amplitude is formed which varies with time as $2\text{A}\cos2\pi\text{v}_\text{m}\text{t}$ where $\text{v}_\text{m}=\frac{\text{v}_1-\text{v}_2}{2}$ Intensity $=4\text{A}^2\cos^22\pi\nu_\text{m}\text{t}$ The intensity pattern can be shown as below.
View full question & answer→Question 385 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Increases with humidity.
AnswerLet $v_m$ and $v_d$ be the speed of sound in moist air and dry air respectively.
Let $\rho_\text{m}$ and $\rho_\text{d}$ be the densities of the moist air and dry air respectively.
Take the relation:
$\text{v}=\sqrt{\frac{\gamma\rho}{\rho}}$
Hence, the speed of sound in moist air is:
$\text{v}_\text{m}=\sqrt{\frac{\gamma\rho}{\rho_\text{m}}}\ \dots(\text{i})$
And the speed of sound in dry air is:
$\text{v}_\text{d}=\sqrt{\frac{\gamma\rho}{\rho_\text{d}}}\ \dots(\text{ii})$
On dividing equations (i) and (ii), we get:
$\frac{\text{v}_\text{m}}{\text{v}_\text{d}}=\sqrt{\frac{\gamma\rho}{\rho_\text{m}}\times\frac{\rho_\text{d}}{\gamma\rho}}=\frac{\rho_\text{d}}{\rho_\text{m}}$
However, the presence of water vapour reduces the density of air, i.e.,
$\rho_\text{d}<\rho_\text{m}$
$\therefore\ \text{v}_\text{m}>\text{v}_\text{d}$
Hence, the speed of sound in mois air is greater than it is in dry air.
Thus, in gaseous medium, the speed of sound increases with humidity.
View full question & answer→Question 395 Marks
An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?
AnswerFor first harmonic of open organ pipe $\text{L}=\frac{\lambda}{2}$ $\Rightarrow\lambda=2\text{L}\Rightarrow\frac{\text{v}}{f_0}=2\text{L}\Rightarrow(\text{f}_0)_\text{open}=\frac{\text{v}}{2\text{L}}$ Where v is speed of the sound wave in air. For first harmonic of closed organ pipe $\text{L}'=\frac{\lambda}{4}$ $\Rightarrow\lambda=4\text{L}'\Rightarrow\frac{\text{v}}{\text{f}_0}=4\text{L}'\Rightarrow(\text{f}_0)_\text{close}=\frac{\text{v}}{4\text{L}'}$ $\Rightarrow\lambda=4\text{L}'\Rightarrow\frac{\text{v}}{2\text{L}}=\frac{\text{v}}{4\text{L}'}$ [$\therefore$ speed remains constant] $\Rightarrow\frac{\text{L}'}{\text{L}}=\frac{2}{4}=\frac{1}{2}\Rightarrow\text{L}'=\frac{\text{L}}{2}$
View full question & answer→Question 405 Marks
What is the amplitude, wavelength and velocity of the wave represented by
$\phi(\text{x, t})=5\sin(6\pi\text{t}-4\text{x})?$
AnswerThe given equation is
$\phi (\text{x, t})= 5\sin(6\pi\text{t}-4\text{x})$
$=5\sin2\pi\Big(3\text{t}-\frac{4\text{x}}{2\pi}\Big)$
Comparing it with standard equation
$\phi (\text{x, t})=\text{r}\sin2\pi\Big(\frac{\text{t}}{\text{T}}-\frac{\text{x}}{\lambda}\Big)$
We obtain, r = 5, i.e., amplitude = 5m
$\frac{1}{\text{T}}=3$ or $\text{v}=3\text{Hz}$
$\frac{1}{\lambda}=\frac{4}{2\pi}$
$\lambda=\frac{\pi}{2},$ i.e., wavelength $=\frac{\pi}{2}\text{m}$
As velocity,
$\nu=\text{v}\lambda$
$\therefore \nu=3\times\frac{\pi}{2}=\frac{3}{2}\times\frac{22}{7}$
$=\frac{33}{7}\text{ms}^{-1}$
$=4.71\text{ms}^{-1}$
View full question & answer→Question 415 Marks
The third overtone of a closed organ pipe is found to be in unison with the first overtone of an open pipe. Find the ratio of the lengths of the pipes.
AnswerLet $n_1$ be the frequency of the closed pipe and $n_2$ of the open pipe and $l_1, l_2$ their corresponding lengths.
$\nu=4\text{l}_1\text{n}_1=2\text{l}_2\text{n}_2$
$\text{n}_1=\frac{\nu}{4\text{l}_1}$ and $\text{n}_2=\frac{\nu}{2\text{l}_2}$
Third overtone of the closed pipe (seventh harmonic) = $7n_1$
First oYertone of the open pipe = $2n_2$
Given: $7\text{n}_1=2\text{n}_2$ or $\frac{7\nu}{4\text{l}_1}=\frac{2\nu}{2\text{l}_2}$ or $\frac{\text{l}_1}{\text{l}_2}=\frac{7}{4}$
View full question & answer→Question 425 Marks
A sitar wire is under a tension of 40N and the length between the bridges is 70cm. A 5m sample of the wire has a mass of 1.0g. Deduce the speed of transverse waves on the wire, frequency of the fundamental and the frequency of the first two harmonics.
Answer$\text{T}=40\text{N},$
$\mu=\frac{1.0\text{gm}}{5\text{m}}=\frac{10^{-3}\text{kg}}{5\text{m}}$
$=0.0002\text{kg/m}$
$\text{l}=70\text{cm}=0.7\text{m}$
$\text{c}=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{40}{0.0002}}=447.2\text{ms}^{-1}$
$\text{v}_1=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}=\frac{447.2}{2\times0.7}=\frac{447.2}{1.4}$
$=319.4\text{Hz}$
$\text{v}_2=\frac{2}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}=638.8\text{Hz}$
$\text{v}_3=\frac{3}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}=958.2\text{Hz}$
View full question & answer→Question 435 Marks
Wavelength of two notes in air are $\frac{80}{195}\text{m}$ and $\frac{80}{193}\text{m}.$ Each note produces five beats per second with a third note of a fixed frequency. Calculate the velocity of sound in air.
AnswerHere given that, $\lambda_1=\frac{80}{195}\text{m}$ and $\text{l}_2=\frac{80}{193}\text{m}.$
If v, and v, be the corresponding frequencies and $v$ be the velocity of sound in air, then
$\text{v}_1=\frac{v}{\lambda_1}=\frac{195v}{80}$
$\text{v}_2=\frac{v}{\lambda_2}=\frac{193v}{80}$
This shown that $v_1 > v_2$.
Let the frequency of third note be v, then
$\text{v}_1-\text{v}=5$ and $\text{v}-\text{v}_2=5$
$\therefore \text{v}_1-\text{v}_2=10$
$\frac{195v}{80}-\frac{193v}{80}=10$
$\Rightarrow20v=80\times10$
$v=400\text{m/ sec}.$
View full question & answer→Question 445 Marks
Show that a function
$\text{y}(\text{x, t})=\text{A}\sin(\text{kx}-\omega\text{t})+\text{B}\cos(\text{kx}-\omega\text{t})$
represents a progressive wave. What is the amplitude, wavelength, velocity and initial phase angle of the wave?
AnswerThe given function is
$\text{y}(\text{x, t})=\text{A}\sin(\text{kx}-\omega\text{t})+\text{B}\cos(\text{kx}-\omega\text{t}).$
Let us put $\text{A}=\text{a}\cos\phi$ and $\text{B}=\text{a}\sin\phi,$ where $\text{a}=\sqrt{\text{A}^2+\text{B}^2}$ and $\tan \phi=\frac{\text{B}}{\text{A}}.$ Then, the above function may be expressed as
$\text{y}(\text{x, t})=\text{a}\cos\phi\sin(\text{kx}-\omega\text{t})\\+\text{a}\sin\phi\cos(\text{kx}-\omega\text{t})$
$\text{y}(\text{x, t})=\text{a}\sin(\text{kx}-\omega\text{t}+\phi)$
As it is a single sinusoidal function of space and time, it is representing the equation of a harmonic progressive wave.
Amplitude of wave $=\text{a}=\sqrt{\text{A}^2+\text{B}^2}$
Wavelength of wave $=\lambda=\frac{2\pi}{\text{k}}.$
wave velocity $=v=\text{v}\lambda=\frac{\text{v}}{2\pi}.\frac{2\pi}{\text{k}}=\frac{\text{v}}{\text{k}}.$
and initial phase angle of the wave $=\phi,$ where $\phi=\tan^{-1}\Big(\frac{\text{B}}{\text{A}}\Big).$
View full question & answer→Question 455 Marks
Show that the velocity of sound increases by $61cm/s$ for every $1^\circ C$.
Answer$\frac{\nu\text{t}}{\nu_0}=\sqrt{\frac{\text{T}}{\text{T}_0}}=\sqrt{\frac{273+\text{t}}{273+0}},$
where $v_t, v_0$ are the velocities of sound at $T$ and $T_0$ respectively.
$\therefore\frac{\nu_\text{t}}{\nu_0}=\Big(1+\frac{\text{t}}{273}\Big)^{\frac{1}{2}}=1+\frac{1}{2}\times \frac{\text{t}}{273}$ (neglecting higher powers)
$\therefore \nu_\text{t}=\nu_0\Big(1+\frac{\text{t}}{546}\Big)=\nu_0+\nu_0\frac{\text{t}}{546}$
$\therefore \nu_\text{t}-\nu_0=\frac{\nu_0\text{t}}{546}=\frac{332\times1}{546}$
$=0.608=61\text{cm/s}$
Thus, the velocity of sound increases by 61cm/s for every 1°C (or 1K) rise in the temperature.
View full question & answer→Question 465 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Increases with temperature,
AnswerTake the relation:
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
or one mole of any ideal gas, the equation can be written as:
PV = RT
$\text{P}=\frac{\text{RT}}{\text{V}}\ \dots(\text{ii})$
Substituting equation (ii) in equation (i), we get:
$\text{v}=\sqrt{\frac{\gamma\text{RT}}{\text{VP}}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}\ \dots(\text{iii})$
where,
mass, $\text{M}=\rho\text{V}$ is a constant
γ and R are also constants
We conclude from equation (iii) that $\text{v}\propto\sqrt{\text{T}}$
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
View full question & answer→Question 475 Marks
A copper wire is held at the two ends by rigid supports. At 30°C, the wire is just taut, with a negligible tension, find the speed of the transverse waves in this wire at 10°C.
$\alpha=1.7\times10^{-5}/^{\circ}\text{C},$
$\text{Y}=1.3\times10^{11}\text{N/m}^2,$
$\rho=9\times10^3\text{kg/m}^3.$
AnswerLet $\text{dl}=l \alpha\theta\text{ or }\frac{\text{dl}}{\text{l}}=\alpha\theta$
$\text{Y}=\frac{\frac{\text{F}}{\text{A}}}{\frac{\text{dl}}{\text{l}}}=\frac{\text{fl}}{\text{Adl}}$
$\frac{\text{F}}{\text{A}}=\frac{\text{Ydl}}{\text{l}}=\text{Y}\alpha \theta$
$=1.3\times10^{11}\times1.7\times10^{-5}\times20$
$\therefore \frac{\text{F}}{\text{A}}=4.42\times10^7$
$\nu=\sqrt{\frac{\text{T}}{\text{m}}}=\sqrt{\frac{\text{F}}{\text{A}\rho}},$ where F = T
and $=\sqrt{\frac{4.42\times10^7}{9\times10^3}}=70\text{m/s}$
View full question & answer→Question 485 Marks
From the equation $\text{y}=\text{r} \sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$ establish the relation between particle velocity, and wave velocity.
Answer$\text{y}=\text{r} \sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$
Velocity of particle
$\text{u}(\text{x},\text{t})=\frac{\text{d}}{\text{dt}}[\text{y}(\text{x,}\text{t})]$
$=\frac{\text{d}}{\text{dt}}\Big[\text{r}\sin\big\{\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})\big\}\Big]$
$\text{u}(\text{x,t})=\frac{2\pi}{\lambda}\nu\Big[\text{r}\cos\big\{\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})\big\}\Big]\dots(1)$
Also, $\frac{\text{d}}{\text{dx}}[\text{y}(\text{x},\text{t})]=\text{r}\cos\Big\{\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})\Big\}\dots(2)$
Dividing (1) and (2), we get
$\frac{\text{u}(\text{x},\text{t})}{\frac{\text{d}}{\text{dx}}\{\text{y}(\text{x},\text{t}\})}=-\nu$
$\text{or }\text{u}(\text{x},\text{t})=-\nu\frac{\text{d}}{\text{dx}}\{\text{y}(\text{x},\text{t})\}$
View full question & answer→Question 495 Marks
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Is independent of pressure,
AnswerTake the relation:
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
where,
Density, $\rho=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{M}}{\text{V}}$
M = Molecular weight of the gas
V = Volume of the gas
Hence, equation (i) reduces to:
$\text{v}=\sqrt{\frac{\gamma\text{PV}}{\text{M}}}\ \dots(\text{ii})$
Now from the ideal gas equation for n = 1:
PV = RT
For constant T, PV = Constant
Since both M and $\gamma$ are constants, v = Constant
Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.
View full question & answer→Question 505 Marks
The air columns (of resonance tubes) $100cm$ and $101cm$ long give $17$ beats in $20$ seconds, when each is sounding its fundamental note. Calculate velocity of sound.
AnswerHere, length of first air column,
$l_1 = 100cm = 1m$
length of other air column
$l_2 = 101cm = 1.01m$
No. of beats per second, $\text{n}=\frac{17}{20}$
Suppose v is the velocity of sound.
Let $v_1, v_2$ be the fundamental frequencies of vibration of two air columns
$\therefore \text{V}_1=\frac{\nu}{4\text{l}_1}=\frac{\nu}{4\times1}=\frac{\nu}{4}$
and $\text{V}_2=\frac{\nu}{4\times1.01}=\frac{\nu}{4.04}$
As per question $\text{V}_1-\text{V}_2=\text{n}$
$\therefore \frac{\nu}{4}-\frac{\nu}{4.04}=\frac{17}{20}$
$\nu=343.4\text{ms}^{-1}$
View full question & answer→Question 515 Marks
A steel rod $100cm$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53$ kHz. What is the speed of sound in steel?
AnswerLength of the steel rod, $l = 100cm = 1m$ Fundamental frequency of vibration, $ν = 2.53 kHz = 2.53 \times 10^3Hz$ When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.
The distance between two successive nodes is $\frac{\lambda}{2} .$
$\therefore\ \text{I}=\frac{\lambda}{2}$ $\lambda=2\text{l}=2\times1=2\text{m}$
The speed of sound in steel is given by the relation: $\text{v}=\text{v}\lambda$
$=5.06\times10^3\text{m/s}$ $=2.53\times10^3\times2$ $=5.06\text{km/s}$ View full question & answer→Question 525 Marks
Write Newton's formula for the speed of sound in air. What was wrong with this formula? What correction was made by Laplace in this formula?
AnswerAccording to Newton, as wave propagates through a medium, temperature remain constant thus wave propagation is an isothermal process, satisfying PV = Constant. Differentiating, we get
PDV + VAP = 0
$\Rightarrow \text{P}=-\frac{(\text{dP})}{\Big(\frac{\text{dV}}{\text{V}}\Big)}=\beta$ (Bulk modulas)
Since velocity of sound waves is $\text{v}=\sqrt{\frac{\beta}{\rho}}$
We have, $\text{v}=\sqrt{\frac{\text{P}}{\rho}}$
According to Laplace, temperature can change, but heat energy change should be zero and so wave propagation is adiabatic. Hence Laplace corrected it as,
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}},$ where $\gamma$ is the ratio of molar specific heat capacities at constant pressure and volume, i.e. $\gamma =\frac{\text{C}_\text{p}}{\text{C}_\text{v}}.$
View full question & answer→Question 535 Marks
In a laboratory experiment (room temperature being $15°C$), the wavelength of a note of sound of frequency $500Hz$ is found to be $0.68m$. If the density of air at STP is $1.29kg m^{-3}$ calculate the ratio of the specific heat of air.
Answer$\nu_1 \text{ at }15^\circ\text{C}=\text{v}\lambda=500\times0.68=340\text{ms}^{-1}$
Velocity $(\nu'_\text{S})$ at $0^\circ\text{C}$
$=340\times\sqrt{\frac{273}{273+15}}=331\text{ms}^{-1}$
As STP, $\rho_0=1.27\text{kg m}^{-3}$
and $\text{P}_0=0.76\times136\times10^3\times9.8\text{Nm}^{-2}$
Since $\nu_0\sqrt{\frac{\gamma\text{P}_0}{\rho_0}},\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{v}}$
$\gamma=\nu^2_0\frac{\rho_0}{\text{P}_0}$
$=\frac{331\times331\times1.27}{0.76\times13.6\times10^3\times9.8}=1.37$
View full question & answer→Question 545 Marks
Calculate the velocity of transverse waves in a copper wire $1mm^2$ in cross section, under the tension produced by $1kg$ wt. The relative density of copper = $8.93$.
Answer$\nu=?, \text{a}=1\text{mm}^2=10^{-2}\text{cm}^2,$
$\rho=8.93\text{g/cm}^3$
$\text{T}=1\text{kg wt.}=9.8\text{N}$
$=9.8\times10^5\text{dyne}$
Mass of unit length of wire,
$\mu=\text{a}\times1\times\rho=10^{-22}\times8.93\text{g/cm}$
As $\nu=\sqrt{\frac{\text{T}}{\mu}}$
$\therefore \nu=\sqrt{\frac{9.8\times10^5}{8.93\times10^{-2}}}$
$=3.312\times10^3\text{cm/s}$
$=33.12\text{m/s}$
View full question & answer→Question 555 Marks
A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
AnswerThewire is stretched both and so frequency of stretched wure is $\text{v}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}$
As number of harmonic n, lengthL and tensuin (T) are kept same in both cases.
$\therefore\text{v}\propto\frac{1}{\sqrt{\text{m}}}$
$\frac{\text{v}_1}{\text{v}_2}=\frac{\sqrt{\text{m}_2}}{\sqrt{\text{m}_1}}\ ...(\text{i})$
Mass per unit length $=\frac{\text{mass of wire}}{\text{length}}=\frac{(\pi\text{r}^2\text{l})\rho}{\text{l}}$
$\text{m}=\pi\text{r}^2\rho$
As matterial of wire is same.
$\frac{\text{m}_2}{\text{m}_1}=\frac{\pi\text{r}_2^2\rho}{\pi\text{r}^2_1\rho}=\frac{(3\text{r})^2}{\text{r}^2}=\frac{9}{1}$
$\therefore\frac{\text{v}_1}{\text{v}_2}==\sqrt{\frac{9}{1}}=\frac{3}{1}$
$\therefore\text{v}_2=\frac{1}{3}\text{v}_1$
So the frequency of sitar reduced by $\frac{1}{2}$ of previous value.
View full question & answer→Question 565 Marks
A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?
AnswerWire of sonometer is twice the length which it vibrates in its second harmonic. Thus, if the tuning fork resonates at L, it will resonate at 2L. This can be explained as below:
The frequency of sonometer is given by
$\text{f}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\mu}}=\frac{\text{nv}}{2\text{L}}$ (n = number of loops)
For a given sonometer velocity of wave will be constant. if after chaning the leggth of wire the tuing fork still be in resonance witrh the wire. then, $\frac{\text{n}}{\text{L}}=\text{constant}\Rightarrow\frac{\text{n}^2}{\text{L}^2}$
$\frac{\text{n}^1}{\text{L}^1}=\frac{\text{n}^2}{2\text{L}^2}\Rightarrow\text{n}_2=2\text{n}_1$
Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.
View full question & answer→Question 575 Marks
A tuning fork is in unison with a sonometer wire $73cm$ long. If $4$ beats are heard on shortening the wire by $5mm$, find the frequency of the fork.
Answer$l_1 = 73cm, n = 4,$
$l_2 = 73 - 0.5 = 72.5cm$
$\frac{\text{V}_2}{\text{V}_1}=\frac{\text{l}_1}{\text{l}_2}=\frac{73}{72.5},$
$\text{V}_=\frac{73\text{V}_1}{72.5}$
Alsio, $\text{n}=\text{V}_2-\text{V} _1$
$4=\frac{73}{72.5}\text{V}_1-\text{V}_1=\frac{0.5\text{V}_1}{72.5}$
$\text{V}_1=\frac{4\times72.5}{0.5}=580\text{Hz}$
View full question & answer→Question 585 Marks
Calculate the number of beats heard per second is there are three sources of sound of frequencies 400, 401 and 402 of equal intensity sounded together.
AnswerLet us consider the case of three disturbances each of amplitude a and frequencies (n - 1), and (n + 1) respectively. The resultant displacement is given by
$\text{y}=\text{a}\sin2\pi(\text{n}-1)\text{t}+\text{a}\sin2\pi\text{nt}\\+\text{a}\sin2\pi(\text{n}+1)\text{t}$
$=2\text{a}\sin2\pi\text{nt}\cos2\pi\text{t}+\text{a}\sin2\pi\text{nt}$
$=\text{a}(1+2\cos2\pi\text{t})\sin2\pi\text{nt}$
So the resultant amplitude is a $(1+2\cos2\pi\text{t})$
which is maximum when $\cos2\pi\text{t}=+1$
$\therefore 2\pi\text{t}=2\text{k}$ where $\text{k}=0,1,2,3....$
$\text{t}=0, 1, 2, 3....$
Thus the time interval between two consecutive maxima is one. This shows that the frequency of maxima is one.
Similarly, the amplitude is minimum when
$1+2\cos2\pi\text{t}=0$ or $\cos2\pi\text{t}=-\frac{1}{2}$
$2\pi\text{t}=2\text{k}\pi+\frac{2\pi}{3}$ (where k = 0, 1, 2, ....)
$\text{t}=\Big(\text{k}+\frac{1}{3}\Big)=\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3}$
Thus the minima occur after an interval of one second, i.e., the frequency of minima is also one. Hence, the frequency of beats is also one.
Thus, one beat is heard per second.
View full question & answer→Question 595 Marks
Find the velocity of source of sound, when the frequency appears to be:
- Double.
- Half, the original frequency to a stationary listener.
Answer
- v' = 2v.
It is possible is source is approaching the stationary listener i.e., $v_s$ is +
As $\text{v}'=\frac{v-v_\text{L}}{u-u_\text{s}}\text{v}$
$\therefore 2\text{v}=\frac{v}{v-v_\text{s}}\text{v}$
$2=\frac{v}{v-v_\text{s}}$
$2v-2v_\text{s}=v$
$2v_\text{s}=v$
$v_\text{s}=\frac{v}{2}$
Therefore, source should approach the listener with half the velocity of sound propagation.
- $\text{v}'=\frac{v}{2}$
It is possible if source is receding away from the stationary listener i.e., $v_\text{s}$ is negative and $v_\text{L}=0$
$\therefore \text{v}'=\frac{v}{v+v_\text{s}}\text{v}$
$\frac{v}{2}=\frac{v}{v+v_\text{s}}\text{v}$
$\frac{1}{2}=\frac{v}{v+v_\text{s}}$
$v+v_\text{s}=2v$
$v_\text{s}=v$
Therefore, source should recede away from the listener with the velocity of sound propagation. View full question & answer→Question 605 Marks
In the given progressive wave $\text{y}=5\sin(100\pi\text{t+0.4x})$ where y and x are in m, t is in s. What is the:
Particle velocity amplitude.
AnswerStandard form of progressive wave travelling in $+\text{x}$ direction (kx and $\omega\text{}t$ have opposite sign is given) Eqn. is $\text{y}=\text{a}\sin(\omega\text{t}-\text{kx}+\phi)$ $\text{y}=5\sin(100\pi\text{t}-0.4\pi\text{t}+0)$Particle (medium) velocity in the direction of amplitude at a distance $\text{x}$ from source.
$\text{y}=5\sin(100\pi\text{t}-0.4\pi\text{x})$ $\frac{\text{dy}}{\text{dt}}=5\times100\pi\cos(100\pi\text{t}-0.4\pi\text{x})$For maximum velocity of particle is at its mean position
$\cos(100\pi\text{t}-0.4\pi\text{x})=1$ $\Rightarrow100\pi\text{t}-0.4\pi\text{t=0}$ $\therefore\Big(\frac{\text{dy}}{\text{dt}}\Big)_\text{max}=5\times100\pi\times1$$\text{v}_\text{max}$ of medium particle $=500\pi\text{m/ s}$
View full question & answer→Question 615 Marks
Find at what temperature the velocity of sound is air will be $1\frac{1}{2}$ times the velocity at 11°C.
AnswerSuppose velocity of sound in air at t°C $1\frac{1}{2}$ times the velocity at 11°C.
i.e., $v_\text{t}=\frac{3}{2}v_{11}$
As $v_\text{t}=v_0\sqrt{\frac{273+\text{t}}{273}}$
$\therefore$ from (i), $v_0=\sqrt{\frac{273+\text{t}}{273}}=\frac{3}{2}v_{11}$
$v_0=\sqrt{\frac{273+\text{t}}{273}}=\frac{3}{2}v_0\frac{284}{273}$
Squaring both sides, we get
$\sqrt{\frac{273+\text{t}}{273}}=\frac{9}{4}\times\frac{284}{273}$
$1092+4\text{t}=2556$
$4\text{t}=2556-1092=1464$
$\text{t}=\frac{1464}{4}=366^\circ\text{C}$
View full question & answer→Question 625 Marks
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency $6Hz$. The tension in the string A is slightly reduced and the beat frequency is found to reduce to $3Hz$. If the original frequency of A is $324Hz$, what is the frequency of B?
AnswerFrequency of string $A, f_A=324 \mathrm{~Hz}$
Frequency of string $B=f_B$
Beat's frequency, $n=6 \mathrm{~Hz}$
Beat's frequency is given as:
$=\left|f_A+-f_B\right|$
$6=324+-f_B$
$f_B=330 H z \text { or } 318 H z$
Frequency decreases with a decrease in the tension in a string. This is because frequency is irectly proportional to the square root of tension. It is given as:
$\mathrm{v} \propto \text { Underroot } T$
Hence, the beat frequency cannot be 330 Hz
$\therefore \mathrm{f}_{\mathrm{B}}=318 \mathrm{~Hz}$
View full question & answer→Question 635 Marks
A wave $\text{Y} = \text{A}\sin(\omega\text{t} - \text{kx})$ allowed through a string gets reflected from a rigid support and forms stationary wave. Derive the expression for the standing wave.
AnswerGiven wave: $\text{Y}_\text{i}=\text{A}\sin(\omega\text{t}-\text{Kx})$
Reflected wave:
$\text{Y}_\text{r}=\text{A}\sin (\omega\text{t}+\text{Kx}+\pi)$
$=-\text{A}\sin(\omega\text{t}+\text{Kx})$
Applying superposition principle,
$\text{y}=\text{Y}_\text{i}+\text{Y}_\text{r}$
$=\text{A}\sin(\omega\text{t}-\text{Kx})-\text{A}\sin(\omega\text{t}+\text{Kx})$
$\text{y}=2\text{A}\sin\text{Kx}\cos\omega\text{t}$
Since amplitude $2\text{A}\sin\text{Kx}$ varies with position, it represents a standing wave. View full question & answer→Question 645 Marks
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340Hz) when the tube length is 25.5cm or 79.3cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
AnswerFrequency of the turning fork, ν = 340Hz Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.
Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation: $\text{l}_1=\frac{\pi}{4}$ where, length of pipe, $\text{l}_1=25.5\text{cm}=0.255\text{m}$ $\therefore\ \lambda=4\text{l}_1=4\times0.255=1.02\text{m}$ The speed of the sound is given by the relation: $\text{v}=\text{v}\lambda=340\times1.02=346.8\text{m/s}$ View full question & answer→Question 655 Marks
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
(a) Does the pulse have a definite,
- frequency,
- wavelength,
- speed of propagation?
(b) If the pulse rate is 1 after every 20s, (that is the whistle is blown for a split of second after every 20s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05Hz?
Answer(a)
- No
- No
- Yes
(b) No
Explanation:
- The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.
- The short pip produced after every 20s does not mean that the frequency of the whistle is 1/20 or 0.05Hz. It means that 0.05Hz is the frequency of the repetition of the pip of the whistle.
View full question & answer→Question 665 Marks
What are beats? Prove that the number of beats produced per second by the two sound sources is equal to the difference between their frequencies.
AnswerBeats: The phenomenon of alternate variation in the intensity of sound with time at a particular position, when two sound waves of nearly same frequencies and amplitudes superimpose on each other is called beats. If $v_1$ and $v_2$ are the two frequencies, $n = |v_1 - v_2|$, Beats are heard only when $|v_1 - v_2| < 10$, since the sound persist in our ear for $\frac{1}{10}\text{th}$ of a second.
Let us consider two wave trains of equal amplitude ‘A’ and slightly different frequencies $v_1$ and $v_2$ travelling in a medium in the same direction. Displacement of waves are $\text{y}_1=\text{A}\sin2\pi\text{v}_1\text{t}$ and $\text{y}_2=\text{A}\sin2\pi\text{v}_2\text{t},$ on superposition, resultant displacement of waves are:
$\text{y}=\text{y}_1+\text{y}_2=\text{A}\sin2\pi\text{v}_1\text{t}+\text{A}\sin2\pi\text{v}_2\text{t}$
$=\text{A}[\sin2\pi\text{v}_1\text{t}+\sin2\pi\text{v}_2\text{t}]$
$=2\text{A}\cos2\pi\Big(\frac{\text{v}_1-\text{v}_2}{2}\Big)\text{t}\sin2\pi\Big(\frac{\text{v}_1+\text{v}_2}{2}\Big)$
$\Big[\text{Using}\sin\text{A}+\sin\text{B}=2 \cos\frac{\text{A}-\text{B}}{2}\sin\frac{\text{A}+\text{B}}{2}\Big]$
Amplitude $=2\text{A}\cos2\pi\Big(\frac{\text{v}_1\text{v}_2}{2}\Big)\text{t}$ becomes maximum,
when $2\pi\text{t}\Big(\frac{\text{v}_1-\text{v}_2}{2}\Big)=0,\pi,2\pi,...=\text{N}\pi$
i.e. $\text{v}_1\text{v}_2=\frac{2\text{N}\pi}{2\pi\text{t}}=\frac{\text{N}}{\text{t}}=\text{n}$
Where n is the beat frequency – the number of times the maxima and minima is repeated in one second. Thus number of beats produced by two superimposing waves is equal to difference of their frequency.
View full question & answer→Question 675 Marks
For the harmonic travelling wave $\text{y}=2\cos2\pi(10\text{t}-0.0080\text{x}+3.5)$ where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of:
What is the phase difference between the oscillation of a particle located at x = 100cm at t = T and t = 5s?
Answer$\text{y}=2\cos2\pi(10\text{t}-0.0080\text{x+3.5})$ $\text{y}=2\cos(20\pi\text{t}-0.0016\pi\text{x}+7.0\pi)$ Wave is propagated in $+\text{x}$ direction because $\omega\text{t}$ and kx are in with opposite sign standard equation $\text{y}=\text{a}\cos(\omega\text{t}-\text{kx}+\phi)$ $\text{a}=2\ \omega=20\pi,\ \text{k}=0.016\pi$ and $\phi=7\pi$ $\text{T}=\frac{2\pi}{\omega}=\frac{2pi}{20\pi}=\frac{1}{10}\sec$ $\text{x}=100\text{cm}$ At $\text{x}=100,\ \text{t}=\text{T}$ $\phi_120\pi\text{T}-0.016\pi(100)+7\pi=20\pi\times\frac{1}{10}-1.6\pi+7\pi=7.4\pi$At $\text{t}=5\text{s}$
$\phi_120\pi\text{(5)}-0.016\pi(100)+7\pi=100\pi-1.6\pi+7\pi=105.4\pi$
$\phi_2-\phi_1=105.4\pi-7.4\pi=98\pi\ \text{radian}$
View full question & answer→Question 685 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is $1.5m$ and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Determine the tension in the string.
AnswerThe velocity of a transverse wave travelling in a string is given by the relation:
$\text{v}=\sqrt{\frac{\text{T}}{\mu}}\ \dots(\text{i})$
Where,
Velocity of the transverse wave, v = 180m/ s
Mass of the string, m = $3.0 \times 10^{-2}kg$
Length of the string, l = 1.5m
Mass per unit length of the string, $\mu=\frac{\text{m}}{\text{l}}$
$=\frac{3.0}{1.5}\times10^{-2}$
$=2\times10^{-2}\text{kg m}^{-1}$
Tension in the string = T
From equation (i), tension can be obtained as:
$\text{T}=\text{v}^2\mu$
$=(180)^2\times2\times10^{-2}$
$=648\text{N}$
View full question & answer→Question 695 Marks
A stone hangs in air from one end of a wire which is stretched over a sonometer. The wire is in unison with a certain tuning fork when the bridges of the sonometer are 45cm apart. Now the stone hangs immersed in water at 4°C and the distance between the bridges has to be altered by 9cm to re-establish unison of the wire with the same fork. Calculate the density of the stone.
Answer$\text{T}_1=\text{mg}$
$\text{T}_2=\text{mg}-\frac{\text{m}}{\rho}\text{g}$
Since the tension has decreased, the length should also be decreased to re-establish unison. Therefore as,
1 = 45cm,
12 = 45 - 9 = 36cm
Since in both the cases, the wire resonates with the same tuning fork, the frequency of the wire in both cases should be equal, i.e.,
$\frac{1}{2\text{l}_1}\sqrt{\frac{\text{T}_1}{\mu}}=\frac{1}{2\text{l}_2}\sqrt{\frac{\text{T}_2}{\mu}}$
$\frac{\text{T}_1}{\text{T}_2}=\Big(\frac{\text{l}_1}{\text{l}_2}\Big)^2$
$\frac{\text{mg}}{\text{mg}-\frac{\text{mg}}{\rho}}=\Big(\frac{45}{36}\Big)^2=\frac{25}{16}$
$\rho=\frac{25}{9}=2.778\text{g/cc}.$
View full question & answer→Question 705 Marks
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is $40 kHz$. During one fast swoop directly toward a flat wall surface, the bat is moving at $0.03$ times the speed of sound in air. What frequency does the bat hear reflected off the wall?
AnswerUltrasonic beep frequency emitted by the bat, ν = 40 kHz
Velocity of the bat, $v_b = 0.03v$
where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as:
$\text{v}'=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{b}}\Big)\text{v}$
$=\Big(\frac{\text{v}}{\text{v}-0.03\text{v}}\Big)\times40$
$=\frac{40}{0.97}\text{kHz}$
This frequency is refiected by the stationary wall (vs) toward the bat.
The frequency (v'') of the received sound is given by the relation:
$\text{v}''=\Big(\frac{\text{v}+\text{v}_0}{\text{v}}\Big)\text{v}'$
$=\Big(\frac{\text{v}+0.03\text{v}}{\text{v}}\Big)\times\frac{40}{0.97}$
$=\frac{1.03\times40}{0.97}=42.47\text{kHz}$
View full question & answer→Question 715 Marks
Write Newton's formula for the speed of sound in air. Discuss the correction made by Laplace in this formula.
AnswerAccording to Newton, as wave propagates through a medium, temperature is a constant and so propagation is an isothermal process, satisfying PV = Constant.
Differentiating, we get.
$\text{PdV}+\text{VdP}=0$
$\Rightarrow \text{P}=-\frac{(\text{dP})}{\Big(\frac{\text{dV}}{V}\Big)}=\text{B}$
Since velocity of sound waves is $\text{v}=\sqrt{\frac{\text{B}}{\rho}}$
we have, $\text{v}=\sqrt{\frac{\text{P}}{\rho}}$
According to Laplace sound wae propagation in air is an adiabatic process in which temperature can change, but heat energy change should be zero. Hence Laplace corrected it as,
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}},$
where $\gamma$ is the ratio of molar-specific heat capacities at constant pressure and volume (i.e.,) $\gamma =\frac{\text{C}_\text{p}}{\text{C}_\text{v}}.$
View full question & answer→Question 725 Marks
The wave pattern on a stretched string is shown in Interpret what kind of wave this is and find its wavelength.
AnswerThe displacement of medium particlles at distance 10, 20, 30, 40, and 50cm are always rest which is the property of nodes in stationary wave.
$\text{AT}\ \text{t}=\frac{\text{T}}{4}$ and $\frac{\text{3T}}{4}$ all particle are at rest wgich is in stationary wave when the particle crossrs its mean position.
so thet praph of wave shos stationaty wave.
The wave at $\text{x}=10,\ 20,\ 30,\ 40\text{cm}$ there are nodes and distance between successive nodes is $\frac{\lambda}{2}$
$\therefore\frac{\lambda}{2}=(30-20)$ or $\lambda=20\text{cm.}$
View full question & answer→Question 735 Marks
For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same,
- Frequency,
- Phase,
- Amplitude?
AnswerFor the wave on the string described in questions we have seen that l = 1.5m and $\lambda=3\text{m}.$ So, it is clear that $\lambda=\frac{\lambda}{2}$ and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.
- Yes, all the sring particles, except nodes, vibrate with the same frequency v = 60Hz.
- As all string particles lie in one segment, all of them are in same phase.
- Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06m. It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.
View full question & answer→Question 745 Marks
Prove that if a, and a, are the amplitudes of two interfering waves $\phi$ and is their phase difference, the amplitude of the resultant wave is given by,
$\text{a}^2=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi$
AnswerLet $\text{y}_1=\text{a}_1\cos\ \omega\text{t},$
$\text{y}_2=\text{a}_52\cos\ (\omega\text{t}+\phi)$
be the two displacements.
On superposition,
$\text{y}=\text{y}_1+\text{y}_2$
$=\text{a}_1\cos\ \omega\text{t}+\text{a}_2\cos(\omega\text{t}+\phi)$
$=\text{a}_1\cos\omega\text{t}+\text{a}_2\cos\ \omega\text{t}\cos\phi$
$=-\text{a}_2\sin\omega\text{t}\sin\phi$
$=-(\text{a}_2\sin \phi)\sin\omega\text{t}\dots(1)$
$\text{put}\text{ a}_1+\text{a}_2\cos\phi=\text{a}\cos\theta\dots(2)$
$\text{a}_2\sin\phi=\text{a}\sin\theta\dots(3)$
Squaring and adding (2) and (3), we have
$\text{y}=\text{a}\cos\theta\cos\omega\text{t}-\text{a}\sin\theta\sin\omega\text{t}$
$=\text{a}\cos(\omega\text{t}+\theta)$
$\therefore\text{y}=\Big(\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi\Big)^{\frac{1}{2}}\cos(\omega\text{t}+\theta)$
where $\theta=\tan^{-1}\Big(\frac{\text{a}_2\sin\phi}{\text{a}_1+\text{a}_2\cos\phi}\Big)$
View full question & answer→Question 755 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is $1.5m$ and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
AnswerA wave travelling along the positive x-direction is given as: $\text{y}_1=\text{a}\sin(\omega\text{t}-\text{kx})$The wave travelling along the negative x-direction is given as:
$\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})$ The superposition of these two waves yields: $\text{y}=\text{y}_1+\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})-\text{a}\sin(\omega\text{t}-\text{kx})$
$=\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})-\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})$
$=-2\text{a}\sin(\text{kx})\cos(\omega\text{t})$
$=-2\text{a}\sin\Big(\frac{2\pi}{\lambda}\text{x}\Big)\cos(2\pi\text{ vt})\ \dots(\text{i})$ The transverse displacement of the string is given as: $\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})\ \dots(\text{ii})$ Comparing equations (i) and (ii), we have: $\frac{2\pi}{\lambda}=\frac{2\pi}{3}$
$\therefore$ Wavelength, $\lambda=3\text{m}$ It is given that: $120\pi=2\pi\text{v}$ Frequency, $\text{v} = 60\text{Hz}$ Wave speed, $\text{v}=\text{v}\lambda$
$=60\times3=180\text{m/s}$
View full question & answer→Question 765 Marks
Give any three differences between progressive waves and stationary waves. A stationary wave is $\text{y}=12\sin300\text{t}\cos^2\text{x}.$ What is the distance between two nearest nodes?
Answer
|
S. No.
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Progressive Wave
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Stationary Wave
|
|
1.
|
All particles have same phase and amplitude.
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Amplitude varies with position.
|
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2.
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Speed of motion is same.
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Speed varies with position.
|
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3.
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Energy is transported.
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Energy is not transported.
|
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4.
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Same change in pressure and density is with every point.
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Pressure and density varies with point.
|
$\text{y}=12\sin300\text{t}\cos2\text{x}$
Comparing with equation of stationary wave
$\text{y}=2\text{A}\sin\omega\text{t}\cos\text{kx}$
$\text{k}=2$
Distance between two consecutive nodes $=\frac{\lambda}{2}$ Where $\lambda$ is wavelength
$\text{k}=\frac{2\pi}{\lambda}$
$\Rightarrow\frac{\pi}{\Big(\frac{\lambda}{2}\Big)}=2$
$\therefore \frac{\lambda}{2}=\frac{\pi}{2}$
So, the distance between two nearest nodes is $\frac{\pi}{2}.$ View full question & answer→Question 775 Marks
- What is beat phenomenon?
-
A whistle revolve in a circle with angular velocity of o =$20 rad s^{-1}$. If the frequency of the sound is $385Hz$ and speed is $340ms^{-1}$, than find the frequency heard by the observer when the whistle is at B.Answer
- Beat phenomenon: When two sound waves of nearly same frequencies and amplitudes travelling in a medium along the same direction, super impose on each other, the intensity of resultant sound at a particular position rises and falls alternatively with time.
This phenomenon of alternate variation in the intensity of sound with time at a particular position, when two sound waves of nearly same frequencies and amplitudes superimpose on each other is called beats.
-
$W = 20 rad s^{-1}$
Frequency of sound $\gamma =385\text{Hz}$
$\nu=\text{Speed}=340\text{ms}^{-1}$
$\nu_\text{s}=\text{rw}=0.50\times20=10\text{m/s}$
Using formula,
$\gamma'=\frac{\text{v}}{\text{v}-\text{v}_\text{s}}\times\gamma$
$=\frac{340\times385}{340-10}=\frac{340\times285}{330}$
$=395\text{Hz}$
Hence, the frequency heard by an observer is 397Hz when the whistle is at B. View full question & answer→Question 785 Marks
What do you mean by interference of waves? Distinguish between constructive and destructive interference.
Standing waves are produced by the superposition of two waves
$\text{y}_1=0.05\sin(3\pi\text{t}-2\text{x})$ and $\text{y}_2=0.05\sin(3\pi\text{t}+2\text{x})$
where y and x are measured in metres and t in seconds. Find the amplitude of a particle at x = 0.5m.
AnswerInterference of waves is the phenomenon of redistribution of energy in space on account of superposition of two waves of same nature, same frequency and equal or comparable amplitudes and travelling in the given medium in the same direction.
Constructive interference takes place when the two superposing waves are in same phase i.e., crest of one wave (in transverse waves) coincides with crest of another wave and vice-versa. As a result, the resultant amplitude and hence intensity of the resultant wave is maximum. Thus, for constructive interference, the phase difference between the superposing waves $\Delta\phi=0$ or $2\text{n}\pi,$ where n is an integer i.e., n = 1, 2, 3.....
Destructive interference takes place when two superposing waves are in mutually opposite phase i.e., in superposing of two transverse waves crest of one wave exactly coincides with trough of another wave. As a result, the resultant amplitude and hence intensity of the resultant wave is minimum. For destructive interference, the phase difference $\Delta\phi=(2\text{n}-1)\pi,$ where n = 1, 2, 3....
Numerical:
The resultant displacement is given by
$\text{y}=\text{y}_1+\text{y}_2=0.05\{\sin(3\pi\text{t}-2\text{x})+\sin(3\pi\text{t}+2\text{x})\}$
Using trigonometric relation
$\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta,$ we have
$\text{y}=0.1\cos2\text{x}\sin3\pi\text{t}$ or $\text{y}=\text{A}\sin3\pi\text{t}$
where A, the amplitude of standing waves, is
given by $\text{A}=0.1\cos2\text{x}$ with
$\text{x}=0.5\text{m}$
$\cos2\text{x}=\cos(2\times0.5\text{ rad})$
$=\cos(1\text{ rad})=\cos\Big(\frac{\pi}{3.142}\Big)=\cos57.3^\circ$
$=0.54$
Amplitude A at (x = 0.5) = 0.1 × 0.54
= 0.054m.
View full question & answer→Question 795 Marks
For the travelling harmonic wave, y = 2.0 cos 2π (10t - 0.0080x + 0.35), where x and y are in cm and t in s. What is the phase difference between oscillatory motion at two points separated by a distance of,
- $4\text{m}$
- $0.5\text{m}$
- $\frac{\lambda}{2}$
- $\frac{3\lambda}{4}$
Answer$\text{y}=2.0\cos[2\pi(10\text{t}-0.0080\text{x})+2\pi\times0.35]$ $\text{y}=2.0\cos\Big[2\pi\times0.0080\big(\frac{101\text{t}}{0.0080}-\text{x}\big)+0.7\pi\Big]$ Compare it with the standard equation of a travelling harmonic wave, $\text{y}=\text{r}\cos\Big[\frac{2\pi}{\lambda}\big(\text{V}_{\text{t}}-\text{x}\big)+\phi\Big]$we get, $\frac{2\pi}{\lambda}=2\pi\times0.0080$
phase difference$\phi=\frac{2\pi}{\lambda}\text{x}$
- When $\text{x}=4\text{m}=400\text{cm,}$
$\phi=6.4\pi\text{ rad.}$
- When $\text{x}=0.5\text{m}=50\text{cm}$
$\phi=2\pi\times0.0080\times50=0.8\pi\text{ rad.}$
- When $\text{x}=\frac{\lambda}{2}$
$\phi=\frac{2\pi}{\lambda}\times\frac{\lambda}{2}=\phi\text{ rad}$
- When $\text{x}=\frac{3\lambda}{4}$
$\phi=\frac{2\pi}{\lambda}\times\frac{3\lambda}{4}=\frac{3\pi}{52}\text{ rad}$ View full question & answer→Question 805 Marks
Given below are some functions of x and t to represent the displacement of an elastic wave.
$\text{y}=100\cos(100\pi\text{t+0.5x})$
Answer$\text{y}=4\sin\Big(5\text{x}-\frac{\text{t}}{2}\Big)+3\cos\Big(5\text{x}-\frac{\text{t}}{2}\Big)$
$\text{Let}4=\text{a}\cos\phi\ ...(\text{ii})$and $3=\text{a}\sin\phi\ ...(\text{iii})$
$\text{a}^2\cos^2\phi+\text{a}^2\sin^2\phi=4^2+3^2$ Squaring and adding (ii), (iii)
$\text{a}^2=25\text{K}\Rightarrow\text{a}=5$
Substituting (ii), (iii) in (i)
$\text{y}=\text{a}\cos\phi\sin\Big(5\text{x}-\frac{\text{t}}{2}\Big)+\text{a}\sin\phi\cos\Big(5\text{x}-\frac{\text{t}}{2}\Big)$
$\text{y}=\text{a}\sin\Big(5\text{x}-\frac{\text{t}}{2}+\phi\Big)$
$\text{y}=\text{5}\sin\Big(5\text{x}-\frac{\text{t}}{2}+\phi\Big)$
Which represents the progressive wave in $+\text{x}$ direction as the sign of Kx (or5x) and $\omega\text{t}\Big(\frac{1}{2}\text{t}\Big)$ are opposite so it travels in $+\text{x}$ direction. So (d) (ii)
View full question & answer→Question 815 Marks
Explain how stationary waves are formed in open and closed pipes. Compare the first three harmonics produced in them.
AnswerStationary or Standing Waves: Formed by two waves moving in opposite directions interacting. They may have equal or unequal amplitudes and generally equal frequencies are $\text{Y}=\pm2\text{A}\sin\text{kx}\cos\omega\text{t}$ refers to a standing wave, where nodes and antinodes are alternatively formed with a separation $\frac{\lambda}{2}.$ In closed pipes, there will be a node at the closed end and antinode at open end while both ends will be antinodes in an open pipe. The frequency in them are given by:
This type of pattern is being formed. On observation, one can conclude that only odd harmonics are available in a closed pipe, while all harmonics are available in an open pipe. View full question & answer→Question 825 Marks
A metallic rod of length $1m$ is rigidly clamped at its midpoint. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the midpoint. The amplitude of an antinode is $2 \times 10^{-6}m$. Write the equation of motion at a point $2cm$ from the midpoint and those of the constituent waves in the rod. (Young's modulus = $2 \times 10^{11}Nm^{-2}$ and density = $8000kg m^{-3}$)
AnswerThe equation of standing wave can be written as
$\text{y}=1\text{A}\sin\text{kx}\cos\omega\text{t}$
where $\text{k}=\frac{2\pi}{\lambda}$ and $\omega=\frac{2\pi\text{V}}{\lambda}.$
The standing wave is obtained by adding the equation of two identical progressive waves travelling in opposite directions
$\text{y}_1=\text{A}\sin(\text{kx}-\omega\text{t});$
$\text{y}_2=\text{A}\sin(\text{kx}+\omega\text{t})$
In the present problem the length L of the rod = 1 metre.
i.e., $\text{}=\frac{5\lambda}{2}$ or $\frac{2}{5}\text{ metre}.$
Velocity of longitudinal wave is given by
$\text{V}=\sqrt{\frac{\text{Y}}{\rho}}=\sqrt{\frac{2\times10^{11}}{8000}}$
$=5\times10^3\text{ms}^{-1}$
$\text{k}=\frac{2\pi}{\lambda}=\frac{2\pi}{\frac{2}{5}}=5\pi\text{ metre}^{-1}$
$\omega=\frac{2\pi\text{V}}{\lambda}$
$=\frac{2\pi\times5\times10^3}{\frac{2}{5}}=(25\times10^3\pi)\text{s}^{-1}$
Hence equation of standing wave is
$\text{y}=(2\times10^{-6})\sin5\pi\text{x}\cos25\times10^3\pi\text{t}$
Equations of component waves are
$\text{y}_1=(1\times10^{-6})\sin(5\pi\text{x}-25\times10^3\pi\text{t})$
$\text{y}_2=(1\times10^{-6})\sin(5\pi\text{x}+25\times10^3\pi\text{t})$ View full question & answer→Question 835 Marks
Derive an expression for nth mode of vibration in case of a closed end organ pipe. Hence, give the value of $V_1 : V_3: V_5$.
AnswerDisplacement at position x and time t in incident wave,
$\text{y}_1=\text{r}\sin\frac{2\pi}{\lambda}(\text{vt}+\text{x})$
Wave reflected at closed end of pipe suffers a phase reversal of $\pi.$
$\therefore$ For reflected wave $\text{y}_2=\text{r}\sin\Big[\frac{2\pi}{\lambda}(\text{vt}-\text{x})+\pi\Big]$
$=-\text{r}\sin\frac{2\pi}{\lambda}(\text{vt}-\text{x})$
According to superposition principle,
$\text{y}=\text{y}_1+\text{y}_2$
$\text{y}=\text{r}$
$\Big[\sin\frac{2\pi}{\lambda}(\text{vt}+\text{x})-\sin\frac{2\pi}{\lambda}(\text{vt}-\text{x})\Big]$
$\text{y}=2\text{r}\cos\frac{2\pi}{\lambda}\text{vt}.\sin\frac{2\pi}{\lambda}\text{x}.$
At closed end of pipe x = 0
$\therefore \sin \frac{2\pi}{\lambda}\text{x}=\sin \theta=0$
y = 0, i.e., a node is formed.
At the open end of the pipe of length L,X = L and antinode is to be formed, i.e., y = max
$\text{y}=2\text{r}\cos\frac{2\pi}{\lambda}\text{vt}.\sin\frac{2\pi}{\lambda}\text{L}$
y will be max, when
$\sin \frac{2\pi\text{L}}{\lambda}=\text{max}= \pm1$
$=\sin(2\text{n}-1)\frac{\pi}{2}$
Where n = 1, 2, 3,....
$\frac{2\pi\text{L}}{\lambda}=(2\text{n}-1)\frac{\pi}{2}$
$\lambda=\frac{4\text{L}}{(2\text{n}-1)}$
- For first normal mode of vibration,
Let $\lambda_1=$ wavelength corresponding tp n = 1
$\lambda_1=\frac{4\text{L}}{(2\times1-1)}=4\text{L}$
$\text{L}=\frac{\lambda_1}{4}$
Frequency, $\text{v}_1\frac{\text{v}}{\lambda_1}=\frac{\text{v}}{4\text{L}}$
$\text{v}_1=\frac{\text{v}}{4\text{L}}$ fundametal frequency
For second normal mode of vibration,
$\lambda_2=$ wavelength of standing waves corresponding to n = 2
$\lambda_2=\frac{4\text{L}}{2\times2-1}=\frac{4\text{L}}{3}$
$\text{v}_2=\frac{\text{v}}{\lambda_2}=\frac{\text{v}}{\frac{4\text{L}}{3}}$
$=\frac{3\text{v}}{4\text{L}}=3\text{v}_1$
For third normal mode of vibration.
$\lambda_3=$ wavelength of standing waves corresponding to n = 3
$\lambda_3=\frac{4\text{L}}{2\times3-1}=\frac{4\text{L}}{5}$
$\text{v}_3=\frac{\text{v}}{\lambda_3}=\frac{\text{v}}{\frac{4\text{L}}{5}}=5\frac{\text{v}}{4\text{L}}$
$\text{v}_3=5\text{v}_1$ fifth harmonic
In genetal, $\text{v}_\text{n}=\frac{(2\text{n}-1)\text{v}}{4\text{L}}=(2\text{n}-1)\text{v}_1$
$\therefore$ ratio of frequencies of harmonics is 1 : 3 : 5 : 7. View full question & answer→Question 845 Marks
- What are beats? Name the basic phenomenon due to which beats are produced.
- Two sources of sound are producing waves of frequency $n_1$ and $n_2$, where $(n_1 - n_2)$ is small, show mathematically that the beat frequency is $(n_1 - n_2)$.
Answer
- Beats: The waxing and waning of sound due to interaction between two slightly different frequencies. If y, and v, are the two frequencies $v_b = |v_1 - v_2|$.
Beats are heard only when $|v_1 - v_2l< 10$, since the sound persist in our ears for $\frac{1}{10}\text{th}$ of a second.
- When n, and n, are two frequencies represented by $\text{y}_1=\text{A}\sin2\pi\text{n}_1\text{t}$ and $\text{y}_2=\text{A}\sin2\pi\text{n}_2\text{t},$ we get on superposition,
$\text{y}=\text{y}_1+\text{y}_2$
$=2\text{A}\cos\pi\Big(\frac{\text{n}_1-\text{n}_2}{2}\Big)\text{t}\sin2\pi\Big(\frac{\text{n}_1\text{n}_2}+{2}\Big)\text{t}$
Amplitude $=2\text{A}\cos2\pi\Big(\frac{\text{n}_1-\text{n}_2}{2}\Big)\text{t}$ becomes maximum, when $2\pi\text{t}\Big(\frac{\text{n}_1-\text{n}_2}{2}\Big)=0,2\pi...\text{N}\pi\text{ i.e., }\text{n}_1-\text{n}_2=\frac{2\text{N}\pi}{2\pi\text{t}}=\frac{\text{N}}{\text{t}}=\text{N}.$ n where n is the beat frequency-the number of times the maxima and minima is repeated in one second. View full question & answer→Question 855 Marks
The displacement of an elastic wave is given by the function $\text{y}=3\sin\omega\text{t}+4\cos\omega\text{t}$ where y is in cm and t is in second. Calculate the resultant amplitude.
Answer$\because\text{y}=3\sin\omega\text{t}+4\cos\omega\text{t}\ ...(\text{i})$
Let $3=\text{a}\cos\phi\ ...(\text{ii})$
$4=\text{a}\sin\phi\ ...{\text{iii}}$
Then $\text{y}=\text{a}\cos\phi\sin\omega\text{t}+\text{a}\sin\phi\cos\omega\text{t}$
$\text{y}=\text{a}\sin(\omega\text{t}+\phi) $
From (ii) and(iii)
$\tan \phi=\frac{4}{3}$ or $\phi=\tan^{-1}\frac{4}{3}$
On squaring and adding (ii) and (iii) equations
$\text{a}_2\cos^2\phi+\text{a}^2\sin^2\phi=3^2+4^2$
$\text{a}^2(\cos^2\phi+\sin^2\phi)=9+16$
$\text{a}^2=25\Rightarrow\text{a}=5$
$\text{y}'=5\sin(\omega\text{t}+\phi)$ when $\phi=\tan^{-1}\frac{4}{3}$
Hence, New amplitude is 5 cm.
View full question & answer→Question 865 Marks
A pipe $20cm$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430Hz$ source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340m s^{–1}$).
AnswerFirst (Fundamental); No
Length of the pipe, l = 20cm = 0.2m
Source frequency = $n^{th}$ normal mode of frequency, $ν_n= 430Hz$
Speed of sound, v = 340m/ s
In a closed pipe, the $n^{th}$ normal mode of frequency is given by the relation:
$\text{v}_\text{n}=(2\text{n}-1)\frac{\text{v}}{4\text{l}}$ n is an interger = 0, 1, 2, 3
$430=(2\text{n}-1)\frac{340}{4\times0.2}$
$2\text{n}-1=\frac{430\times4\times0.2}{340}=1.01$
$2\text{n}=2.01$
$\text{n}\sim1$
Hence, the first mode of vibration frequency is resonantly excited by the given source. In a pipe open at both ends, the $n^{th}$ mode of vibration frequency is given by the relation:
$\text{v}_\text{n}=\frac{\text{nv}}{2\text{l}}$
$\text{n}=\frac{2\text{lv} _n}{\text{v}}$
$=\frac{2\times0.2\times430}{340}=0.5$
View full question & answer→Question 875 Marks
Explain Doppler effect in sound. Obtain an expression for apparent frequency of sound when source moves and listener is at rest.
Answer
- Doppler effect: The phenomena of apparent change in pitch of sound caused due to relative motion between a source and an observer is called Doppler effect.
- Let S and O be the source and observer. If v is the frequency of sound with velocity v released by the source, then a number of waves will be received by the observer at rest.
- When the source approaches the stationary listener, the number of waves received, increases due to the apparent shortening of the wavelength.
Wavelength perceived $\lambda'=\frac{\text{velocity of sound w.r. to moving source}}{\text{frequency}}$
$\therefore \lambda'=\frac{\nu-\nu_\text{s}}{\text{V}}$
Using $\lambda'=\frac{\nu}{\text{V}'}$ we get,
and $\frac{\nu}{\text{V}'}=\frac{\nu-\nu_\text{s}}{\text{V}}$
$\text{V}'=\text{V}\Big(\frac{\nu}{\nu-\nu_\text{s}}\Big)$
- When the source is moving away from the listener who is at rest, then velocity of source is negative.
$\therefore \text{V}'=\frac{\nu}{\nu-(-\nu_\text{s})}\text{V}=\frac{\nu}{\nu+\nu_\text{s}}\text{V}$ View full question & answer→Question 885 Marks
A drop of water, $2 mm$ in diameter, falling from a height of $50 cm$ in a bucket generates sound which can be heard from the $5 m$ distance. Take all the gravitational energy difference as going into the sound form, the transformation being spread in time over $0.2 s$ . Deduce the average intensity and the amplitude of vibration at the listener's end. Given: density of air $=1.3 \mathrm{~kg} \mathrm{~m}^{-3}$, frequency of wave $=1000 \mathrm{~Hz}$ and $\mathrm{c}=350 \mathrm{~ms}^{-1}$ ?
AnswerMass of drop, $\text{m}=\frac{4}{3}\pi\text{r}^3\times\rho,$ where $\rho$ is the density of water.
Loss in gravitational energy when the drop falls through a height h,
$\text{E}=\frac{4}{3}\pi\text{r}^3\times\rho\times\text{gh}$
Ift be the time during which this energy is fully convened into sound energy, then the intensity of sound at a distance R is given by
$\text{I}=\frac{\text{E}}{4\pi\text{R}^2\times\text{t}}=\frac{\frac{4}{3}\pi\text{r}^3\times\rho\text{gh}}{4\pi\text{R}^2\text{t}}$
$\text{I}=\frac{\text{r}^3\rho\text{gh}}{3\text{R}^2\text{t}}$
Now, $d = 2mm = 2 \times 10^{-3}m,$
$r = 10^{-3}m$
$\rho = 10^3\text{kg m}^{-3},$
$\text{g}=9.8\text{m s}^{-2},$
$\text{h}=50\text{cm}=0.5\text{m},$
$\text{R}=5\text{m}$ and $\text{t}=0.2\text{s}$
$\therefore \text{I}=\frac{(10^{-3})^3\times1000\times9.8\times0.5}{3\times(5)^2\times0.2}\text{Wm}^{-2}$
$=3.267\times10^{-7}\text{Wm}^{-2}$
$\Rightarrow \text{amp}=\sqrt{\text{I}}=\sqrt{3.267\times10^{-7}}$
$=5.71\times10^{-4}\text{m}$
View full question & answer→Question 895 Marks
A simple harmonic wave of amplitude 1cm and frequency 100Hz is travelling along positive x-direction with a velocity of 15m/s. Calculate the displacement y, particle velocity and particle acceleration at x = 180 cm from the origin at t = 5s.
AnswerHere, $\text{a}=1\text{cm}$
$\text{v}=100\text{Hz}$
$\nu=15\text{m/s}$
$\text{y}=?,\frac{\text{dy}}{\text{dt}}=? \frac{\text{d}^2\text{y}}{\text{dt}^2}=?\text{x}=180\text{cm}$
$\text{t}=5\text{s}$
Now $\lambda=\frac{\nu}{\text{v}}=\frac{1500}{100}=15\text{cm}$
$\text{T}=\frac{1}{\text{v}}=\frac{1}{100}=0.01\text{sec.}$
The equation of the wave is
$\text{y}=\text{a}\sin\Big[\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\Big]$
$=1\sin\Big[\frac{2\pi\times5}{0.01}-\frac{2\pi\times180}{15}\big]$
$\text{y}=1\sin(1000\pi-24\pi)$
$=\sin976\pi=0$
As $\sin\Big[\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\Big]=0;\cos\Big[\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\Big]=1$
$\therefore \frac{\text{dy}}{\text{dt}}=\text{a}\times\frac{2\pi}{\text{T}}\cos\Big(\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\Big)$
$=1\times\frac{2\pi}{0.01}\times1=200\pi\text{cm/s}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{a}\times\Big(\frac{2\pi}{\text{T}}\Big)^2\sin\Big[\frac{2\pi\text{t}}{\text{T}}-\frac{2\pi\text{x}}{\lambda}\Big]$
$=\text{zero}$
View full question & answer→Question 905 Marks
An incident wave and a reflected wave are represented by:
$\xi_1=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$
$\xi_2=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$
Derive the equation of the stationary wave and calculate the position of the nodes and antinodes.
Answer$\xi_1=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$ [Incident wave]
$\xi_2=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$ [Reflected wave]
As there is a phase change of a radian on reflection at the rigid boundary, then
$\xi_2=\text{a}\sin\Big[\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})+\pi\Big]$
$\therefore \xi_2=-\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$
According to the superposition principle, the resultant displacement y at time t and position x is given by
$\xi=\xi_1+\xi_2$
$\xi=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})-\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$
$\therefore \xi=2\cos\frac{2\pi}{\lambda}\nu\text{t}.\sin\frac{2\pi}{\lambda}(-\text{x})$
$\xi=-2\sin\frac{2\pi}{\lambda}\text{x}\cos\frac{2\pi}{\lambda}\nu\text{t}$
Here $2\sin\frac{2\pi}{\lambda}\text{x}=\text{Amplitude}$
Posotion of Nodes:
At nodes, amplitude = 0
From the figures, we see that there are two nodes in the first normal mode of vibration, then three nodes in the second normal mode; and so on, therefore in the nth normal mode of vibration, there will be (n + 1) nodes.
These nodes are located at $\text{x}=0,\frac{\text{L}}{\text{n}},\frac{3\text{L}}{\text{n}},...\text{L}.$
Position of Antinodes: At antinodes, displacement is maximum. As antinodes are located in between the nodes; therefore, their position will be given by
$\text{x}=\frac{\text{L}}{2\text{n}},\frac{3\text{L}}{2\text{n}},\frac{5\text{L}}{2\text{n}},...,\frac{(2\text{n}-1)\text{L}}{2\text{n}}$
where n= 1, 2, ... View full question & answer→Question 915 Marks
Discuss Newton's formula for the velocity of longitudinal waves in air. What correction was applied by Laplace and why?
AnswerAccording to Newton, as wave propagates through a medium, temperature is a constant and so propagation is an isothermal process, satisfying PV = Constant. Differentiating, we get.PdV + VdP = 0
$\Rightarrow\text{P}=-\frac{(\text{dP})}{\Big(\frac{\text{dV}}{\text{V}}\Big)}=\text{B}$
Since velocity of sound waves is $\nu=\sqrt{\frac{\text{B}}{\rho}}$
we have, $\nu=\sqrt{\frac{\text{P}}{\rho}}$
According to Laplace, temperature can change, but heat energy change should be zero and so the process is adiabatic. Hence Laplace corrected it as,
$\nu=\sqrt{\frac{\gamma\text{P}}{\rho}},$ where $\gamma$ is the ratio of molar-specific heat capacities at constant pressure and volume (i.e.,) $\gamma =\frac{\text{C}_\text{p}}{\text{C}_\nu}.$
View full question & answer→Question 925 Marks
If c is r.m.s. speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/ v is constant and independent of temperature for all diatomic gases.
AnswerWe know that $\text{c}\sqrt{\frac{3\text{p}}{\rho}}$ for molecules.
$\text{c}=\sqrt{\frac{3\text{RT}}{\text{M}}}$
$\therefore\frac{\text{p}}{\rho}=\frac{\text{PT}}{\text{M}}\because\frac{\text{P}}{\rho}=\frac{\text{RT}/\text{V}}{\text{M/V}}$
M = molar mass of gas
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}$
$\therefore\text{RV}=\text{nRT}$
$\text{n}=1$
$\text{p}\frac{\text{PT}}{\text{V}}$
$\frac{\text{c}}{\text{v}}=\frac{\sqrt{\frac{3\text{RT}}{\text{M}}}}{\sqrt{}\frac{\gamma\text{RT}}{\text{M}}}=\sqrt{\frac{3}{\gamma}}$
$\gamma=\frac{\text{C}_\text{P}}{\text{C}_\text{v}}=$ adiabatic constant for diatomic gas
$\gamma=\frac{7}{5}$
$\therefore\frac{\text{c}}{\text{v}}=\sqrt{\frac{3}{7/5}}=\sqrt{\frac{15}{7}}=$ constant.
View full question & answer→Question 935 Marks
Explain Doppler effect in sound. Obtain an expression for apparent frequency of sound when source and listener are approaching each other.
AnswerWhenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by listener is different from actual frequency of sound emitted by the source.
Let S be a source of sound and L, the listener of sound, both initially at rest. Let v be the actual frequency of sound emitted by the source and a be the actual wavelength of sound emitted. If v is velocity of sound in still air, then
$\lambda=\frac{\nu}{\text{V}}$
Let the distance between source and listener be V, so that v waves from the source reach listener in 1 second.
$V_m$= Velocity of medium, $V_s$ = Velocity of source, $V_L$ = Velocity of listener
Resultant velocity of sound along
$SL = (V + V_m)$
SS; = distance proved by source in 1sec.
= $V_s$ along SL
$\therefore$ Relative velecity of sound w.r.t. source = $[(V + V_m) - V_s]$
As the frequency remains unchanged.
$\therefore$ V waves emitted in one second occupy the distance $[(V + V_m) - V_s]$
$\lambda'=\frac{[\text{V}+\text{V}_\text{m}-\text{V}_\text{s}]}{\nu}$
LL; = VL Relative vel. of sound waves w.r.t. listener $[(V + V_m)_- VL]$
Apparent frequency of sound waves heard by listener is $\text{v}'=\frac{(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}}{\lambda'}$
$\text{v}'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}]}{(\text{V}+\text{V}_\text{m}-\text{V}_\text{S})}$
when both approach each other $\text{V}_\text{S}=(+),\text{V}_\text{L}=(-)$
$\text{v}'=\frac{\text{V}-(-\text{V}_\text{L})}{\text{V}-\text{V}_\text{S}}\text{v}=\Big(\frac{\text{V}+\text{V}_\text{L}}{\text{V}-\text{V}_\text{S}}\Big)\text{v}.$ View full question & answer→Question 945 Marks
What are stationary waves? How are they formed in strings? Draw the various modes of vibration in them.
AnswerStationary or Standing Waves:
Formed by two waves moving in opposite directions interacting. They may have equal or unequal amplitudes and generally equal frequencies. $\text{Y}=\pm2\text{A}\sin\text{kx}\cos\omega\text{t}$ refers to a standing wave, where nodes and antinodes are alternatively formed with a separation $\frac{\lambda}{2}.$
Given wave: $\text{Y}_\text{i}=\text{A}\sin(\omega\text{t}-\text{kx})$
Reflacted wave; $\text{Y}\text{r}=\text{A}\sin(\omega\text{t}+\text{kx}+\pi)$
In strings, stationary waves formed profuce frequencies, multiple of $\Big(\frac{\nu}{2\text{l}}\Big)$ or harmonics of $\Big(\frac{\nu}{2\text{l}}\Big)$ .i.e., $\text{v}=\frac{\text{n}\nu}{2\text{l}}=\frac{\text{n}}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}.$
Since $\text{v}=\frac{\text{n}\nu}{2\text{l}},$ the pattern can be shown as below:
Applying superposition principle,
$\text{y}=\text{Y}_\text{i}+\text{Y}_\text{r}$
$=\text{A}\sin(\omega\text{t}-\text{kx})-\text{A}\sin(\omega\text{t}+\text{kx})$
$\text{y}=2\text{A}\sin\text{kx}\cos\omega\text{t}.$
Since amplitude $2\text{A}\sin\text{kx}$ varies with position, it represerts a standing wave. View full question & answer→Question 955 Marks
Stationary waves are set up by the superposition of two waves given by $\text{y}_1=0.05\sin(5\pi\text{t}-\text{x})$ and $\text{y}_2=0.05\sin(5\pi\text{t}),$ where x and y are in metre and t in sec. Calculate the displacement of a particle at a distance of x = 1m.
AnswerUsing superposition principle, the resultant displacement at time t is given by
$\text{y}=\text{y}_1+\text{y}_2$
$=0.05\sin(5\pi\text{t}-\text{x})+0.05\sin(5\pi\text{t}+\text{x})$
$=0.05\times2\sin\Big(\frac{5\pi\text{t}-\text{x}+5\pi\text{t}+\text{x}}{2}\Big)$
$\cos\frac{5\pi\text{t}+\text{x}-5\pi\text{t}+\text{x}}{2}$
$=0.1\sin5\pi\text{t}\cos\text{x}$
$\text{y}=[0.1\cos\text{x}]\sin5\pi\text{t}$
Amplitude $\text{r}=0.1\cos\text{x}$
At $\text{x}=1\text{m, r}=0.1\cos1=0.1\cos\frac{180}{\pi}$
$=0.1\cos57.3^\circ$
$=0.1\times0.5406$
$=0.054\text{m}.$
View full question & answer→Question 965 Marks
Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4.
AnswerLet n be the number of loop in the string.The length of each loop is $\frac{\lambda}{2}$
$\therefore\text{L}=\frac{\text{n}\lambda}{2}$ or $\lambda=\frac{2\text{L}}{\text{n}}$
$\text{v}=\text{v}\lambda$ and $\lambda=\frac{\upsilon}{\text{v}}.$
so $\frac{\upsilon}{\text{v}}=\frac{2\text{L}}{\text{n}}$
$\text{v}=\frac{\text{n}}{2\text{L}}.\text{v}$ v is stretch string $=\sqrt{\frac{\text{T}}{\text{m}}}$
$\therefore\text{v}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}$
For n =1, $\text{v}_1=\frac{1}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=\text{v}_0$
If n = 2 then $\text{v}_2=\frac{2}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=2\text{v}_0$
n = 3 then $\text{v}_3=\frac{3}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=3\text{v}_0$
$\therefore\text{v}_1:\text{v}_2:\text{v}_3:\text{v}_4:=\text{n}_1:\text{n}_2:\text{n}_3:\text{n}_4:=1:2:3:4$ View full question & answer→Question 975 Marks
A simple harmonic wave is expressed by equation:
$\text{y}=7\times10^{-6}\sin\Big(800\pi\text{t}-\frac{\pi}{42.5}\text{x}\Big)$
where y and x are in cm. and t in seconds. Calculate the following:
- Amplitude.
- Frequency.
- Wavelength.
- Wave velocity.
- Phase difference between two particles separated by 17.0cm.
AnswerComparing the given equation with
$\text{Y}=\text{A}\sin(\omega\text{t}-\text{kx}),$ we get
- Amplitude $=\text{A}=7\times10^{-6}\text{cm}$
- Frequency $=\text{v}=\frac{\omega}{2\pi}=\frac{800\pi}{2\pi}=400\text{Hz}$
- Wavelength $=\lambda=\frac{2\pi}{\text{k}}=\frac{2\pi}{\Big(\frac{\pi}{42.5}\Big)}=85\text{cm}$
- Wave velocity $=\nu=\frac{\omega}{\text{k}}=\frac{800\pi}{\Big(\frac{\pi}{42.5}\Big)}$
$=3400\text{cm s}^{-1}$
$=340\text{ms}^{-1}$
- Using $=\frac{\phi}{2\pi}=\frac{\text{x}}{\lambda},$ we get
Phase difference $=\phi=\frac{2\pi}{85}\times17$
$=\frac{2\pi}{5}\text{radian}$ View full question & answer→Question 985 Marks
Discuss the various factors influencing velocity of sound. A sonometer wire of length $110cm$ is stretched with a tension T and fixed at its ends. The wire is divided into three segments by placing two bridges below it. Where should the bridges be placed so that the fundamental frequencies of the segments are in the ratio $1: 2: 3$?
AnswerFor factors influencing velocity of sound, see text.
Numerical: Let $L_1, L_2$, and $L_3$ be the lengths of the segments of wire AB (Fig.).
Then
$L_1 + L_2 + L_3 = 110cm$ ...(1)
Let $n_1 n_2$ and $n_3$ be their respective fundamental frequecies, Thus
$\text{n}_1=\frac{1}{2\text{L}_1}\sqrt{\frac{\text{T}}{\text{m}}}$
$\text{n}_2=\frac{1}{2\text{L}_2}\sqrt{\frac{\text{T}}{\text{m}}}$ and $\text{n}_3=\frac{1}{2\text{L}_3}\sqrt{\frac{\text{T}}{\text{m}}}$
Hence $\text{n}_1\text{L}_1=\text{n}_2\text{L}_2=\text{n}_3\text{L}_3\ \dots(2)$
But $\text{n}_1:\text{n}_2:\text{n}_3=1:2:3$
$\therefore \text{n}_2=2\text{n}_1$ and $\text{n}_3=3\text{n}_1\ \dots(3)$
From (2) and (3) we have
$\text{L}_1=2\text{L}_2=3\text{L}_3\ \dots(4)$
Substituting (4) in (1) we get
$\text{L}_1+\frac{1}{2}\text{L}_1+\frac{1}{3}\text{L}_1=110$
$\text{L}_1=60\text{cm}$
Hence $\text{L}_2=30\text{cm}$ and $\text{L}_3=20\text{cm}$
Thus, the bridges should be placed at distances of 60cm and 90cm from end A. View full question & answer→Question 995 Marks
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45Hz$. The mass of the wire is $3.5 \times 10^{–2}kg$ and its linear mass density is $4.0 \times 10^{–2}kg m^{–1}$. What is
- The speed of a transverse wave on the string,
- The tension in the string?
Answer
- Mass of the wire, $m = 3.5 \times 10^{–2}kg$
Linear mass density, $\mu=\frac{\text{m}}{\text{l}}=4.0\times10^2\text{kg m}^{-1}$
Frequency of vibration, v = 45Hz
$\therefore$ length of the wire, $\text{l}=\frac{\text{m}}{\mu}=\frac{3.5\times10^{-2}}{4. 0\times10^{-2}}=0.875\text{m}$
The wavelength of the stationary wave $(\lambda)$ is related to the length of the wire by the relation:
$\lambda=\frac{2\text{l}}{\text{m}}$
where,
n = Number of nodes in the wire
For fundamental node, n = 1:
$\lambda=2\text{l}$
$\lambda=2\times0.875=1.75\text{m}$
The speed of the transverse wave in the string is given as:
$\text{v}=\text{v}\lambda=45\times1.75=78.75\text{m/s}$
- The tension produced in the string is given by the relation:
$\text{T}=\text{v}^2\mu$
$=(78.75)^2\times4.0\times10^{-2}=248.06\text{N}$ View full question & answer→Question 1005 Marks
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
AnswerAll the waves have different phases. The given transverse harmonic wave is: $\text{y}(\text{x, t})=3.0\sin\Big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\Big)\ \dots(\text{i})$ For x = 0, the equation reduces to: $\text{y}(\text{x, t})3.0\sin\Big(36\text{t}+\frac{\pi}{4}\Big)$ Also, $\omega=\frac{2\pi}{\text{t}}=36\text{ rad/s}^{-1}$ $\therefore\ \text{t}=\frac{\pi}{18}\text{s}$ Now, plotting y vs. t graphs using the different values of t, as listed in the given table
| t (s) |
0 |
T/8 |
2T/7 |
3T/8 |
4T/8 |
5T/8 |
6T/8 |
7T/8 |
| y (cm) |
$\frac{3}{\sqrt{2}}$ |
3 |
$\frac{3}{\sqrt{2}}$ |
0 |
$-\frac{3}{\sqrt{2}}$ |
–3 |
$-\frac{3}{\sqrt{2}}$ |
0 |
View full question & answer→Question 1015 Marks
A wave travelling along a string is described by equation y (x, t) = 0.05 sin(40x - 5t) in which the numerical constants are in SI units ($0.05m, 40 radm^{-1}$ and $5 rad s^{-1}$).
Calculate the (a) amplitude (b) wavelength (c) time period (d) frequency of wave. Also calculate the displacement at distance $35cm$ and time $10 sec$.
Answer$\text{y}(\text{x, t})=0.05\ \sin(40\text{x}-5\text{t})$
Comparing with standard equation,
$\text{y}(\text{x, t})=\text{A}\sin\Big(\frac{2\pi\text{x}}{\lambda}-\frac{2\pi\text{t}}{\text{T}}\Big)$
- Amplitude = 0.05m
- $\frac{2\pi}{\lambda}=40$
wavelength $\lambda=+\frac{\pi}{20}\text{m}$
- Time period $=-\frac{2\pi\text{t}}{\text{T}}=-5\text{t}$
$\text{T}=\frac{2\pi}{5}\text{seconds}$
- Frequency of wave $\text{v}=\frac{1}{\text{T}}=\frac{5}{2\pi}=0.8\text{Hz}$
$\text{y}=0.05\sin(+40\times0.35-5\times10)\$\because \text{x}=0.35\text{m},\text{t}=10\text{s})$
$=0.05\sin(-36^\circ)$
$=-0.05\sin36^\circ$ View full question & answer→Question 1025 Marks
Explain Doppler effect in sound. Obtain an expression for apparent frequency of sound when source and listener are approaching each other.
AnswerWhenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by listener is different from actual frequency of sound emitted by the source.
Let S be a source of sound and L, the listener of sound, both initially at rest. Let v be the actual frequency of sound emitted by the source and $\lambda$ be the actual wavelength of sound emitted. If v is velocity of sound in still medium, then
$\lambda=\frac{v}{\text{v}}$
Let the distance between source and listener be V, so that v waves from the source reach listener in 1 second.
$V_m$ = Velocity of medium, $V_s$ = Velocity of source,
$V_L$ = Velocity of listener
Resultant velocity of sound along SL = (V + Vm)
SS' = Distance moved by source in 1 sec. = $V_s$ along SL
$\therefore$ Relative velocity of sound w.r.t. source SS' = $V_s' = [(V + V_m) - V_s]$
As the frequency remains unchanged.
$\therefore$ V waves emitted in one second occupy the distance $[(V + V_m) - V_s]$ apparent wavelength $\lambda'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}]}{\text{V}}$
Assuming both listener and source are moving in same direction, i.e. toward right.
$LL' = V_L’$ relative velocity of sound wave w.r.t. listener $(V + V_m) - V_L$
Apparent frequency of sound waves heard by listener is
$\text{v}'=\frac{(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}}{\lambda'}$
$\text{v}'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}]\text{v}}{(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}}$
When both approach each other, i.e. listener move towards the source:
$V_s$ = Positive, $V_L$ = Negative.
$\text{v}'=\frac{\text{V}-(-\text{V}_\text{L})}{\text{V}-\text{v}_\text{s}}\text{v}=\Big(\frac{\text{V}+\text{V}_\text{L}}{\text{V}-\text{V}_\text{s}}\Big)\text{v}$
Apparent frequency (v') is greater than actual frequency (v). View full question & answer→Question 1035 Marks
A tube of certain diameter and of length 48cm is open at both ends. Its fundamental frequency of resonance is found to be 320Hz. The velocity of sound in air is 320m/s. Estimate the diameter of the tube. One end of the tube is now closed, calculate the lowest frequency of resonance for the tube.
AnswerFor the open tube (i)
$\frac{\lambda}{2}=48+2\times0.3\text{D}$
But $\lambda=\frac{\nu}{\text{V}}=\frac{32000}{320}=100\text{cm}$
$\therefore \frac{100}{2}=48+2.3\times\text{D or D}$
$=\frac{20}{6}=3.33\text{cm}$
When one end of the tube is closed (ii)
$\frac{\lambda}{4}=48+0.3\text{D}$
$\lambda=[48+0.3\times3.33]\times4$
$=196\text{cm}$
$\therefore$ Lowest frequency,
$\text{V}=\frac{\nu}{\lambda}=\frac{32000}{196}=163.3\text{Hz}$ View full question & answer→Question 1045 Marks
An incident wave and a reflected wave are represented by
$\xi_1=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$
$\xi_2=\text{a}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$
Derive the equation of the stationary wave, and calculate the position of the nodes and antinodes.
AnswerApplying superposition principle,
$\xi=\xi_1+\xi_2$
$=\text{a}\Big[.\sin\frac{2\pi}{\lambda}\nu\text{t}\cos\frac{2\pi}{\lambda}\text{x}-\cos\frac{2\pi}{\lambda}\nu\text{t}\\ \sin\frac{2\pi}{\lambda}\text{x}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x}+\pi)\Big]$
$\xi=2\text{a}\sin\frac{2\pi}{\lambda}\text{x}\cos\frac{2\pi}{\lambda}\nu\text{t}$ as the reflected waYe will be out of phase by $\pi$ radians besides oppositely directed.
Amplitude $=2\text{a}\sin\frac{\pi}{\lambda}\text{x}.$
Amplitude is minimum at nodes. Therefore at
x = 0 and x = l, for nodes, $\frac{2\pi}{\lambda}\text{x}=0,\pi,2\pi...\text{N}\pi$
$\text{x}=\frac{\text{N}\lambda}{2}$ or $\text{l}=\frac{\text{N}\lambda}{2}.$
So at all points separated by $\frac{\lambda}{2}$ from one end nodes are formed. The pattern at the fundamental mode and II harmonic can be shown as below.
View full question & answer→Question 1055 Marks
One end of a long string of linear mass density $8.0 \times 10^{–3}kg m^{–1}$ is connected to an electrically driven tuning fork of frequency $256Hz$. The other end passes over a pulley and is tied to a pan containing a mass of $90kg$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At $t = 0$, the left end (fork end) of the string $x = 0$ has zero transverse displacement $(y = 0)$ and is moving along positive y-direction. The amplitude of the wave is $5.0cm$. Write down the transverse displacement y as function of x and t that describes the wave on the string.
AnswerThe equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: $\text{y}(\text{x, t})=\text{a}\sin(\text{wt}-\text{kx})\ \dots(\text{i})$ Linear mass density, $\mu=8.0\times10^{-3}\text{kg m}^{-1}$ Frequency of the tuning fork, ν = 256Hz Amplitude of the wave, a = 5.0cm = 0.05m …(ii) Mass of the pan, m = 90kg Tension in the string, $T = mg = 90 \times 9.8 = 882N$ The velocity of the transverse wave v, is given by the relation: $\text{v}=\sqrt{\frac{\text{t}}{\mu}}$
$=\frac{882}{8.0\times10^{-3}}=332\text{m/s}$ Angular Frequency, $\omega=2\pi\text{v}$
$=2\times3.14\times256$
$=1608.5=1.6\times10^3\text{ rad/s}\ \dots(\text{iii})$ Wanelength, $\lambda=\frac{\text{v}}{\text{v}}=\frac{332}{256}\text{m}$
$\therefore$ Propagation constant, $\text{k}=\frac{2\pi}{\lambda}$
$=\frac{2\times3.14}{\frac{332}{256}}=4.84\text{m}^{-1}\ \dots(\text{iv})$Substituting the values from equations (ii), (iii) and (iv) in equatio (i) we get the displacement equation:
$\text{y}(\text{x, t})=0.05\sin(1.6\times10^3\text{t}-4.84\text{x})\text{m}.$
View full question & answer→Question 1065 Marks
A man standing in front of a mountain at a certain distance beats a drum at regular intervals. The drumming rate is gradually increased, and he finds that the echo is not heard distinctly, when the rate becomes 40 per minute. He then moves nearer to the mountain by 90 metres, and finds what the echo is again not heard when the drumming rates becomes 60 per minute. Calculate
- The distance between the mountain and the initial position of the man.
- The velocity of sound.
AnswerLet d be the distance between the man and the mountain and v be the velocity of sound.
$\therefore$ Distance covered by the echo = 2d
$\text{Time}=\frac{2\text{d}}{\nu}$
Interval between the successive beats $=\frac{60}{40}=1.5/\text{sec}$
[$\because$ Drumming rate = 40 per minute]
According to the conditions given,
$\frac{2\text{d}}{\nu}=1.5\dots{\text{(i})}$
Again, 2(d - 90) = Distance covered by the echo
Or $\frac{2\text{d}-180}{\nu}=1$
[$\because$ Drumming rate = 60 per minute] ...(ii)
From (i) and (ii), we have
$1.5-\frac{180}{\nu}=1$
$\nu=\frac{180}{0.5}=360\text{m/s}$
$\therefore \text{d}=170\text{m}$
So,
- Distance = 270m
- Velocity = 360m/s
View full question & answer→Question 1075 Marks
A source of frequency $250Hz$ produces sound waves of wavelength $1.32m$ in a gas at STP. Calculate the change in the wavelength, when temperature of the gas is $40°C$.
AnswerWe have, $v_0 = 250Hz$, T0 $= 273K$
$T_1 = 273 + 40 = 313K; \lambda_0=132\text{m}$
$\therefore$ Speed of sound, $\text{v}_0=\text{v}_0.\lambda_0=250\times1.32$
$=330\text{m/s}$
As we know that,
Speed of sound, $\text{v}\propto\sqrt{\text{T}}$
Thus, $\frac{\text{v}_1}{\text{v}_0}=\sqrt{\frac{\text{T}_1}{\text{T}_0}}$
$\text{v}_1=\text{v}_0\sqrt{\frac{\text{T}_1}{\text{T}_0}}$
$=330\sqrt{\frac{313}{273}}=353.34\text{m/s}$
$\because \text{v}_1=\text{v}_0\lambda_1$
$\lambda_1=\frac{353.34}{250}=1.14\text{m}$
$\therefore$ Change in the wavelenth,
$\Delta \lambda=\lambda-\lambda_0$
$=1.14-1.32=0.09\text{m}$
View full question & answer→Question 1085 Marks
A progressive wave is given by $\text{y}(\text{x, t})=8\cos(300\text{t}-0.15\text{x})$ where x in m, y in cm and t in second. What is the-
- Direction of propagation?
- Wavelength?
- Frequency?
- Wave speed?
- Phase difference between two points 0.2m apart?
Answer$\text{y}(\text{x, t})=8\cos(300\text{t}-0.15\text{x})$
On comparing it with $\text{y}=\text{a}\cos2\pi\Big(\frac{\text{t}}{\text{T}}-\frac{\text{x}}{\lambda}\Big)$
- Direction of propogation is + x-axis.
- $\frac{2\pi}{\lambda}=0.15$
$\Rightarrow \lambda=\frac{2\pi}{0.15}$
$=41.87\text{m}$
- $\frac{2\pi}{\text{T}}=300$
$2\pi\text{v}=300$
$\text{v}=\frac{300}{2\pi}$
$=47.78\text{Hz}$
- $\nu=\lambda\text{v}$
$=\frac{2\pi}{0.15}\times\frac{300}{2\pi}$
$=2000\text{m/s}$
- $\Delta\phi=\frac{2\pi}{\lambda}\Delta\text{x}=\frac{2\pi}{\lambda}\times0.2$
$=\frac{2\pi\times0.2\times0.15}{2\pi}=0.03\text{ radian}$ View full question & answer→Question 1095 Marks
The earth has a radius of $6400km$. The inner core of $1000km$ radius is solid. Outside it, there is a region from $1000km$ to a radius of $3500km$ which is in molten state. Then again from $3500km$ to $6400km$ the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of $8km s^{–1}$ in solid parts and of $5km s^{–1}$ in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?
Answer$\text{r}_1=1000\text{km}$
$\text{r}_2=3500\text{km}$
$\text{r}_3=6400\text{km}$
$\text{d}_1=1000\text{km}$
$\text{d}_2=3500-1000=2500\text{km}$
$\text{d}_3=6400-3500=2900\text{km}$
Solid distance diametrically $=2(\text{d}_1+\text{d}_3)=(1000+2900)$
$2\times3900\text{km}$ Time taken by wave produced by earthquake in solid part $=\frac{3900\times2}{8}\sec$ Liquid part along diametrically $2\text{d}_2=2\times2500$
$\therefore$ Time taken by seismic wave in liquid part $=\frac{2\times2500}{5}$ Total time $\frac{2\times3900}{8}+\frac{2\times2500}{5}=2\Big[\frac{3900}{8}+\frac{2500}{5}\Big]$
$=2[487.5+500]=2\times987.5=1975\sec.$
$=32\ \text{min}\ 55\sec.$ View full question & answer→Question 1105 Marks
What do you understand by beat? Explain beats analytically.
AnswerFor definition, see text:
Consider two simple harmonic progressive waves travelling simultaneously in the same direction and in the same medium. Let
'A' be the amplitude of each wave.
There is no initial phase difference between them.
$v_1$ and $v_2$ be their frequencies.
If $y_1$ and $y_2$ be displacements of the two waves, then
$\text{y}_1=\text{A}\sin2\pi\text{v}_1\text{t}$ and $\text{y}_2=\text{A}\sin2\pi\text{v}_2\text{t}$
If y be the result and displacement at any instant, then
$\text{y}=\text{y}_1+\text{y}_2=\text{A}\sin(2\pi\text{v}_1\text{t})+\sin(2\pi\text{v}_2\text{t})$
$=\text{A}\Bigg[2\sin\Big(\frac{2\pi(\text{v}_1+\text{v}_2)\text{t}}{2}\Big)\cos\Big(\frac{2\pi(\text{v}_1-\text{v}_2)\text{t}}{2}\Big)\Bigg]$
$=2\text{A}\cos\pi(\text{v}_1\text{v}_2)\text{t}\sin \pi(\text{v}_1+\text{v}_2)\text{t}$
$=\text{R}\sin\pi(\text{v}_1+\text{v}_2)\text{t}\ \dots(1)$
where $\text{R}=2\text{A}\cos\pi(\text{v}_1-\text{v}_2)\text{t}\ \dots(2)$
is the amplitude of the resultant displacement and depends upont. The following cases arise.
- If R is maximum, then
$\cos\pi(\text{v}_1-\text{v}_2)\text{t}=\text{max}.=\pm1=\cos\text{n}\pi$
$\therefore \pi(\text{v}_1-\text{v}_2)\text{t}=\text{n}\pi$
$\text{t}=\frac{\text{n}}{\text{v}_1-\text{v}_2}\ \dots(3)$
where n = 0, 1, 2...
$\therefore$ Amplitude becomes maximum at times given by
$\text{t}=0,\frac{1}{\text{v}_1-\text{v}_2},\frac{2}{\text{v}_1-\text{v}_2}=\frac{3}{\text{v}_1-\text{v}_2},...$
$\therefore$ Time interval between two consecutive maxima is
$=\frac{1}{\text{v}_1-\text{v}_2}$
$\therefore$ Beat period $=\frac{1}{\text{v}_1-\text{v}_2}$
$\therefore$ Beat frequency $=\text{v}_1-\text{v}_2$
$\therefore$ no. of beats formed per sec. $=\text{v}_1-\text{v}_2$
- If R is minimum, then
$\cos\pi(\text{v}_1-\text{v}_2)\text{t}=\text{min}=0=\cos(2\text{n}+1)\frac{\pi}{2}$
$\therefore \pi(\text{v}_1-\text{v}_2)\text{t}=(2\text{n}+1)\frac{\pi}{2}$
$\text{t}=\frac{(2\text{n}+1)}{2(\text{v}_1-\text{v}_2)},$ where n = 0, 1, 2,...
$\therefore$ Amplitude becomes minimum at times grven by
$\text{t}=\frac{1}{2(\text{v}_1-\text{v}_2)},\frac{3}{2(\text{v}_1-\text{v}_2)},\frac{5}{2(\text{v}_1-\text{v}_2)},...$
$\therefore$ Time interval between two consecutive minima is $=\frac{1}{\text{v}_1-\text{v}_2}.$
$\therefore$ Beat period $=\frac{1}{\text{v}_1-\text{v}_2}.$
$\therefore$ Beat requency = $v_1 - v_2$.
$\therefore$ No. of beats formed per sec = $v_1 - v_2$.
Hence the number of beats formed per second is equal to the difference between the frequencies of two component waves. View full question & answer→Question 1115 Marks
Show that in the case of a closed organ pipe, the ratio of the frequencies of the harmonics is 1 : 3 : 5 : 7.
AnswerDisplacement at position x and time t in incident wave,
$\text{y}_1=\text{r}\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})$
Wave reflected at closed end of pipe suffers a phase reversal of $\pi.$
$\therefore$ For reflected wave
$\text{y}_2=\text{r}\sin\Big[\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})+\pi\Big]$
$=-\text{r}\sin\frac{2\pi}{\lambda}(\nu\text{t}-\text{x})$
According to superposition principle,
$\text{y}=\text{y}_1+\text{y}_2$
$\text{y}=\text{r}\Big[\sin\frac{2\pi}{\lambda}(\nu\text{t}+\text{x})-\sin\frac{12\pi}{\lambda}(\nu\text{t}-\text{x})\Big]$
$\text{y}=2\text{r}\cos\frac{2\pi}{\lambda}\nu\text{t}.\sin\frac{2\pi}{\lambda}\text{x}.$
At closed end of pipe x = 0
$\therefore \sin \frac{2\pi}{\lambda}\text{x}=\sin\theta=0$
y = 0, i. e., a node is formed.
At the open end of the pipe of length, Lx = L an antinode is to formed, i. e., y = max
$\text{y}=2\text{r}\cos\frac{2\pi}{\lambda}\nu\text{t}.\sin\frac{2\pi}{\lambda}\text{L}$
y will be max, when
$\sin \frac{2\pi\text{L}}{\lambda}=\text{max}=\pm1$
$=\sin (2\text{n}-1)\frac{\pi}{2}$
where n = 1, 2, 3.......
$\frac{2\pi\text{L}}{\lambda}=(2\text{n}-1)\frac{\pi}{2}$
$\lambda=\frac{4\text{L}}{(2\text{n}-1)}$
For first normal mode of vibration,
Let $\lambda_1=$ wavelength corresponding to n = 1
$\lambda_1=\frac{4\text{L}}{(2\times1-1)}=4\text{L}$
$\text{L}=\frac{\lambda_1}{4}$
Frequency
$\text{V}_1=\frac{\nu}{\lambda_1}=\frac{\nu}{4\text{L}}$
$\text{V}_1=\frac{\nu}{4\text{L}}$ fundamental frequency
For second normal mode of vibration,
$\lambda_2=$ wavelength of standing waves correspondig to n = 2
$\lambda_2=\frac{4\text{L}}{2\times2-1}=\frac{4\text{L}}{3}$
$\text{V}_2=\frac{\nu}{\lambda_2}=\frac{\nu}{\frac{4\text{L}}{3}}$
$=\frac{3\nu}{4\text{L}}=3\text{V}_1$
$\text{V}_2=3\text{V}_1$ third harmonic.
$\lambda_3=$ wavelength of standing waves corresponding to n = 3
$\lambda_3=\frac{4\text{L}}{2\times3-1}=\frac{4\text{L}}{5}$
$\text{V}_3=\frac{\nu}{\lambda_3}=\frac{\nu}{\frac{4\text{L}}{5}}=5\frac{\nu}{4\text{L}}$
$\text{V}_3=5\text{V}_1$ fifth harmonic
In general
$\text{V}_\text{n}=\frac{(2\text{n}-1)\nu}{4\text{L}}=(2\text{n}-1)\text{V}_1$
$\therefore$ Ratio of frequencies of harmonics is 1 : 3 : 5 : 7
View full question & answer→Question 1125 Marks
A train, standing at the outer signal of a railway station blows a whistle of frequency $400Hz$ in still air.
- What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of $10m s^{–1}$, (b) recedes from the platform with a speed of $10m s^{–1}$?
- What is the speed of sound in each case? The speed of sound in still air can be taken as $340m s^{–1}$.
Answer
- (a) Frequency of the whistle, $ν = 400Hz$
Speed of the train, $v_T= 10m/ s$
Speed of sound, v = 340m/ s
The apparent frequency (v') of the whistle as the train approaches the platform is given by the
relation:
$\text{v}'=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{r}}\Big)\text{v}$
$=\Big(\frac{340}{340-10}\Big)\times400=412.12\text{Hz}$
(b) The apparent frequency (v') of the whistle as the train recedes from the platform is given by the relation:
$\text{v}''=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{r}}\Big)\text{v}$
$=\Big(\frac{340}{340+10}\Big)\times400=388.57\text{Hz}$
- The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340m/ s.
View full question & answer→Question 1135 Marks
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse $(S)$ and longitudinal ( $P$ ) sound waves. Typically the speed of $S$ wave is about $4.0 km s ^{-1}$, and that of $P$ wave is $8.0 km s ^{-1}$. A seismograph records $P$ and $S$ waves from an earthquake. The first $P$ wave arrives $4$ min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
AnswerLet $v_S$ and $v_P$ be the velocities of S and P waves respectively.
Let L be the distance between the epicentre and the seismograph.
We have:
$L = v_St_S …(i)$
$L = v_Pt_P …(ii)$
Where,
$t_S $and $t_P $are the respective times taken by the S and P waves to reach the seismograph from the epicentre
It is given that:
$v_P = 8km/ s$
$v_S = 4km/ s$
From equations (i) and (ii), we have:
$v_S t_S = v_P t_P$
$4t_S = 8t_P$
$t_S = 2 t_P …(iii)$
It is also given that:
$t_S – t_P = 4 min = 240s$
$2t_P – t_P = 240$
$t_P = 240$
And $t_S = 2 \times 240 = 480s$
From equation (ii), we get:
$L = 8 \times 240$
$= 1920km$
Hence, the earthquake occurs at a distance of $1920\ km$ from the seismograph.
View full question & answer→Question 1145 Marks
What are the characteristics of stationary waves? Distinguish between stationary waves and progressive waves.
AnswerCharacteristics of Stationary waves:
- Stationary waves are produced in a bounded medium. A medium whose boundaries are separated from other media by distinct surfaces is called bounded medium. The boundaries of a bounded medium may be rigid or free. For example, string fixed at both the ends (i.e., string of a guitar), closed and open organ pipes.
- There are certain points in the bounded medium (in which stationary waves are formed) which are always in the state of rest. These points are called nodes. If the stationary waves are longitudinal, then the change in pressure and density is maximum at nodes as compared to the other points.
- There are points in between the nodes whose displacement is maximum as compared to other points. These points are called anti-nodes. In the longitudinal stationary waves, there is no change in pressure and density of the medium at anti-nodes.
- The distance between any two successive nodes or antinodes is $\frac{\lambda}{2}.$ The distance between a node and the neighbouring anti-node is $\frac{\lambda}{4}.$
- All particles of the medium lying between two successive nodes vibrate but the amplitude of vibration is different for different particles. The amplitude of vibration is zero at nodes and maximum at anti-nodes.
- All particles between two successives nodes vibrate in the same phase. They pass simultaneously through their mean positions and also pass simultaneously through their positions of maximum displacement.
- At any instant, the phase of vibration of the particles on one side of a node is opposite from the phase of vibration of the particles on the other side.
- All particles of the medium pass through their equilibrium positions (i.e., mean positions) simultaneously twice in each period. That is, the stationary wave takes the form of a straight line twice.
- In a stationary wave, the medium splits up into a number of segments. Each segment vibrates up and down as a whole.
- All the particles except those at nodes, execute simple harmonic motion about their mean positions with the same time period.
- In a stationary wave, there is no onward motion of the disturbance from one particle to the other particle.
- Stationary wave does not advance in the medium, but remains steady at its place. In other words, stationary wave does not transmit energy in the medium.
For distinction between progressive and stationary waves, see text. View full question & answer→Question 1155 Marks
A train, standing in a station-yard, blows a whistle of frequency $400Hz$ in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10m s^{–1}$. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of $10m s^{–1}$? The speed of sound in still air can be taken as $340m s^{–1}$.
AnswerFor the stationary observer: 400Hz; 0.875m; 350m/ s
For the running observer: Not exactly identical
For the stationary observer:
Frequency of the sound produced by the whistle, ν = 400Hz
Speed of sound = 340m/ s
Velocity of the wind, v = 10m/ s
As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, $v_e = 340 + 10 = 350m/ s$
The wavelength $(\lambda)$ of the sound heard by the observer is given by the relation:
$\lambda=\Big(\frac{\text{v}_\text{e}}{\text{V}}\Big)-=\frac{350}{400}=0.875\text{m}$
For the running observer:
Velocity of the observer, $v_0 = 10m/ s$
The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (V').
This is given by the relation:
$\text{v}'=\Big(\frac{\text{v}+\text{v}_0}{\text{v}}\Big)\text{v}$
$=\Big(\frac{340+10}{340}\Big)\times400=411.76\text{Hz}$
Since the air is still, the effective speed of sound = 340 + 0 = 340m/s
The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875m.
Hence, the given two situations are not exactly identical.
View full question & answer→Question 1165 Marks
A travelling harmonic wave on a string is described by
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
What are the displacement and velocity of oscillation of a point at x = 1cm, and t = 1s? Is this velocity equal to the velocity of wave propagation?
AnswerThe given harmonic wave is:
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
For x = 1cm and t = 1s,
$\text{y}=(1,1)=7.5\sin\Big[0.0050+12+\frac{\pi}{4}\Big]$
$=7.5\sin\Big[12.0050+\frac{\pi}{4}\Big]$
$=7.5\sin\theta$
Where, $\theta=12.0050+\frac{\pi}{4}=12.0050+\frac{3.14}{4}=12.79\text{ rad}$
$=\frac{180}{3.14\times12.79}=732.81^\circ$
$\therefore\ \text{y}=(1,1)=7.5\sin[732.81^\circ]$
$=7.5\sin(90\times8+12.81^\circ)$
$=7.5\sin(12.81^\circ)$
$=7.5\times0.2217$
$=1.6629\approx1.663\text {cm}$
The velocity of the oscillation at a given point and time is given as:
$\text{v}=\frac{\text{d}}{\text{dt}}\text{y}(\text{x, t})=\frac{\text{d}}{\text{dt}}\Big[7.5\sin\big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\big)\Big]$
$=7.5\times12\cos\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
At x = 1cm and t = 1s:
$\text{v}=\text{y}(1, 1)=90\cos\Big(12.005+\frac{\pi}{4}\Big)$
$=90\cos(732.81^\circ)=90\cos(90\times8+12.81^\circ)$
$=90\cos(12.81^\circ)$
$=90\times0.975=87.75\text{cm/s}$
Now, the equation of a propagating wave is given by:
$\text{y}(\text{x, t})=\text{a}\sin(\text{kx}+\text{wt}+\phi)$
Where,
$\text{k}=\frac{2\pi}{\lambda}$
$\therefore\ \lambda=\frac{2\pi}{\text{k}}$
And $\omega=2\pi\text{ v}$
$\therefore\ \text{v}=\frac{\omega}{2\pi}$
Speed $=\text{v}=\text{v}\lambda=\frac{\omega}{\text{k}}$
Where
$\omega=12\text{ rad/s}$
$\text{k}=0.0050\text{m} ^{-1}$
$\therefore\ \text{v}=\frac{12}{0.0050}=2400\text{cm/s}$
$\therefore$ Hence, the velocity of the wave oscillation at x = 1cm and t = 1s is not equal to the velocity of the wave propagation.
View full question & answer→Question 1175 Marks
A pipe, $30.0 cm$ long, is open at both ends. Which harmonic mode of the pipe resonates a $1.1 kHz$ source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as $330 m s ^{-1}$.
AnswerThe first harmonic frequency is given by
$
v_1=\frac{v}{\lambda_1}=\frac{v}{2 L}
$
(open pipe)
where $L$ is the length of the pipe. The frequency of its $n$th harmonic is:
$
v_{ n }=\frac{n v}{2 L}, \text { for } n=1,2,3, \ldots \text { (open pipe) }
$
First few modes of an open pipe are shown in Fig. 14.15.
For $L=30.0 cm , V=330 m s ^{-1}$,
$
v_{ n }=\frac{n 330\left( m s ^{-1}\right)}{0.6( m )}=550 n s ^{-1}
$
Clearly, a source of frequency $1.1 kHz$ will resonate at $v_2$, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig. 14.15), it follows from Eq. (14.15) that the fundamental frequency is
$
v_I=\frac{v}{\lambda_1}=\frac{v}{4 L} \text { (pipe closed at one end) }
$
and only the odd numbered harmonics are present :
$
v_3=\frac{3 v}{4 L}, v_5=\frac{5 v}{4 L} \text {, and so on. }
$
For $L=30 cm$ and $v=330 m s ^{-1}$, the fundamental frequency of the pipe closed at one end is $275 Hz$ and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed. View full question & answer→