Question 13 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)$
Answer$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)=+2\cos\Big(3\text{t}+\frac{\pi}{3}+\frac{\pi}{2}\Big)$$=2\cos\Big(3\text{t}+\frac{5\pi}{6}\Big)$
If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 2cm
Phase angle, $\phi=\frac{5\pi}{6}=150^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=3\text{rad/sec.}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 23 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=3\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$
Answer$\text{x}=3\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$$=-3\cos\bigg[\Big(2\pi\text{t}+\frac{\pi}{4}\Big)+\frac{\pi}{2}\bigg]=-3\cos\Big(2\pi\text{t}+\frac{3\pi}{4}\Big)$
If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 3cm
Phase angle, $\phi=\frac{3\pi}{4}=135^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=2\pi\text{ rad/s}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 33 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=\cos\Big(\frac{\pi}{6}-\text{t}\Big)$
Answer$\text{x}=\cos\Big(\frac{\pi}{6}-\text{t}\Big)=\cos\Big(\text{t}-\frac{\pi}{6}\Big)$If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 1
Phase angle, $\phi=-\frac{\pi}{6}=-30^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=1\text{rad/s}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 43 Marks
Answer the following questions:The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than $2\pi\sqrt{\frac{\text{l}}{\text{g}}}.$ Think of a qualitative argument to appreciate this result.
AnswerIn the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
$\text{F}=-\text{mg}\sin\theta$
Where,
F = Restoring force
m = Mass of the bob
g = Acceleration due to gravity
$\theta=$ Angle of displacement
For small $\theta,\ \sin\theta\simeq\theta$
For large $\theta,\ \sin\theta$ is greater than $\theta.$
This decreases the effective value of g.
Hence, the time period increases as:
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
Where, l is the length of the simple pendulum.
View full question & answer→Question 53 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
$\text{x}=2\cos\pi\text{t}$
Answer$\text{x}=2\cos\pi\text{t}$If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 2cm Phase angle, $\phi=0$ Angular velocity, $\omega=\pi\text{ rad/s}$The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 63 Marks
A simple pendulum of length $l$ and having a bob of mass $M$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
AnswerThe bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = $g$
Centripetal acceleration = $v^2/R$
where,
v is the uniform speed of the car
R is the radius of the track
Effective acceleration (g') is given as:
$\text{g}'=\sqrt{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$
$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$
$=2\pi\frac{\text{l}}{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$
View full question & answer→Question 73 Marks
Derive the expression for resultant spring constant when two springs having constants $k_1$ and $k_2$ are connected
- In parallel.
- In series.
Answer
- When the springs are connected in parallel, the extension in them will be same and the total restoring force is the sum of their restoring forces.
$\therefore\text{F = F}_1+\text{F}_2$
$-\text{k}_{\text{eq}}\text{x}=-\text{k}_1\text{x}-\text{k}_2\text{x}$
$\text{k}_{\text{eq}}=\text{k}_1+\text{k}_2$
- When the springs are connected in series, the restoring force is same in both the springs and the extensions will be different so the not extension
i.e., $\text{x = x}_1+\text{x}_2$
$=\frac{\text{F}}{-\text{k}_{\text{eq}}}=\frac{-\text{F}}{\text{k}_1}-\frac{\text{F}}{\text{k}_2}$
$\therefore\frac{1}{\text{k}_{\text{eq}}}=\frac{1}{\text{k}_1}+\frac{1}{\text{k}_2}$
when connected in series.
View full question & answer→Question 83 Marks
The displacement of a particle having $\text{S.H.M}$. is $x = 10\sin\Big[10\pi\text{t}+\frac{\pi}{4}\Big]\text{m}.$
- Amplitude.
- Angular frequency.
- Epoch.
- Time period.
- Frequency.
- Maximum velocity.
AnswerGiven equation is, $\text{x}=10\sin\Big(10\pi\text{t}+\frac{\pi}{4}\Big)\text{m},$ comparing with $\text{x = A}\sin(\omega\text{t}+\phi)$ we have
- Amplitude, $\text{A = 10m.}$
- Angular frequency, $\omega=10\pi.$
- Epoch $=$ initial phase, $\frac{\pi}{4}.$
- Time period, $\text{T}=\frac{1}{5}\text{sec}.$
- Frequency, $\text{f = 5Hz.}$
- Maximum velocity $\omega\text{A}=100\pi\text{ms}^{-1}.$
View full question & answer→Question 93 Marks
What is Simple Harmonic Motion? Show that in $\text{S.H.M}$., acceleration is directly proportional to its displacement at a given instant.
AnswerSimple Harmonic Motion :
- Motion is always directed towards a fixed point or equilibrium point.
- Motion being represented by bounded trigonometric functions.
- Acceleration is directly proportional to negative of displacement, i.e., $\text{a}\propto-\text{x}$
Equation for $\text{S.H.M.}$
Acceleration $=-\omega^2\text{x}$
$\Rightarrow\frac{\text{d}^2\text{x}}{\text{dt}^2}+\omega^2\text{x}=0,\omega=2\pi\text{f}$
$\omega$ is angular frequency $($radian$/$ sec$), f$ is linear fiequency $(s^{-1}) $or $($hertz$)$ View full question & answer→Question 103 Marks
Find the expression for the total energy of a particle executing S.H.M.
AnswerP.E. with a S.H.M.$=\frac{1}2{}\text{Kx}^2=\frac{1}{2}\text{m}\omega^2\text{x}^2$
K.E. with a S.H.M.$=\frac{1}2{}\text{mv}^2=\frac{1}2{}\text{m}\big[\omega\sqrt{\text{A}^2-\text{x}^2}\big]^2$
$=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$
Where m is mass, A is amplitude, x is any position and $\omega$ is the angular fiequency,$\therefore$ Total energy $=\frac{1}{2}\text{m}\omega^2\text{x}^2+\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$
$=\frac{1}{2}\text{m}\omega^2\text{A}^2$
View full question & answer→Question 113 Marks
A particle starts S.H.M. from the mean position. Its amplitude is A and its time period is T. At one time its speed is half that of the maximum speed. What is this displacement?
AnswerLet the particle be at R when its velocity $\text{v}=\frac{\text{v}_{\text{max}}}{2}=\frac{\text{A}\omega}{2}$ and its displacement from the mean position O be y.
As $\text{v}=\omega\sqrt{\text{A}^2-\text{y}^2},$
So $\text{y}=\sqrt{\text{A}^2-\frac{\text{v}^2}{\omega^2}}$
Given $\text{v}=\frac{\text{A}\omega}{2},$
then $\text{y}=\sqrt{\text{A}^2-\frac{\text{A}^2\omega^2}{4\omega^2}}=\frac{\sqrt{3}}{2}\text{A}$
View full question & answer→Question 123 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)$
Answer$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)=+2\cos\Big(3\text{t}+\frac{\pi}{3}+\frac{\pi}{2}\Big)$$=2\cos\Big(3\text{t}+\frac{5\pi}{6}\Big)$
If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 2cm
Phase angle, $\phi=\frac{5\pi}{6}=150^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=3\text{rad/sec.}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 133 Marks
In a HCl molecule, we may treat Cl to be of infinite mass and H alone be oscillating. If the oscillation of HCl molecule shows a frequency of $9 \times 10^{13}s^{-1}$, deduce the force constant. [Given: Avogadro's number = $6 \times 10^{26}$ per kg mole.)
Answer$1kg$ of $H$ has $6 \times 10^{26}$ atoms.
$\therefore\text{m}=\frac{1}{6\times10^{26}},\text{v}=9\times10^{13}\text{s}^{-1}$
$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$\text{v}^2=\frac{1}{4\pi^2}\times\frac{\text{k}}{\text{m}}$
$\text{k}=4\pi^2\text{v}^2\text{m}$
$=4(3.14)^2(9\times10^{13})\times\frac{1}{6\times10^{26}}\text{Nm}^{-1}$
$=5.32\times10^2\text{Nm}^{-1}$
View full question & answer→Question 143 Marks
A harmonic oscillation is represented by $\text{y}=0.34\cos(3000\text{t}+0.74),$ where $y$ and $t$ are in $m$ and $s$ respectively. Deduce: $(i)$ the amplitude, $(ii)$ the frequency and angular frequency, $(iii)$ the period, and $(iv)$ the initial phase.
AnswerGiven that, $\text{y}=0.34\cos(3000\text{t}+0.74)$
While the general expression for displacement is, $\text{y = a}\cos(\omega\text{t}+\phi_0)$
Comparing these two expressions,
- Amplitude,$ a = 0.34m.$
- Angular frequency
$\omega=3000\text{ radian/ sec}^{-1}$
Frequency $\text{v}=\frac{\omega}{2\pi}=\frac{3000}2\pi{}=\frac{1500}{\pi}\text{Hz.}$
- Period $\text{T}=\frac{1}{\text{v}}=\frac{1}{\frac{1500}{\pi}}=\frac{\pi}{1500}\text{s}.$
- lnitial phase $\phi_0=0.74\text{ rad.}$
View full question & answer→Question 153 Marks
A body of mass $5kg$ executes $\text{S.H.M}$. of amplitude of $0.5m$. If the force constant is $100Nm^{-1}$, calculate
- Its time period.
- Its maximum kinetic energy, maximum potential energy and total energy.
AnswerGiven: $m = 5\ kg, k = 100N/m; A = 0.5m$
- Time period is given by
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{5}{100}}=0.41\text{s}$
- Angular velocity,
$\omega=\frac{2\pi}{\text{T}}=\frac{2\pi}{0.41}=4.5\text{ rad/ s}$
Maximum $\text{K.E.}=\text{E}_{\text{Kmax}}=\frac{1}{2}\text{mv}^2_0$
$=\frac{1}2{}\text{m}(\omega\text{A})^2=12.50\text{J}$
Maximum $\text{P.E. = E}_{\text{Pmax}}$
$=\frac{1}{2}\text{k}\text{A}^2=12.50\text{J}$
Total energy $\text{E = E}_{\text{Kmax}}$
$=\text{E}_{\text{Pmax}}=12.50\text{J}$ View full question & answer→Question 163 Marks
A particle is subjected to two simple harmonic motions in the same direction having equal amplitude and equal frequency. If the resultant amplitude is equal to the amplitude of individual motions, what is the phase difference between the motions?
AnswerHere, $\text{a}_1=\text{r};\text{a}_2=\text{r}$ and $\text{R = r},\theta=?$ As $\text{R}^2=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\theta$$\therefore\text{r}^2=\text{r}^2+\text{r}^2+2\text{r.r}\cos\theta$
$=2\text{r}^2(1+\cos\theta)$
$1+\cos\theta=\frac{1}{2}$
$\cos\theta=-\frac{1}{2},$
$=\cos120^{\circ}$
$\theta=120^{\circ}$
$=\frac{2\pi}{3}\text{radian}$
View full question & answer→Question 173 Marks
A body oscillates with $\text{S.H.M}$. according to the equation:$\text{x(t)}=5\cos\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$
where $x$ is in meters and $t$ is in seconds. Calculate the following:
- Displacement at $t = 0$
- Angular frequency
- Magnitude of velocity $($Maximum$)$.
Answer$\text{x(t)}=5\cos\Big(2\pi\text{t}+\frac{\pi}{4}\Big)\text{m}$
- $\text{x}(0)=5\cos\Big(0+\frac{\pi}{4}\Big)=5\times\frac{1}{\sqrt{2}}\text{m}=\frac{5}{\sqrt{2}}\text{m}$
- Angular frequency $=\omega=2\pi\text{v}\text{ rad/sec}.$
- Maximum velocity $=\omega\text{A}=2\pi\times5=10\pi\text{ ms}^{-1}$
View full question & answer→Question 183 Marks
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density $\rho_{\text{l}}.$ The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.$\text{T}=2\pi\sqrt{\frac{\text{h}\rho}{\rho_{\text{l}}\text{g}}}$
where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid).
AnswerGiven, Area of cross-section of cork = A, Height of cork = h Density of liquid $=\rho_{\text{l}}$ In equilibrium position of the weight of liquid displacement by cork = weight of cork. When cork is further depressed by $\xi,$ than it further displaces liquid, an extra upthrust acts upwards, which provides restoring force to the cork. Restoring force = extra upthrust = weight of extra displaced water.$\text{F}=-(\text{volume}\times\text{density}\times\text{g})$
$\text{F}=-\text{A}\times\text{y}\times\rho_{\text{l}}\times\text{g} \ ...(\text{i})$
$\text{k}=\frac{\text{F}}{\text{y}}=-\text{A}\rho_{\text{l}}\text{g}$
The period of oscillation of cork is given by$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
where m = mass of cork = volume of cork × density of cork$=\text{A}\times\text{h}\times\rho$
$\text{T}=2\pi\sqrt{\frac{\text{A}\times\text{h}\times\rho}{\text{A}\rho_{\text{l}}\text{g}}}=2\pi\sqrt{\frac{\text{h}\rho}{\rho_{\text{l}}\text{g}}}$
thus, $\text{T}=2\pi\sqrt{\frac{\text{h}\rho}{\rho_{\text{l}}\text{g}}}$
View full question & answer→Question 193 Marks
A particle is in linear simple harmonic motion between two points $A$ and $B, 10\ cm$ apart. Take the direction from $A$ to $B$ as positive direction and give the signs of velocity and acceleration on the particle when it is :
- At the end $B$.
- At $3\ cm$ away from $A$ going towards $B$.
Answer
- At the end $B$ velocity is zero. Here acceleration and force are negative as they are directed along $BR$ i.e., along negative direction.
- At $3\ cm$ away from $A$ going towards $B$, the particle is at $R$, with a tendency to move along $RP$ which is positive direction, here velocity, acceleration are all positive.
View full question & answer→Question 203 Marks
A body of mass $12kg$ is suspended by a coil spring of natural length $50cm$ and force constant $2.0 \times 10Nm^3$. What is the stretched length of the spring? If the body is pulled down further stretching the spring to a length of $5.9cm$ and then released, what is the frequency of oscillation of the suspended mass?
(Neglect the mass of the spring.)
Answer$m = 12 kg$; Original length $l = 50 cm; k = 2.0 \times 10^3Nm^{-1}$, As $\text{F = mg}$$\therefore\frac{\text{F}}{\text{k}}=\frac{\text{mg}}{\text{k}}=\frac{12\times9.8}{2\times10^3}$
$=5.9\times10^{-2}\text{m}=5.9\text{cm}$
$\therefore$ Stretched length of the spring
$=\text{l + y}=50+5.9=55.9\text{cm}$
$\text{v}=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$=\frac{1}{2\times3.14}\sqrt{\frac{2\times10^3}{12}}$
$=2.06\text{s}^{-1}$
View full question & answer→Question 213 Marks
Define the restoring force and it characteristic in case of an oscillating body.
AnswerA force which takes the body back towards the mean position in oscillation is called restoring force.
Characteristic of restoring force: The restoring force is always directed towards the mean position and its magnitude of any instant is directly proportional to the displacement of the particle from its mean position of that instance.
View full question & answer→Question 223 Marks
The potential energy of a particle of mass 1kg in motion along the x-axis is given by $\text{U}=4(1-\cos2\text{x})$ Here x is in metres. Find the period of small oscillations.
Answer$\text{F}=-\frac{\text{dU}}{\text{dx}}$$=-\frac{\text{d}}{\text{dx}}[4(1-\cos2\text{x})]-8\sin2\text{x}$
$\text{F}=-8\times2\text{x}$
[When x is small, $\sin2\text{x}=2\text{x}$]
$\text{F}=-16\text{x}$
As $\text{F}\propto\text{x}$ and -ve sign shows that x is directed towards equilibrium position hence the particle will execute SHM.
Here spring factor, k = 16N/ m
Inertia factor m = 1kg
$\therefore$ Time period $\text{T}=32\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$=2\pi\sqrt{\frac{1}{16}}=\frac{\pi}{2}\text{s}$
View full question & answer→Question 233 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=3\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$
Answer$\text{x}=3\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$$=-3\cos\bigg[\Big(2\pi\text{t}+\frac{\pi}{4}\Big)+\frac{\pi}{2}\bigg]=-3\cos\Big(2\pi\text{t}+\frac{3\pi}{4}\Big)$
If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 3cm
Phase angle, $\phi=\frac{3\pi}{4}=135^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=2\pi\text{ rad/s}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 243 Marks
If the acceleration due to gravity on moon is one sixth that on the earth what will be the length and time period of a second's pendulum there? (g = 9.8 ms?)
AnswerOn moon $\text{g}_{\text{m}}=\frac{\text{g}}{6}=\frac{9.8}{6};\text{T}=2\text{s}$$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}_{\text{m}}}}$
$\text{l}=\frac{\text{T}^2\text{g}_{\text{m}}}{4\pi^2}=\frac{2^2\times\big(\frac{9.8}{6}\big)}{4\times\big(\frac{22}{7}\big)^2}$
$=0.165\text{m}=16.5\text{cm}$
View full question & answer→Question 253 Marks
Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.
AnswerLet us assume that the required displacement where PE is half of the maximum energy of the oscillator be x. The potential energy of the oscillator at this position,$\text{PE}=\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{m}\omega^2\text{x}^2$
maximum energy of the oscillator = maximum potential energy = Total energy$\text{TE}=\frac{1}{2}\text{m}\omega^2\text{A}^2$
Where, A = amplitude of motion. We are given, $\text{PE}=\frac{1}{2}\text{TE}$$\Rightarrow\frac{1}{2}\text{m}\omega^2\text{x}^2=\frac{1}{2}\Big[\frac{1}{2}\text{m}\omega^2\text{A}^2\Big]$
$\Rightarrow\text{x}^2=\frac{\text{A}^2}{2}\ \text{or}\ \text{x}=\sqrt{\frac{\text{A}^2}{2}}=\pm\frac{\text{A}}{\sqrt{2}}$
View full question & answer→Question 263 Marks
Two simple harmonic motions are represented by:$\text{x}_1=10\sin\Big(4\pi\text{t}+\frac{\pi}{4}\Big)$
$\text{x}_2=5(\sin4\pi\text{t}+\sqrt{3}\cos4\pi\text{t})$
What is the ratio of the amplitudes?
Answer$\text{x}_2=5\sin4\pi\text{t}+5\sqrt{3}\cos4\pi\text{t}$Amplitude of $\text{x}_2=\sqrt{5^2+(5\sqrt{3})^2}=10$
Since the $\sin\pi\text{t}$ and $\cos4\pi\text{t}$ functions are out ofphase by $\frac{\pi}{2}.$
Amplitude of $x_2 = 10$
$\therefore$ Ratio of amplitudes is $1 : 1$
View full question & answer→Question 273 Marks
A man stands on a weighing machine placed on a horizontal platform. The machine reads $50kg$. By means of a suitable mechanism the platform is made to execute harmonic vibrations up and down with a frequency of $2$ vibrations per second. What will be the effect on the reading of the weighing machine? The amplitude of vibrations of platform is 5cm. Take $g = 10ms^{-2}$?
AnswerHere, $m = 50kg, v = 2s^{-1} a = 5cm = 0.05m$ Max. acceleration$\text{a}_{\text{max}}=\omega^2\text{A}=(2\pi\text{v})^2\text{A}=4\pi^2\text{v}^2\text{A}$
$=4\times\Big(\frac{2}{7}\Big)^2\times(2)^2\times0.05$
$=7.9\text{ms}^{-2}$
$\therefore$ Maximum force felt by the man $= m(g + a_{max}) = 50(10 + 7.9) = 895.0N = 89.5kg f$
Minimum force felt by the man = $m(g - a_{max}) = 50(10 - 7.9) = 105.0 = 10.5kg\ f$
Hence, the reading of the weighing machine varies between 10.5kg f and 89.5kg f.
View full question & answer→Question 283 Marks
- What is meant by Simple Harmonic Motion $\text{(S.H.M)}$?
- At what points is the energy entirely kinetic and potential in $\text{S.H.M}$?
- What is the total distance travelled by a body executing $\text{S.H.M}$ in a time equal to its time period, if its amplitude is $A$?
Answer
- Simple harmonic motion is the projection of uniform circular motion on a diameter of a circle of reference.
- At mean position $- \text{K.E.}$
At extreme position $\text{- P.E.}$
- $4A$ because in completing one oscillation it crosses mean position $2$ times so total dist. is $4A.$
View full question & answer→Question 293 Marks
A spring of constant k is attached with a mass m and is made to oscillate. What is its time period?
AnswerThe mass when displaced will stretch or compress the spring. If x is the displacement, the restoring force will be F = -kx. This makes the mass to oscillate. 
$\therefore\text{ma}=-\text{kx,a}=\frac{-\text{k}}{\text{m}}\text{x}$
We know, $\text{T}=2\pi\sqrt{\frac{\text{Displacement}}{\text{Acceleration}}}$$2\pi\sqrt{\frac{\text{Inertial factor}}{\text{Spring factor}}}$
$\text{T}=2\pi\sqrt{-\frac{\text{x}}{\text{a}}}$
$2\pi\sqrt{\frac{\text{m}}{\text{k}}}$ View full question & answer→Question 303 Marks
What is the ratio of maxmimum acceleration to the maximum velocity of a simple harmonic oscillator?
AnswerConsider a SHM. $\text{x}=\text{A}\sin\omega\text{t}$$\text{v}=\frac{\text{dx}}{\text{dt}}=\text{A}\omega\cos\omega\text{t}$
For $\text{v}_\text{max}\cos\omega\text{t}=-1$$\therefore\text{v}_\text{max}=\text{A}\omega$
$\text{a}=\frac{\text{dv}}{\text{dt}}=-\text{A}\omega^2\sin\omega\text{t}$
For $\text{a}_\text{max}\sin\omega\text{t}=-1 $$\text{a}_\text{max}=\text{A}\omega^2$
$\therefore\frac{\text{a}_\text{max}}{\text{v}_\text{max}}=\frac{\text{A}\omega^2}{\text{A}\omega}=\frac{\omega}{1}$
View full question & answer→Question 313 Marks
Two identical springs of spring constant k each are attached to a block of mass m as shown in figure:
Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.
Answer
Let the mass be displaced by a small distance x to the right of equilibrium position. Due to this, spring on left gets elongated by length equal to x and that on the right side gets compressed by same length. Then force acting on masses are
$F_1 = -kx$ (force acting on left side and trying to pull the mass towards the mean position.)
$F_2 = -kx$ (force exerted by spring on right side trying to push the mass towards mean position.)
Net force F, acting on the mass
$F = -2kx$
$\therefore$ Force acting on mass is directly propotional to displacement and it directed towards mean position.
$\therefore$ Motion is simple harmonic and time period of oscillation is
$\text{T}=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}$ View full question & answer→Question 323 Marks
A pendulum of length $l$ is attached with a bob and placed in a lift. What will be time period when the lift is :
- Having uniform motion upwards.
- Accelerated upwards by a.
- Accelerated downward by a?
Answer
- When there is uniform motion there is no change in acceleration.
$\therefore\text{T}=\pi\sqrt{\frac{\text{l}}{\text{g}}}$
- When there is an upward acceleration, $g \rightarrow g + a.$
$\therefore\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g + a}}}$
- When there is a downward acceleration $g \rightarrow g - a.$
$\therefore\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}-\text{a}}}$ View full question & answer→Question 333 Marks
A body oscillates with SHM according to the equation (in SI unit)$\text{x}=5\cos\Big[2\pi\text{t}+\frac{\pi}{4}\Big]$
At t = 1.5 second, calculate (i) displecement, (ii) speed.
Answer$\text{x}=5\cos\Big[2\pi\text{t}+\frac{\pi}{4}\Big]\text{at t}=1.5\text{ sec}$Displacement, $\text{x}=5\cos\Big[2\pi(1.5)+\frac{\pi}{4}\Big]$
$=5\cos\Big[3\pi+\frac{\pi}{4}\Big]$
Velocity of oscillation,
$\text{u}=\frac{\text{dx}}{\text{dt}}=-5\times2\pi\times\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$
$=-10\pi\sin\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$
at $\text{t}=1.5\text{ sec}, \text{u}=-10\pi\sin\Big(2\pi(1.5)+\frac{\pi}{4}\Big)$
$=-10\pi\sin\Big(3\pi+\frac{\pi}{4}\Big)$
View full question & answer→Question 343 Marks
Two simple harmonic motions are represented by the following equations:$\text{y}_1=10\sin\frac{\pi}{4}(12\text{t}+1)$
and $\text{y}_2=5(\sin3\pi\text{t}+\sqrt{3}\cos\pi\text{t}).$ What is the ratio of their amplitudes?
AnswerHere, $\text{y}_1=10\sin\frac{\pi}{4}(12\text{t}+1)$$=10\sin(3\pi\text{t}+\frac{\pi}{4}) \ ...(\text{i})$
and $\text{y}_2=5[\sin3\pi\text{t}+\sqrt{3}\cos3\pi\text{t}]$$=10\Big[\frac{1}{2}\sin3\pi\text{t}+\frac{\sqrt{3}}{2}\cos3\pi\text{t}\Big]$
$=10\Big[\cos\frac{\pi}{3}\sin3\pi\text{t}+\sin\frac{\pi}{3}\cos3\pi\text{t}\Big]$
$=10\sin\big(3\pi\text{t}+\frac{\pi}{3}\big) \ ...(\text{ii})$
Thus, from (i) and (ii), the amplitude ratio of motion $=\frac{10}{10}=\frac{1}{1}=1:1$
View full question & answer→Question 353 Marks
The angular velocity and amplitude of a simple pendulum is $\omega$ and r respectively. At a displacement x from the mean position, if its kinetic energy is T and potential energy is V, find the ratio of T to V.
AnswerKinetic energy at. x is$\text{T}=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$
Potential energy at x is$\text{V}=\frac{1}{2}\text{m}\omega^2\text{x}^2$
$\frac{\text{T}}{\text{V}}=\frac{\text{A}^2-\text{x}^2}{\text{x}^2}=\Big(\frac{\text{A}^2}{\text{x}^2}-1\Big)$
View full question & answer→Question 363 Marks
A force of 6.4N stretches a vertical spring by 0.1m. Find the mass that must be suspended from the spring so that it oscillates with the period of $\Big(\frac{\pi}{4}\Big)$ second.
Answerspring factor, $\text{k}=\frac{\text{f}}{\text{m}}=\frac{6.4}{0.1}=64\text{Nm}^{-1};\text{T}=\frac{\pi}{4}\text{s};$ Inertial factor = mass suspended = m. ln S.H.M the time period is given by$\text{T}=2\pi\sqrt{\frac{\text{Inertial factor}}{\text{Spring factor}}}$
$\therefore\frac{\pi}{4}=2\pi\sqrt{\frac{\text{m}}{64}}$ or $\text{m}=1\text{kg}$
View full question & answer→Question 373 Marks
A body of mass ‘m' suspended from a spring executes S.H.M. Calculate the ratio of the kinetic energy and potential energy of the body when it is at half the amplitude far from the mean position.
Answer$\text{y}=\frac{\text{a}}{2}$$\text{K.E.}=\frac{1}2{}\text{m}\omega^2(\text{a}^2-\text{y}^2)$
$\text{P.E.}=\frac{1}2{}\text{m}\omega^2\text{y}^2$
$\frac{\text{K.E.}}{\text{P.E.}}=\frac{\frac{\text{a}^2-\text{a}^2}{4}}{\frac{\text{a}^2}{4}}=\frac{3\text{a}^2}{4}\times\frac{4}{\text{a}^2}$
$\frac{\text{K.E.}}{\text{P.E.}}=\frac{3}{1}$
View full question & answer→Question 383 Marks
Show that the motion of a particle represented by $\text{y}=\sin\text{ax}-\cos\cot$ is simple harmonic with a period of $\frac{2\pi}{\omega}.$
AnswerA function will represent S.H.M. if it can be written uniquely in the form of a or a sin$\Big(\frac{2\pi}{\text{T}}\text{t}+\phi\Big)$
Now $\text{y}=\sin\omega\text{t}-\cos\omega\text{t}$$\text{y}=\sqrt{2}\Big[\sin\omega\text{t}\frac{1}{\sqrt{2}}-\cos\omega\text{t}\sin\frac{\pi}{4}\Big]$
$\text{y}=\sqrt{2}\sin\Big(\omega\text{t}-\frac{\pi}{4}\Big)$
Comparing with standard SHM $\text{y}=\text{a}\sin\Big(\frac{2\pi}{\text{T}}\text{t}+\phi\Big)$$\text{w}=\frac{2\pi}{\text{T}}\ \text{or}\ \text{T}=\frac{2\pi}{\omega}.$
View full question & answer→Question 393 Marks
A cylindrical wooden block of cross-section $15.0cm$? and $230$ grams is floated over water with an extra weight $50$ grams attached to its bottom. The cylinder floats vertically. From the state of equilibrium, it is slightly depressed and released. If the specific gravity of wood is $0.3$ and $g = 9.8ms^{-2}$, deduce the frequency of oscillation of the block.
Answer$\text{mg = V}\rho\text{g}$$(230+50)\times980=15.0\times\text{l}\times1.0\times980$
$\text{l}=\frac{280}{15\times1.0}=18.66\text{cm}$
$\text{T}=2\pi\sqrt{\frac{18.66}{980}}=0.8\text{sec}$
View full question & answer→Question 403 Marks
Displacement versus time curve for a particle executing $\text{S.H.M}$. is shown in Fig. Identify the points marked at which,
- Velocity of the oscillator is zero,
- Speed of the oscillator is maximum.
Answer
Key concept: In displacement$-$time graph of $\text{SHM}$, zero displacement values correspond to mean position; where velocity of the oscillator is maximum. Whereas the crest and troughs represent amplitude positions, where displacement is maximum and velocity of the oscillator is zero.
- The points $\text{A, C, E, G}$ lie at extreme positions $($maximum displacement, $y = A).$ Hence the velocity of the oscillator is zero.
- The points $\text{B, D, F, H}$ lie at mean position $($zero displacement, $y = 0)$. We know the speed is maximum at mean position.
View full question & answer→Question 413 Marks
A uniform U-tube has a liquid of density $\rho$ and a length L. The cross-sectional area is A. If it is made to oscillate, show that it will be S.H.M. and find its frequency.
AnswerLet L be the length of the liquid column in the U-tube of uniform cross section A. The liquid density be $\rho.$ If by tilting, the level of liquid in the limbs differ by 2x, the excess pressure on the higher level side from the same height as the other equalling $2\text{x}\rho\text{g}$ provides restoring force. The entire liquid oscillates as a result.
Since, restoring force = mass × acceleration.$=-2\text{x}\rho\text{gA = AL}\rho\text{a,a}=-\frac{2\text{gx}}{\text{L}}$
The frequency $=\text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{acceleration}}{\text{displacement}}}$$=\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}$ View full question & answer→Question 423 Marks
A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is $2.5cm$. What must be the least period of these oscillations so that the object is not detached from the platform? Take $g = 10ms^{-2}$?
AnswerThe object will not detach from the platform, if the angular frequency $\omega$is such that, during the downward motion, the maximum acceleration equals the acceleration due to gravity, i.e.,$\omega^2_\text{max}\text{A}=\text{g}$
$\omega_\text{max}=\sqrt{\frac{\text{g}}{\text{A}}}$
$\text{T}_\text{min}=\frac{2\pi}{\omega_\text{max}}$
$=2\pi\sqrt{\frac{\text{A}}{\text{g}}}$
Now $\text{A}=2.5\text{cm}$$=2.5\times10^{-2}\text{m}$
$\text{g}=10\text{ms}^{-2}$
Substituting thesa values we get $\text{T}_\text{min}=\frac{\pi}{10}$
View full question & answer→Question 433 Marks
A trolley of mass $3.0\ kg$, as shown in Figure, is connected to two springs, each of spring constant $600Nm^{-1}$. If the trolley is displaced from its equilibrium position by $5.0\ cm$ and released, what is $(a)$ the period of ensuing oscillations, and $(b)$ the maximum speed of the trolley? How much energy is dissipated as heat by the time trolley comes to rest due damping forces?
AnswerEquivalent spring constant: $k\ ' = 2k$
$= 1200Nm^{-1}, m = 3kg$
- $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{3}{120}}$
$=\frac{2\pi}{20}=\frac{\pi}{10}\text{sec}.$
- Maximum speed
$\text{v}=\omega\text{A}=20\times5\times10^{-2}=1\text{ms}^{-1}$
- Energy dissipated $=$ Maximum energy
$=\frac{1}{2}\text{m}\omega^2\text{A}^2$
$=\frac{1}{2}\times3\times20^2\times25\times10^{-4}$
$=600\times25\times10^{-4}$
$=150\times10^{-2}\text{Joule}$
$=1.5\text{ Jule}$ View full question & answer→Question 443 Marks
A particle of mass $10g$ is placed in a potential field given by $U = 50x^2 + 100 erg/gm$. Calculate the frequency of oscillation.
AnswerP.E. of 10 gram particle is $U = 10(50x^2 + 100) erg$. The force acting on the particle is given by$\text{F}=\frac{-\text{dU}}{\text{dx}}=\frac{-\text{d}}{\text{dx}}(500\text{x}^2+1000)$
$=-1000\text{x}$
But $\text{F = m}\frac{\text{d}^2\text{x}}{\text{dt}^2}$$\therefore\text{m}\frac{\text{d}^2\text{x}}{\text{dt}^2}=-1000\text{x}$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=\frac{-1000\text{x}}{\text{m}}=\frac{-1000\text{x}}{10}$
$=-100\text{x}$
As $\frac{\text{d}^2\text{x}}{\text{dt}^2}=\omega^2\text{x};$ so $\omega^2=100$$\omega=\sqrt{100}=10$
Frequency of oscillation,$\text{v}=\frac{\omega}{2\pi}=\frac{\sqrt{100}}{2\pi}=1.58\text{s}^{-1}$
View full question & answer→Question 453 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=\cos\Big(\frac{\pi}{6}-\text{t}\Big)$
Answer$\text{x}=\cos\Big(\frac{\pi}{6}-\text{t}\Big)=\cos\Big(\text{t}-\frac{\pi}{6}\Big)$If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 1
Phase angle, $\phi=-\frac{\pi}{6}=-30^\circ$
Angular velocity, $\omega=\frac{2\pi}{\text{T}}=1\text{rad/s}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 463 Marks
The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.
AnswerLet R be the radius of the dip, and O be its centre. Let the rickshaw of mass M be at P at any instant. This case is similar to that of a simple pendulum. The force that produces oscillations in the rickshaw is $\text{F}=\text{Mg}\sin\theta.$ If $\theta$ is small and is measured in radian then, $\sin\theta=\theta,$
$\therefore\text{F}=-\text{Mg}\theta$ ($\because$ force acts to reduce $\theta$)
Displacement of the rickshaw OP $=\text{y}=\text{R}\theta$$\therefore$ Force constant,
$\text{k}=\frac{-\text{Force}}{\text{Displacement}}$
$=-\Big(\frac{-\text{Mg}\theta}{\text{R}\theta}\Big)=\frac{\text{Mg}}{\text{R}}$
$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{Inertial factor}}{\text{Spring factor}}}$
$=2\pi\sqrt{\frac{\text{MR}}{\text{Mg}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$
$\therefore\text{T}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$ View full question & answer→Question 473 Marks
A cylinder of length l, cross-sectional area A is floating on a liquid of density $\sigma.$ If the cylinder (material density $\rho$) is depressed by a length x further by an external force acting for a short while, estimate the time period of S.H.M.
AnswerLet y be the length immersed in the liquid as it floats. The weight was balanced by upthrust $(\text{Ay}\sigma\text{g})$ while floating. If further displacement x is brought, the upthrust increases and so the oscillation is made possible, i.e., 
Restoring force $=\text{excess upthrust}=-\text{x}\text{A}\sigma\text{g}.$ Also, $\text{ma = Al}\rho\text{a}$$\therefore\text{Al}\rho\text{a}=-\text{A}\sigma\text{gx}$
$\text{a}=\frac{-\sigma\text{g}}{\rho\text{l}}\text{x}$
$\text{T}=2\pi\sqrt{\frac{\rho\text{l}}{\sigma\text{l}}}$ View full question & answer→Question 483 Marks
A body of mass $1.0kg$ is suspended from a weightless spring having force constant $600Nm^{-1}$. Another body of mass $0.5kg$ moving vertically upwards hits the suspended body with a velocity of $3.0ms^{-1}$ and gets embedded in it. Find the frequency of oscillations and amplitude of motion.
AnswerHere, inertia factor = Oscillating mass $= (m + m_1) = 1 + 0.5 = 1.5kg$ Spring factor = Force conslant $= K = 600Nm^{-1}$ Frequency of oscillation,$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{Spring factor}}{\text{Inertial factor}}}$
$=\frac{10}{\pi}\text{Hz}$
Let $v_1$ be the velocity ofthe mass $(m + m_1)$ after collision. According to law of conservation of linear momentum, we have$\text{m}_1\text{v}=(\text{m + m}_1)\text{v}_1$
$\text{v}_1=\frac{\text{m}_1\text{v}}{\text{m + m}_1}$
$\frac{\text{m}_1\text{v}}{\text{m + m}_1}=\frac{0.5\times3}{1+0.5}=1\text{ms}^{-1}$
Here the collision is inelastic. According to the law of conservation of mechanical energy, we have$(\text{K.E}.)_{\text{max}}=(\text{P.E.})_{\text{max}}$
i.e., $\frac{1}{2}(\text{m + m}_1)\text{v}^2_1=\frac{1}{2}\text{kA}^2$$\text{A}=\text{v}_1\sqrt{\frac{\text{m + m}_1}{\text{k}}}$
$=\sqrt{\frac{1.5}{600}}=\frac{1}{20}=5\text{cm}$
View full question & answer→Question 493 Marks
Answer the following questions:The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than $2\pi\sqrt{\frac{\text{l}}{\text{g}}}.$ Think of a qualitative argument to appreciate this result.
AnswerIn the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:$\text{F}=-\text{mg}\sin\theta$
Where, F = Restoring force m = Mass of the bob g = Acceleration due to gravity$\theta=$ Angle of displacement
For small $\theta,\ \sin\theta\simeq\theta$ For large $\theta,\ \sin\theta$ is greater than $\theta.$ This decreases the effective value of g. Hence, the time period increases as: $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ Where, l is the length of the simple pendulum.
View full question & answer→Question 503 Marks
The displacement of a particle executing periodic motion is given by:$\text{y}=4\cos^2\Big(\frac{\text{t}}{2}\Big)\sin(1000\text{t}).$ Find independent constituent simple harmonic motion.
Answer$\text{y}=4\cos^2\Big(\frac{\text{t}}{2}\Big)\sin(1000\text{t})$$=2(1+\cos\text{t})\sin(1000\text{t})$ $[\because2\cos^2\theta=1+\cos2\theta]$
$=2\sin1000\text{t}+2\sin1000\text{t}\times\cos\text{t}$
$=2\sin1000\text{t}+\sin(1000+1)\text{t}+\sin(1000-1)\text{t}$ $[\because2\sin\text{A}\cos\text{B}=\sin(\text{A + B})+\sin(\text{A}-\text{B})]$
$=2\sin1000\text{t}+\sin1001\text{t}+\sin999\text{t}$
$=\sin1000\text{t}+\sin1000\text{t}+\sin1001\text{t}+\sin999\text{t}$
View full question & answer→Question 513 Marks
Prove that if a liquid taken in a U-tube is disturbed from the state of equilibrium, it will oscillate simple harmonically. Find expressions for the angular frequency and time period.
AnswerThe restoring force, F = weight of liquid column of the height 2y$\Rightarrow\text{F}=-(\text{volume})\times\text{density}\times\text{g}$
$=-(\text{A}\times-2\text{y}.\rho\text{g})$
$\Rightarrow\text{F}=-2\text{A}\rho\text{g.y} \ ...(\text{i})$
where A = Area of cross-section of the tube 
$\rho$ = density of mercury
$\text{k}=-\frac{\text{F}}{\text{y}}=2\text{A}\rho.\text{g}$
Time period$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$=2\pi\sqrt{\frac{\text{m}}{2\text{A}\rho.\text{g}}}$
Let L = length of the whole mercury column therefore, mass of mercury$\text{m}=\text{volume}\times\text{density}=\text{A.L.}\rho.$
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{\text{A.L.}\rho}{2\text{A}\rho\text{g}}}=2\pi\sqrt{\frac{\text{L}}{2\text{g}}}$ where L is the totat length of mercury column of L = 2h.
Where h is the height of mercury column in U-tube. It shows that mercury column executes S.H.M. Frequency, $\text{v}=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}$ So, angular frequency$\omega=2\pi\text{v}=2\pi\times\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}=\sqrt{\frac{2\text{g}}{\text{L}}}$ View full question & answer→Question 523 Marks
Aman stands on a weighing machine placed on a horizontal platform. The machine reads $50kg$. By means of a suitable mechanism, the platform is made to execute harmonic vibrations up and down with a frequency of two vibrations per second. What will be the effect on the reading of the weighing machine? The amplitude of vibrations of platform is $5cm$. Take, $g = 10ms^{-2}$.
AnswerHere. $m = 50kg. v = 2s^{-1} A = 5cm = 0.05m$ Maximum acceleration$\text{a}_\text{max}=\omega^2\text{A}$
$=(2\pi\text{v}^2)\text{A}=4\pi^2\text{v}^2\text{A}$
$=4\times\Big(\frac{22}{7}\Big)^2\times(2)^2\times0.05$
$=7.9\text{cm}^{-2}$
$\therefore$ Maximum force felt by the man $=\text{m}(\text{g}+\text{a}_\text{max})$
$=50(10+7.9)$
$=895.0\text{N}$
$=89.5\text{kgf}$
Minimum force felt by the man $=\text{m}(\text{g}-\text{a}_\text{max})$$=50(10-7.9)$
$=105.0\text{N}$
$=10.5\text{kgf}$
Hence, the reading of the weighing machine varies between 10.5kgf and 69.5kgf.
View full question & answer→Question 533 Marks
A particle is performing simple harmonic motion along x-axis with amplitude 4.0cm and time period 1.2s. What is the minimum time taken by the particle to move from x = +2cm to x = +4cm and back again?
Answer As $\text{x = a}\sin\omega\text{t}=\text{a}\sin\frac{2\pi\text{t}}{\text{T}}$ So, $\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big),$ where a = 4cm At $\text{x}=2,\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{2}{4}\Big)$$=\frac{\text{T}}{2\pi}\times\frac{\pi}{6}=\frac{\text{T}}{12}=\frac{1.2}{12}=\frac{1}{10}$
At $\text{x}=4,\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{4}{4}\Big)$$=\frac{\text{T}}{2\pi}\times\frac{\pi}{2}=\frac{\text{T}}{4}=\frac{1.2}{4}=\frac{3}{10}$
Total Time taken $=\frac{3}{10}-\frac{1}{10}=\frac{2}{5}$
View full question & answer→Question 543 Marks
At a time when the displacement is half the amplitude, what fraction of the total energy is kinetic and what fraction is the potential in S.H.M?
Answer When $\text{x}=\frac{\text{A}}{2},$ Fraction of energy as potential is$=\frac{\frac{1}{2}\text{m}\omega^2\Big(\text{A}^2-\frac{\text{A}^2}{4}\Big)}{\frac{1}{2}\text{m}\omega^2\text{A}^2}=\frac{3}{4}\text{ or }75\%$
So, fraction of energy as kinetic is$=\frac{\frac{1}{2}\text{m}\omega^2\frac{\text{A}^2}{4}}{\frac{1}{2}\text{m}\omega^2\text{A}^2}=\frac{1}{2}\text{ or }25\%$
View full question & answer→Question 553 Marks
A spring of force constant k has a mass M suspended from it. If the spring is cut into two halves, and the same mass is attached to one of the pieces, what will be the frequencies of oscillation of the mass?
Answer When the spring is cut into two equal halves, the force constant of each part will be doubled. Therefore, the original frequency, $\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{M}}}$ will become $\text{v}'=\frac{1}{2\pi}\sqrt{\frac{2\text{k}}{\text{M}}}=\sqrt{2\text{v}}$
View full question & answer→Question 563 Marks
Calculate the percentage change in time period of a simple pendulum if its length is increased by 8%.
Answer As $\text{T}\propto\sqrt{\text{l}}$$\frac{\Delta\text{T}}{\text{T}}\times100=\frac{1}2{}\frac{\Delta\text{l}}{\text{l}}\times100$
$\therefore\%\text{ change in T}=\frac{1}{2}\times8=4\%$
View full question & answer→Question 573 Marks
A particle of mass $0.1kg$ is held between two rigid supports by two springs of force constants $8N/ m$ and $2N/ m$. If the particle is displaced along the direction of the length of the springs, calculate its frequency of vibration.
AnswerThe situation is shown in the fig.
When the mass is displaced along the direction of the length of the spring, one spring is compressed while the other is extended but the force due to both the springs is in the same direction. Hence the effective force constant $k = k_1 + k_2 = 8N/ m + 2N/ m = 10N/ m$ The frequency of vibration is given by$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$=\frac{1}{2\pi}\sqrt{\frac{10}{0.1}}$
$\text{v}=\frac{10}{2\pi}$
$=\frac{5}{\pi}\text{s}^{-1}$ View full question & answer→Question 583 Marks
Show that when a particle is moving in S.H.M. its velocity at a distance $\frac{\sqrt{3}}{2}$ of its amplitude from the central position is half its velocity in central position.
Answer In a S.H.M. velocity is given by $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ where x is the displacement from mean position. Velocity at $\text{x}=\frac{\sqrt{3}\text{A}}{2}$ is,$\text{v}_1=\omega\sqrt{\text{A}^2-\frac{3}{4}\text{A}^2}=\omega\text{A}\sqrt{\frac{1}{4}}$
$=\frac{\omega\text{A}}{2}$
Velocity at central position$=\omega\sqrt{\text{A}^2-0^2}=\omega\text{A}$
$\therefore$ Velocity at $\frac{\sqrt{3}\text{A}}{2}=\frac{1}{2}$ (velocity at central position)
View full question & answer→Question 593 Marks
A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the left
- Moves up with uniform velocity v.
- Moves down with uniform velocity v.
- Moves up with uniform acceleration a.
- Moves down with uniform acceleration a.
- Beings to fall freely under gravity?
Answer
- And
- Since acceleration of the lift is zero therefore there will be no effect on time period.
- When the lift moves up with uniform acceleration a, the effective value of acceleration due to gravity is g + a
$\therefore\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}+\text{a}}}$
Clearly, $\text{T}'<\text{T}$.
- When the lift moves down with uniform acceleration a, then the effective value of g is g - a.
$\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}+\text{a}}}$
Clearly,$\text{T}'<\text{T}$.
- When the lift begins to fall freely under gravity, the effective value of g becomes zero. So, T is infinite i.e., the simple pendulum shall not oscillate.
View full question & answer→Question 603 Marks
Show that for a particle in linear SHM, the average kinetic energy over a period of oscillation is equal to average potential energy over the same period.
Answer K.E. $=\frac{1}{2}\text{m}\omega^2\text{A}^2\cos^2\omega\text{t}$ P.E. $=\frac{1}{2}\text{m}\omega^2\text{A}^2\sin^2\omega\text{t}$ Average K.E. over a one time period:$(\text{K.E.)}_{\text{avg}}=\frac{1}{\text{T}}\int\limits^{\text{T}}_0(\text{K.E.)dt}$
$=\frac{1}{\text{T}}\int\limits^{\text{T}}_{0}\frac{1}{2}\text{m}\omega^2\text{A}^2\cos^2\omega\text{t dt}$
$=\frac{1}{2\text{T}}\text{m}\omega^2\text{A}^2\int\limits^\text{T}_0\cos^2\omega\text{t dt}$
$=\frac{\text{m}\omega^2\text{A}^2}{2\text{T}}.\frac{\text{T}}{2}=\frac{1}{4}\text{m}\omega^2\text{A}^2$
Similarly, $\text{P.E.}=\frac{1}{2}\text{m}\omega^2\text{A}^2$
View full question & answer→Question 613 Marks
Derive an expression for the potential energy of an elastic stretched spring.
Answer Consider a spring attached with a mass m stretching by a length y after t sec. Restoring force F = mass × acceleration$=-\text{}\omega^2\text{y}=-\text{ky}$
where, $\text{k = spring constant = m}\omega^2$ Work done for an additional displacement dy against restoring force is$\text{dW}=-\text{F dy}$
$=-(-\text{ky})\text{dy = ky dy}$
Total work done$\text{W}=\int\limits^{\text{y}}_0\text{ky dy}=\frac{1}{2}\text{ky}^2$
This work done appears as a PE. 'U' of the particle.$\text{U}=\frac{1}2{}\text{ky}^2=\frac{1}{2}\text{m}\omega^2\text{y}^2$
$=\frac{1}{2}\text{m}\omega^2\text{a}^2\sin^2\omega\text{t}$
View full question & answer→Question 623 Marks
What is a second's pendulum? How much is its length on the surface of moon?
Answer A second's pendulum is one whose time period of e. oscillation is 2 seconds. On the surface of moon,$\text{a}=\frac{\text{g}}{6}.$
$\therefore\text{Using}\text{ T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}},$
We have $\text{l}=\frac{4\text{g}}{6\times4\pi^2}=\frac{1}{6}\text{m}$
View full question & answer→Question 633 Marks
A particle is executing S.H.M. If $v_1$ and $v_2$ are the speeds of the particle at distance $x_1$ and $x_2$ from the equilibrium position, show that the frequency of oscillations is$\text{f}=-\frac{1}{2\pi}\bigg(\frac{\text{v}^2_1-\text{v}^2_2}{\text{x}^2_2-\text{x}^2_1}\bigg)^{\frac{1}{2}}$
AnswerThe displacement of a particle executing S.H.M. is given by$\text{x = a}\cos\omega\text{t}$
$\frac{\text{dx}}{\text{dt}}=\omega\text{a}\sin\omega\text{t}$
$\therefore\text{Velocity},\text{v}=\frac{\text{dx}}{\text{dt}}$ or $\text{v}^2=\text{a}^2\omega^2\sin^2\omega\text{t}.$
$=\text{a}^2\omega^2(1-\cos^2)\omega\text{t}$
$=\omega^2(\text{a}^2-\text{x}^2)$
Hence, $\text{v}^2_1=\omega^2(\text{a}^2-\text{x}^2_1)$ And $\text{v}^2_2=\omega^2(\text{a}^2-\text{x}^2_2)$ Subtracting the two,$\text{v}^2_1-\text{v}^2_2=\omega^2(\text{x}^2_2-\text{x}^2_1)$
$\omega^2=\frac{\text{v}^2_1-\text{v}^2_2}{\text{x}^2_2-\text{x}^2_1}$
$\omega=\bigg(\frac{\text{v}^2_1-\text{v}^2_2}{\text{x}^2_2-\text{x}^2_1}\bigg)^{\frac{1}{2}}$
But $\omega=2\pi\text{f}$$\therefore\text{f}=\frac{1}{2\pi}=\bigg(\frac{\text{v}^2_1-\text{v}^2_2}{\text{x}^2_2-\text{x}^2_1}\bigg)^{\frac{1}{2}}$
View full question & answer→Question 643 Marks
For an oscillating pendulum, establish the relation, $\frac{\text{d}^2\theta}{\text{dr}^2}+\omega^2=\theta=0,$ Where $\omega^2=\frac{\text{g}}{\text{l}}$ and $\theta$ is small angular displacement.
Answer Restoring force is provided by the portion $\text{mg}\sin\theta$ of gravitational force. Since, it acts perpendicular to length l, the restoring torque $=-\text{mg}\sin\theta\text{l}$ Also, $\tau=\text{l}\alpha=\text{m}\text{l}^2\alpha$$\therefore\text{ml}^2\alpha=-\text{mg}\sin\theta.\text{l}$
$\alpha=-\frac{\text{g}\sin\theta}{\text{l}}$
For small angles ofoscillation, $\sin\theta\cong\theta.$$\therefore\alpha=-\frac{\text{g}}{\text{l}}.\theta$
$\frac{\text{d}^2\theta}{\text{dt}^2}=-\frac{\text{g}}{\text{l}}.\theta$
i.e. $\frac{\text{d}^2\theta}{\text{dt}^2}+\omega^2\theta=0.$ giving $\omega=\sqrt{\frac{\text{g}}{\text{l}}}$ and $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
View full question & answer→Question 653 Marks
Explain the relation in phase between displacement, velocity and acceleration in S.H.M., graphically as well as theoretically.
Answer If displacement:$\text{y = A}\sin(\omega\text{t}+\phi_0)$
velocity $\text{v}=\omega\text{A}\cos(\omega\text{t}+\phi_0)$ 
acceleration $\text{a}=-\omega^2\text{A}\sin(\omega\text{t}+\phi_0)$ 
x and v differ in phase by $\frac{\pi}{2}.$ v and a differ in phase by $\frac{\pi}{2}.$ View full question & answer→Question 663 Marks
For a particle in S.H.M., the displacement x of the particle as a function of time t is given as $\text{x = A}\sin(2\pi\text{t}).$ Here x is in cm and t is in seconds.
Let the time taken by the particle to travel from x = 0 to $\text{x}=\frac{\text{A}}{2}$ be $T_1$ and the time taken to travel from $\text{x}=\frac{\text{A}}{2}$ to x = A be $T_2$. Find $\frac{\text{T}_1}{\text{T}_2}.$
Answerx = 0 at t = 0, t = 1s Let $\text{x}=\frac{\text{A}}{2}\text{ at }\text{t = T}_1$ then $\frac{\text{A}}{2}=\text{A}\sin(2\pi\text{T}_1)$$\text{A}.\sin\Big(\frac{\pi}{6}\Big)=\text{A}\sin(2\pi\text{T}_1)$
$\Rightarrow\frac{\pi}{6}=2\pi\text{T}_1$
$\text{T}_1=\frac{1}{12}\text{s}$
Time taken from x = 0 to x = A is $\frac{\text{T}}{4}$$\Rightarrow\text{at}\text{ x}=\text{A}$ and $\text{t = T}$
$\text{A = A}\sin2\pi\text{T}$
$\sin\frac{\pi}{2}=\sin2\pi\text{T}$
$\text{T}=\frac{1}{4}\sec$
$\text{T}_1+\text{T}_2=\frac{1}{4}$
$\text{T}_2=\frac{1}{4}-\frac{1}{12}=\frac{1}{6}\text{s}$
So, $\frac{\text{T}_1}{\text{T}_{2}}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$
View full question & answer→Question 673 Marks
If $\text{x = a}\cos\omega\text{t + b}\sin\omega\text{t},$ show that it represents S.H.M.
Answer $\text{x = a}\cos\omega\text{t + b}\sin\omega\text{t}$$\frac{\text{dx}}{\text{dt}}=-\text{a}\omega\sin\omega\text{t}+\text{b}\omega\cos\omega\text{t}$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\omega^2\text{a}\cos\omega\text{t}-\text{b}\omega^2\sin\omega\text{t}$
$=-\omega^2(\text{a}\cos\omega\text{t}+\text{b}\sin\omega\text{t})$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\omega^2\text{x}$
$\therefore$ It represents a S. H. M.
View full question & answer→Question 683 Marks
When will the motion of a simple pendulum be simple harmonic?
Answer Consider a pendulum of length l and mass of bob m is displaced by angle $\theta$ as shown in fig. 
The restoring force = F $=\text{mg}\sin\theta$ If $\theta$ is small then $\sin\theta=\theta=\frac{\text{arc}}{\text{radius}}=\frac{\text{x}}{\text{l}}$$\therefore\text{F}=-\text{mg}\frac{\text{x}}{1}\ \text{or}\ \text{F}\propto(-\text{x})$
$(\because\text{m}_2\text{g},1\text{are constant})$
Hence, the motion of simple pendulam will be smile harmonic for small angle $\theta.$ View full question & answer→Question 693 Marks
- What is echo?
- If a reflector is situated at a distance of $860m$ from a sound source, what is the time of echo? Speed of sound in air at a room temperature can be taken as $344m/ s.$
Answer
- The sound heard by an observer$/$ listner after the reflection from a surface is called echo.
- Total distance travelled by sound to come back $= 2 \times 860 = 1720m$ speed of sound is $344m/ s.$
$\therefore\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
$=\frac{1720}{344}=5\text{sec}$ View full question & answer→Question 703 Marks
A block with a mass of $3.0\ kg$ is suspended from an ideal spring having negligible mass and stretches the spring by $0.2m$.
- What is the force constant of the spring?
- What is the period of oscillation of the block if it is pulled down and released?
Answer
- Force constant $\text{k}=\frac{\text{F}}{\text{l}}=\frac{\text{mg}}{\text{l}}$
Here $m = 3.0\ kg$ and elongation in length of spring $l = 0.2m$
$\therefore$ Force constant $\text{k}=\frac{3.0\times9.8}{0.2}=174\text{Nm}^{-1}$
- Period of oscillation $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$=2\times3.14\times\sqrt{\frac{3}{147}}=0.9\text{s}$ View full question & answer→Question 713 Marks
Answer(a), (b) In parallel equivalent value of $k = k_1 + k_2$
$\therefore\text{f}=\frac{1}2\pi{}\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}$
(c), (d) In series equivalent $\text{k}=\frac{\text{k}_1\text{k}_2}{\text{k}_1+\text{k}_2}$
$\therefore\text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1\text{k}_2}{(\text{k}_1+\text{k}_2)\text{m}}}$
View full question & answer→Question 723 Marks
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).$\text{x}=2\cos\pi\text{t}$
Answer $\text{x}=2\cos\pi\text{t}$If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:
Amplitude, A = 2cm
Phase angle, $\phi=0$
Angular velocity, $\omega=\pi\text{ rad/s}$
The motion of the particle can be plotted as shown in the following figure.
View full question & answer→Question 733 Marks
A 0.2kg. of mass hangs at the end of a spring. When 0.02kg more mass is added to the end of the spring, it stretches 7cm more. If the 0.02kg mass is removed, what will be the period of vibration of the system?
Answer When 0.02kg is added, there is a stretch of 7cm. Using mg = Kx, we have$\text{K}=\frac{0.02\times10}{7\times10^{-2}}=\frac{20}{7}=2.86\text{N/m}$
Time period $=\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{K}}}$$=2\pi\sqrt{\frac{0.2}{2.86}}=1.66\sec.$
View full question & answer→Question 743 Marks
A spring compressed by $0.1m$ develops a restoring force of $10N$. A body of mass $4\ kg$ is placed on it. Deduce the $(i)$ force constant of the spring $(ii)$ depression of the spring under the weight of the body and $(iii)$ period of oscillation, if the body is disturbed.
AnswerRestoring force,
$F = 10N;$
Mass of body $m = 4 kg$
Displacement $\xi=0.1\text{m}$
- The force constant of spring
$\text{k}=\frac{\text{Force}}{\text{Displacement}}=\frac{10}{0.1}$
$=100\text{Nm}^{-1}$
- Depression due lo weight
$=\frac{\text{Force}}{\text{k}}=\frac{4\times10}{100}=0.4\text{m}$
- Period of oscillation,
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{4}{100}}=\frac{2\pi}{5}\text{s}$ View full question & answer→Question 753 Marks
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
AnswerThe bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity = g Centripetal acceleration = $v^2/R$ where, v is the uniform speed of the car R is the radius of the track Effective acceleration (g') is given as:$\text{g}'=\sqrt{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$
$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$
$=2\pi\frac{\text{l}}{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$
View full question & answer→Question 763 Marks
A particle is subjected to two simple harmonic motions$\text{x}_1=\text{A}_1\sin\omega\text{t}$ $\text{x}_2=\text{A}_2\Big(\omega+\frac{\pi}{3}\Big)$ Find
- The disphcement at $t = 0.$
- The maximum speeil of thc particle and.
- The madmum accelerution of the particle.
Answer
- At $t = 0$
$\text{x}_1=\text{A}_1\sin\omega\text{t}=0$
$\text{x}_2=\text{A}_2\Big(\omega+\frac{\pi}{3}\Big)$
$=\frac{\text{A}_2\sqrt{3}}{2}$
Thus the resultant displacement at $t = 0$ is
$\text{x}=\text{x}_1+\text{x}_2=\frac{\text{A}_2\sqrt{3}}{2}$
- $\text{A}=\sqrt{\text{A}_1^2+{\text{A}_2^2}+2\text{A}_1\text{A}_2^2\cos\frac{\pi}{3}}$
$\text{A}=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$
The maximum speed is
$\text{V}_\text{max}=\omega\text{A}$
$=\omega=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$
- The maximum acceleration is
$\text{a}_\text{max}=\omega^2\text{A}$
$=\omega^2=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$ View full question & answer→Question 773 Marks
A body oscillates with SHM according to the equation $($ in $SI$ units $)$, $x=5 \cos [2 \pi t+\pi / 4] \text {. }$
At $t=1.5 s$, calculate the $(a)$ displacement, $(b)$ speed and $(c)$ acceleration of the body.
AnswerThe angular frequency $\omega$ of the body $=2 \pi s ^{-1}$ and its time period $T=1 s$.
At $t=1.5 s$
(a) displacement $=(5.0 m ) \cos \left[\left(2 \pi s ^{-1}\right) \times\right.1.5 s +\pi / 4] $
$ =(5.0 m ) \cos [(3 \pi+\pi / 4)]$
$ =-5.0 \times 0.707 m$
$ =-3.535 m$
$(b)$ Using Eq. $(13.9),$ the speed of the body
$=-(5.0 m )\left(2 \pi s ^{-1}\right) \sin \left[\left(2 \pi s ^{-1}\right) \times 1.5 s \right.$
$+\pi / 4]$
$=-(5.0 m )\left(2 \pi s ^{-1}\right) \sin [(3 \pi+\pi / 4)]$
$=10 \pi \times 0.707 m s ^{-1}$
$=22 m s ^{-1}$
$(c)$ Using Eq.$(13.10),$ the acceleration of the body
$=-\left(2 \pi s ^{-1}\right)^2 \times \text { displacement }$
$=-\left(2 \pi s ^{-1}\right)^2 \times(-3.535 m )$
$=140 m s ^{-2}$
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