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Physics 5 Marks Questions

PART - 2 CH - 14 Waves · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 22 questions.

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Question 15 Marks
A transverse harmonic wave on a string is described by
$
y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)
$
where $x$ and $y$ are in cm and $t$ in s. The positive direction of $x$ is from left to right.
(a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Answer
The given equation is transverse frquency
$y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)$$\quad\ldots$(1)
Standard equation of progressive wave
$y(x, t) =a \sin \left(\frac{2 \pi}{\lambda}(v t-x)+\phi\right) $
$=a \sin \left(\frac{2 \pi}{T} t-\frac{2 \pi}{\lambda} x+\phi\right)$$\quad$$\quad$$\left[\because \frac{1}{T}=\frac{ v }{\lambda}\right]$$\quad\ldots$(2)
(a) By comparing equation (1) with equation (2)
$\frac{2 \pi}{T}=36$
or $\quad \frac{-2 \pi}{\lambda}=0.018$
or$\quad$$\lambda=\frac{-2 \pi}{0.018}\quad\ldots(3)$
$2 \pi n=36$$\quad\ldots$(4)
Multiply equation (3) and equation (4)
$2 \pi n \lambda=\frac{-2 \pi \times 36}{0.018}$
or $\quad n \lambda=\frac{-36 \times 1000}{18}=-2000 cm / s$
Velocity $(v)=-2000 cm / s =-20 m / s$
Here Negative sign represent the wave propagate in right to left.
(b) Given : $\text { speed } =20 m / s $
$a =3.0 cm=3.0 \times 10^{-2} m $
$\frac{2 \pi}{T} =36 $
$\frac{1}{T} =\frac{36}{2 \pi}=\frac{18}{\pi} $
$\text { frequency } n =\frac{18}{3.14}=5.73 Hz$
(c) Phase angle $(\phi)=\frac{\pi}{4}$ Radian
(d) Distance minimum between two consecutive crests wave length $=\lambda$
$=\frac{2 \pi}{0.018} $
$=\frac{2 \times 3.14}{0.018}=348.9 cm $
$\simeq 349 cm $
$\simeq \frac{349}{100} m=3.49 m

$
$\simeq 3.5 m$
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Question 25 Marks
Use the formula $v =\sqrt{\frac{\gamma P }{\rho}}$ explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Answer
(a) Formula of sound velocity in air
$v =\sqrt{\frac{\gamma P }{\rho}}$$\quad\ldots$(1)
Here, $\gamma=$ constant for given gases
$\rho=$ density of gases
Gas equation
$
PV=RT \quad \therefore P=\frac{RT}{V}
$
Put the value of P in equation (1)
$
v=\sqrt{\frac{\gamma RT}{V \rho}}
$
But mass $M=$ volume $(V) \times$ density $(\rho)$
$\therefore \quad v =\sqrt{\frac{\gamma RT }{ M }}$$\quad\ldots$(2)
For given gas M is constant and R also constant. If T is constant then we conclude from equation 2 that if temeprature is fix then $v$ is also free from pressure.
(b) Effect of temperature :
From gas equation $PV = RT$
$
\begin{array}{l}
P=\frac{RT}{V} \\
v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma RT}{\rho V}}=\sqrt{\frac{\gamma RT}{M}}
\end{array}
$
Here, $M =\rho, V =$ molecular weight of gass
Thus, if $\gamma, R$ and M is constant.
So, $\quad v \propto \sqrt{T}$
Velocity of sound is directly proportional to square root of temperature. Thus we can say that if temperature increase then speed will increases.
(c) Effect of humidity : Presence of water vapour in air changes its density. Thus persence of humidity is changes sound velocity.
Let
Velcoity of dry air $= v _{ d }$
Sound velocity in humid arir $= v _{ m }$
Density of dry air $=\rho_d$
Density of humid air $=\rho_{ m }$
By formula$\quad$$
v=\sqrt{\frac{\gamma P}{\rho}}
$
$\Rightarrow \quad v _{ m }=\sqrt{\frac{\gamma P }{\rho_m}}$$\quad\ldots$(1)
or
$
v_d=\sqrt{\frac{\gamma P}{\rho_d}}\quad\ldots(2)
$
Equation (1) divided by equation (2)
$\frac{ v _m}{ v _d}=\sqrt{\frac{\rho_d}{\rho_m}}$$\quad\ldots$(3)
We know that density of dry air is less then water vapour
$\rho_m<\rho_d$
or $\quad \frac{\rho_d}{\rho_m}>1$$\quad\ldots$(4)
By equation (3) and (4) we gets $\frac{ v _m}{ v _d}>1$
or
$
v_{m}>v_d
$
Means sound's velocity increase with humidity. Sound velocity in dry air is greater in humid air. So during rainy days sound velocity is fast.
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Question 35 Marks
A 1 metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tunning fork of frequency 340 Hz ) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer
Tube with piston work like argon pipe. Let $l_1$ and $l_2, r$ and $(r+1)$ 's resonance consecutive length which close pipe's cycle consecutive.
Given : $\quad l_1=25.5 cm $
$l_2=79.3 cm$
Sound velocity of wave is $v$ and frequency consecutive is $n_1$ and $n_2$ then value of frequency of following mode:
$n_1=(2 r-1) \frac{ v }{4 l_1}$$\quad \ldots$(1)
and $\quad n_2=(2(r+1)-1) \frac{ v }{4 l_2}$
$n_2=(2 r+1) \frac{ v }{4 l_2}$$\quad \ldots$(2)
Given : Both mode resonancete with frequency of 340 Hz.
$(2 r-1) \frac{v}{4 l_1}=(2 r+1) \frac{v}{4 l_2}=340$
or$\quad$$\frac{(2 r-1)}{4 l_1}=\frac{(2 r+1)}{4 l_2}$
$ \frac{(2 r-1)}{2 r+1} =\frac{l_1}{l_2}=\frac{25.5}{79.3}=\frac{1}{3} $
$\Rightarrow 3(2 r-1) =2 r+1 $
$\Rightarrow 6 r-3 =2 r+1 $
$4 r =4 $
$r =1$
$\frac{(2 r-1) v}{4 l_1}=340$
Put the value $r=1$
or $\quad \frac{(2 \times 1-1) v }{4 l_1}=340$
$\frac{v}{4 l_1}=340$
or$\quad$$v=4 \times 340 \times l_1$
Put the value
$ v =4 \times 340 \times 25.5 cm / s $
$=34680 cm / s =346.80 m / s $
$v \simeq 347 m / s $
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Question 45 Marks
The transverse displacement of a string (clamped at its both ends) is given by
$
y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)
$
where $x$ and $y$ are in m and $t$ in s . The length of the string is $1 . 5 ~ m$ and its mass is $3 . 0 \times 1 0 ^{- 2 } ~ k g$.
Answer the following :
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
Answer
Given function
$y(x, t)=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)$$\quad\ldots$(1)
It represent to progressive wave equation
$y(x, t)=a \sin \frac{2 \pi}{\lambda}(v t-x)$
and following equation represent to stationary wave
$y(x, t)=-2 a \sin \left(\frac{2 \pi x}{\lambda}\right) \cos \left(\frac{2 \pi v t}{\lambda}\right)$
(a) Since given function represented to stationary wave because it is simillar to stationary wave equation.
When a wave propogate in positive x-axis direction then
(b) $y_1=a \sin \frac{2 \pi}{\lambda}( v t-x)$
$ y_2 =-a \sin \frac{2 \pi}{\lambda}( v t-x) $
$y =y_1+y_2$
$=a \sin \frac{2 \pi}{\lambda}( v t-x)-a \sin \frac{2 \pi}{\lambda}( v t+x)$
$=a\left[2 \cos \frac{\frac{2 \pi}{\lambda}( v t-x+ v t+x)}{2}\right] \sin \left(\frac{\frac{2 \pi}{\lambda}( v t-x- v t-x)}{2}\right) $
$=a\left[2 \cos \left(\frac{2 \pi}{\lambda} v t\right) \sin \left(\frac{-2 \pi}{\lambda} x\right)\right]$
$=-2 a \sin \left(\frac{2 \pi}{\lambda} x\right) \cos \left(\frac{2 \pi}{\lambda} v t\right)$$\quad\ldots$(2)
By the comparing of eq. (1) and eq. (2)
$\frac{2 \pi}{\lambda} =\frac{2 \pi}{3} \text { or } \lambda=3 m . $
$\frac{2 \pi}{\lambda} v =120 \pi$
or $\quad v=60 \lambda=60 \times 3=180 m / s$
Frequency $\quad n=\frac{ v }{\lambda}=\frac{180}{3}=60 Hz$
Velocity of transverse velocity
$v=\sqrt{\frac{T}{m}} \text { or } v^2=\frac{T}{m}$
$T = v ^2 \times m$$\quad\ldots$(3)
$m=\text { unit length mass }=\frac{ M }{l}=\frac{3 \times 10^{-2} kg}{1.5 m} $
$m=2 \times 10^{-2} kg / m $
Speed of wave $\quad(v)=180 m / s$
(c) T= Tension in string $=$ ?
Put the value in eq. (3)
$ T =(180)^2 \times\left(2 \times 10^{-2}\right) $
$=32400 \times 2 \times 10^{-2} $
$=648 N$
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Question 55 Marks
For the travelling harmonic wave
$
y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35)
$
where $x$ and $y$ are in cm and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m$\quad$$\quad$(b) 0.5 m$\quad$$\quad$(c) $\frac{\lambda}{2}$$\quad$$\quad$(d) $\frac{3 \lambda}{4}$
Answer
Progressive frequency wave equation
$y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35) $
$y(x, t)=2.0 \cos (2 \pi \times 10 t-2 \pi \times 0.0080x +2 \pi \times 0.35)\quad\ldots(1)$
Standard equation of progressive frequency wave
$y(x, t)=a \cos \left(\frac{2 \pi}{\lambda}( v t-x)+\phi\right)$$\quad\ldots$(2)
By comparing coefficient of equation (1) and equation (2)
$\frac{2 \pi}{\lambda}=2 \pi \times 0.0080$$\quad\ldots$(3)
What know that
Phase difference $=\frac{2 \pi}{\lambda} \times$ path difference
(a) When path difference $=4 m=400 cm$
$\text { Phase difference }=\frac{2 \pi}{\lambda} \times 400$
Put the value in equation 3
$=2 \pi \times 0.0080 \times 400 $
$=6.4 \pi rad$
(b) When path difference $=0.5=50 cm$
When phase difference $=2 \pi \times 0.0080 \times 50$
$=0.8 \pi rad$
(c) When path difference $=\frac{\lambda}{2}$
Phase difference $=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi rad$
(d) When path difference $=\frac{3 \lambda}{4}$
$\text { then phase difference } =\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}=\frac{6 \pi}{4} $
$=\frac{3}{2} \pi rad .$
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Question 65 Marks
A progressive harmonic wave on a string is expressed as follow
$
y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)
$
(a) Find the speed at a point $x=1 cm$ on time $t=1$ sec. Is this speed is equal to the speed of wave propagation?
(b) Find the location of those points of the string whose tranverse displacement and speed are the same as that of the point located at $x=1 cm$ at times $t =2 s, 5 s$ and $11s.$
Answer
(a) $y(x, t)=7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)$
On comparing the above equation with the following equation of normal progressive wave.
$y=a \sin [k x+\omega t+Q]$
Hence, $k=0.0050$ and $\omega=12$
$\text {Speed of wave}\quad = v =n \lambda=(2 \pi n) \times \frac{\lambda}{2 \pi}=\frac{\omega}{ k }$
$ \begin{array}{l}\text {Speed of wave}\quad\text { (v) }=\frac{12}{0.0050}=\frac{12000}{5}=2400 cm / s \\v=\frac{2400}{100} m / s=24 m / s\end{array}$
In given equation displacement at $x=1 cm, \pm=1 s$
$y(1,1)=7.5 \sin \left(0.005+12+\frac{\pi}{4}\right)$
Put the value in given equation
$\begin{array}{l}=7.5 \sin \left(12.005+\frac{\pi}{4}\right) \\=7.5 \sin \left(12.005+\frac{3.14}{4}\right) \\=7.5 \sin (12.005+0.785)\end{array}$
$\begin{aligned} & =7.5 \sin (12.79) \\ & =7.5 \sin \left(12.79 \times \frac{180}{\pi}\right) \\ & \quad\left[\because \pi \text { Radian }=180^{\circ}\right] \\ & =7.5 \sin \left(732.83^{\circ}\right) \\ & =7.5 \sin \left(720^{\circ}+12.83^{\circ}\right) \\ & =7.5 \sin 12.83^{\circ} \\ \text { or } \quad y(1,1) & =7.5 \times 0.2215 \\ & =1.67 cm \end{aligned}$
Speed of oscillation on point
$\begin{aligned} v & =\frac{d y}{d t} \\ v & =7.5 \times 12 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\ & =90.0 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\ x & =1 cm, t=1 \text { second } \\ v & =90 \cos (0.005 \times 1+12 \times 1+0.785) \\ & =90 \cos (12.79) \\ & =90 \cos \left(732.83^{\circ}=90 \cos 12.83^{\circ}\right) \\ & =90 \times 0.9751=87.76 cm / s \end{aligned}$
Hence, speed of osicllation point $ \simeq$ 88 cm/s
But speed of wave =24 m/s
Hence, it is clear that the speed of oscillation of the point is different from the speed of wave motion. Hence not equivalent.
(b) Given equation
$y=7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)$
Progressive wave equation
$y=a \sin (\omega t+k x+\phi)$ on compare
$k =0.005$ Radian/cm
$\lambda=\frac{2 \pi}{ k }=\frac{2 \times 3.142}{0.005}=12.57 m$
In a transverse wave the displacement of all points is equal. Which are at distance $\lambda, 2 \lambda, 3 \lambda \ldots$ etc. Therefore displacement of all point $12.57 m, 25.14 m, 37.71 m$ will be same.
Thus, all the points when $n= \pm 1, \pm 2, \pm 3, \pm 4, \ldots x=$ 1 cm at other distance but displacement will be same.
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Question 75 Marks
Define the following :
(i) Frequency (ii) Time period (iii) Amplitude (iv) Wave length (v) Phase angle (vi) Angular frequency (vii) Propagation con (viii) Wave velocity.
Answer
(i) Frequency : The number of vibrations made by a particle in a second is called frequency. Its S.I. unit is Hertz. We can represent it by U.
Image
(ii) Time period : The time raken by a particle for one vibration is called time period. Time period is the reciporcal of frequency. That is, time period
$T =\frac{1}{n}=\frac{2 \pi}{\omega}$
(iii) Amplitude : The maximum displacement from mean position of particle is called amplitude. It is represented by 6a'.
(iv) Wave length : The distance between two nearest particle vibrating in the same phase is called wave length. We represent it by ' $\lambda$ '.
(v) Phase angle : The phase angle of a vibrating particle at any instant reveals the state of motion at that instant. It is represented by $\phi$.
(vi) Angular frequency : The rate of change in the phase angle of the particle with time is called angular frequency. Angular frequency $\omega=2 \pi n=\frac{2 \pi}{T} rad / sec$
(vii) Propogation constant : If the distance between any two vibrating points in the wave is $\lambda$ and phase difference is $2 \pi$. The phase difference between vibrating particles located at unit distance called propagation constant.
(viii) Wave velocity : When there is wave motion the distance the distance covered by the vibrational disturbance in one second is called wave velocity.
Wave velocity $v =\frac{\lambda}{ T }$ or $v = n \lambda$
$\text {or}\quad v=\frac{\omega}{2 \pi} \times \frac{2 \pi}{k}$
$\text {or}\quad v=\frac{\omega}{k} \text { meter } / sec .$
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Question 85 Marks
What are waves and wave speed? For wave propagation. What are the essential properties? How many types of waves are there? Explain by giving examples.
Answer
SELF
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Question 95 Marks
Calculate the frequency of the original tone in stretched cord. Explain the fundamental and overtones of a string and prove that both even and odd harmonics are produced in the string.
Answer
SELF
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Question 115 Marks
Write a labelled diagram of resonance tube and explain it. With its help establish formulas to determine the velocity of sound.
OR
Make a labelled diagram of resonance tube and describe its structure and establish the formula to find the velocity of sound at room temperature.
Answer
SELF
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Question 135 Marks
Prove that closed argon pipe produced only odd harmonics whereas both odd and even harmonic are produced in an open argon pipe.
Answer
SELF
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Question 165 Marks
What is Beat? Analyze it mathematically and prove that the number of beat per second is equal to the difference of the frequencies of both the sound sources. How can the frquency of a tunning fork be clamed down with the help of pulses?
Answer
SELF
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Question 175 Marks
Explain the experimental demonstration of in- terference by making a labelled diagram of a quinke tube.
Answer
SELF
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Question 185 Marks
What is the principle of superposition? On what factors frequency and amplitude of the resultant wave depend? What effect can be produced by the superposi- tion of two sound waves? Exaplain interference of sound wave and analyze it mathematically.
Answer
SELF
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Question 195 Marks
Explain the velocity of sound in different medium and write the factor affecting the velocity of sound in gas.
Answer
SELF
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Question 205 Marks
What are progressive wave? Establish the equation representing progressive wave and explain phase and path difference.
OR
Establish the equation of plane progressive wave and find out phase difference.
Answer
SELF
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Question 225 Marks
What are electromagnetic waves? Explain the characteristics of waves. Prove that $v=n \lambda$ where $V$ is the velocity $n$ is frequency and $\lambda$ is the wavelength.
Answer
SELF
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