Additional Questions - Chapter 10 Oscillations 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
The total energy of a particle vibrating in SHM is proportional to the square of its ........
(a) velocity
(b) acceleration
(c) amplitude
(d) none of these
Answer:
(a) veloctiy
Question 2.
In order to double the period of a simple pendulum
(a) its length should doubled
(b) its length should be quadrupled.
(c) the mass of its bob should be doubled.
(d) the mass of its bob should be quadrupled.
Answer:
(b) its length should be quadrupled.
Question 3.
A simple harmonic oscillator has amplitude $A$ and time period $T$. Its maximum speed is .......
(a) $\frac{4 \mathrm{~A}}{\mathrm{~T}}$
(b) $\frac{2 \mathrm{~A}}{\mathrm{~T}}$
(c) $\frac{4 \pi \mathrm{A}}{\mathrm{T}}$
(d) $\frac{2 \pi \mathrm{A}}{\mathrm{T}}$
Answer:
(d) $\frac{2 \pi \mathrm{A}}{\mathrm{T}}$

Solution:
$
\nu_{\max }=\omega \mathrm{A}=\frac{2 \pi}{\mathrm{T}} \mathrm{A}
$
Question 4.
A simple harmonic oscillator has a period of $0.01 \mathrm{~s}$ and an amplitude of $0.2 \mathrm{~m}$. The magnitude of the velocity in $\mathrm{m} / \mathrm{s}$ at the centre of oscillation is
(a) $20 \pi$
(b) $40 \pi$
(c) $60 \pi$
(d) $80 \pi$
Answer:
(b) $40 \pi$
Solution:
Velocity is maximum at the centre of oscillation and is given by
$
v_0=\frac{2 \pi}{\mathrm{T}} \mathrm{A}=\frac{2 \pi \times 0.2}{0.01}=40 \pi \mathrm{m} / \mathrm{s}
$
Question 5.
A particle is executing SHM. Then the graph of acceleration as a function of displacement is .......
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(a) straight line
Solution:
In SHM, $F \propto \mathrm{y} \Rightarrow \mathrm{a} \propto \mathrm{y}$; Thus the graph is a straight line.
Question 6.
A particle is executing SHM. Then the graph of velocity as a function of displacement is
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(c) ellipse
Solution:
$
=\mathrm{A} \omega \sqrt{1-\frac{y^2}{\mathrm{~A}^2}} \Rightarrow \frac{y^2}{\mathrm{~A}^2}+\frac{v^2}{(\mathrm{~A})^2}=1
$

Question 7.
The amplitude of a vibrating body situated in a resisting medium
(a) decreases linearly with time
(b) decrease exponentially with time
(c) decreases with time in some other manner
(d) remains constant with time
Answer:
(b) decreases exponentially with time
Question 8 .
The frequency of a vibrating body situated in air ......
(a) is the same as its natural frequency
(b) is higher than its natural frequency
(c) is lower than its natural frequency
(d) can have any value
Answer:
(c) is lower than its natural frequency
Question 9.
The equation $\frac{d^2 y}{d t^2}+b \frac{d y}{d t}+\omega^2 y=0$ represents the equation of motion for a vibration
(a) free
(b) damped
(c) forced
(d) resonant
Answer:
(b) damped
Question 10 .
The displacement equation of an oscillator is $\mathrm{y}=5 \sin (0.27 \pi \mathrm{t}+0.5 \pi)$ in SI units. The time period of oscillation is
(a) $10 \mathrm{~s}$
(b) $1 \mathrm{~s}$
(c) $0.2 \mathrm{~s}$
(d) $0.5 \mathrm{~s}$

Answer:
(a) $10 \mathrm{~s}$
Solution:
Comparing with the standard equation $y=A \sin (\omega t+\varphi)$
We have, $\omega=0.2 \pi \Rightarrow \mathrm{T}=\frac{2 \pi}{0.2 \pi}=10 \mathrm{~s}$
Question 11.
A loaded spring vibrates with a period $T$. The spring is divided into four equal parts and the same load is suspended from one as these parts. The new time period is .........
(a) $\frac{\mathrm{T}}{4}$
(b) $\frac{\mathrm{T}}{2}$
(c) $2 \mathrm{~T}$
(d) $4 \mathrm{~T}$
Answer:
(b) $\frac{T}{2}$
Solution:
Let the force constant of the spring be $k$. Then $T=2 \pi \sqrt{\frac{m}{k}}$
If the spring is divided into four equal parts, then the force constant of each part will be $4 \mathrm{k}$.
Therefore, $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\mathrm{M}}{4 k}}=\frac{\mathrm{T}}{2}$
Question 12 .
The vertical extension in a light spring by a weight of $1 \mathrm{~kg}$, in equilibrium is $9.8 \mathrm{~cm}$. The period of oscillation of the spring, in seconds, will be ......
(a) $\frac{2 \pi}{10}$
(b) $\frac{2 \pi}{100}$
(c) $20 \pi$
(d) $200 \pi$
Answer:
(a) $\frac{2 \pi}{10}$
Solution:

$
\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{9.8 \times 10^{-2}}{9.8}}=\frac{2 \pi}{10} \mathrm{~s}
$
Question 13.
A particle executing SHM has an acceleration of $64 \mathrm{~cm} / \mathrm{s}^2$ with its displacement is $4 \mathrm{~cm}$. Its time period, in seconds is ......
(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{4}$
(c) $\pi$
(d) $2 \pi$
Answer:
(a) $\frac{\pi}{2}$
Solution:
$
a=\omega^2 y \Rightarrow \omega=\sqrt{\frac{a}{y}} \Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{y}{a}}=2 \pi \sqrt{\frac{4}{64}}=\frac{\pi}{2}
$
Question 14.
A body executes SHM with an amplitude A. Its energy is half kinetic and half potential when the displacement is .......
(a) $\frac{\mathrm{A}}{3}$
(b) $\frac{\mathrm{A}}{2}$
(c) $\frac{\mathrm{A}}{\sqrt{2}}$
(d) $\frac{\mathrm{A}}{2 \sqrt{2}}$
Answer:
$\frac{\mathrm{A}}{\sqrt{2}}$
Solution:
Potential Energy $=\frac{1}{2}$ (Total energy)
$
\frac{1}{2} m \omega^2 y^2=\frac{1}{2}\left(\frac{1}{2} m \omega^2 \mathrm{~A}^2\right) \Rightarrow y=\frac{\mathrm{A}}{\sqrt{2}} .
$

Answer:
(a) $\frac{2 \pi}{\sqrt{k}}$
Solution:
Here $\mathrm{k}$ is same as $\omega^2$
Question 19.
The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minutes will be $\frac{1}{x}$ time the original, where $\mathrm{x}$ is
(a) $2 \times 3$
(b) $2^3$
(c) $3^2$
(d) $3 \times 2^2$
Answer:
(b) $2^3$
Solution:
The amplitude decreases exponentially with time and becomes half in 1 minute.
Amplitude after 3 minutes $=\left(\frac{1}{2}\right)^3$ of the original value. Thus, $\mathrm{x}=2^3$
Question 20.
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is .......
(a) $\frac{a}{4}$
(b) $\frac{a}{3}$
(c) $\frac{a}{2}$
(d) $\frac{2 a}{3}$
Answer:
(c) $\frac{a}{2}$
Solution:
$
\frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left(\frac{1}{2} m \omega^2 a^2\right) \Rightarrow y=\frac{a}{2}
$

Question 21.
A massless spring, having force constant $\mathrm{k}$, oscillates with a frequency $\mathrm{n}$ when a mass $\mathrm{m}$ is suspended from it. The spring is cut into two equal halves and a mass $2 \mathrm{~m}$ is suspended from one of the parts. The frequency of oscillation will now be
For a simple pendulum the graph between length and time period will be
(a) $\mathrm{n}$
(b) $n \sqrt{2}$
(c) $\frac{n}{\sqrt{2}}$
(d) $2 \mathrm{n}$
Answer:
(a) $n$
Solution:
$
n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} ; \quad n^{\prime}=\frac{1}{2} \sqrt{\frac{2}{2}}=n
$
Question 22 .
For a simple pendulum the graph between the length and time period will be a
(a) hyperbola
(b) Parabola
(c) Straight line
(d) none of these
Answer:
(b) Parabola
Question 23.
$
x=5 \sin \left(4 t-\frac{\pi}{6}\right)
$
A particle is executing simple harmonic motion given by The velocity of the particle when its displacement is 3 units is .......
(a) $\frac{2 \pi}{3}$ units
(b) $\frac{5 \pi}{6}$ units
(c) 20 units
(d) 16 units

Answer:
(a) 16 units
Solution:
$
v=\omega \sqrt{\mathrm{A}^2-y^2}=4 \sqrt{25-9}=4 \times 4=16 \text { units }
$
Question 24.
When a particle oscillates simple harmonically, its potential energy varies periodically. If the frequency of oscillation of the particle is $n$, the frequency of potential energy variation is
(a) $\frac{n}{2}$
(b) $n$
(c) $2 n$
(d) $4 \mathrm{n}$
Answer:
(c) $2 \mathrm{n}$
Question 25.
A particle, moving along the $\mathrm{x}$-axis, executes simple harmonic motion when the force acting on it is given by ( $A$ and $k$ are positive constants.)
(a) $-\mathrm{Akx}$
(b) $\mathrm{A} \cos (\mathrm{kx})$
(c) $\mathrm{A} \exp (-\mathrm{kx})$
(d) $\mathrm{Akx}$
Answer:
(a) $-\mathrm{Akx}$
Question 26.
The motion of a particle is expressed by the equation $\mathrm{a}=-\mathrm{bx}$, where $\mathrm{x}$ is the displacement from the mean position, $\mathrm{a}$ is the acceleration and $\mathrm{b}$ is a constant. The periodic time is
(a) $\frac{2 \pi}{b}$
(b) $\frac{2 \pi}{\sqrt{b}}$
(c) $2 \pi \sqrt{b}$
(d) $2 \sqrt{\frac{\pi}{b}}$
Answer:
(b) $\frac{2 \pi}{\sqrt{b}}$

Solution:
Here $\omega^2=b$. Therefore $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{b}}$
Question 27.
The angular velocity and the amplitude of a simple pendulum are $\omega$ and a, respectively. The ratio of its kinetic and potential energies at a displacement $\mathrm{x}$ from the mean position is
(a) $\frac{x^2 \omega^2}{a^2-x^2 \omega^2}$
(b) $\frac{x^2}{a^2-x^2}$
(c) $\frac{a^2-x^2 \omega^2}{x^2 \omega^2}$
(d) $\frac{a^2-x^2}{x^2}$
Answer:
(d) $\frac{a^2-x^2}{x^2}$
Solution:
Kinetic energy $\mathrm{K}=\frac{1}{2} m \omega^2\left(a^2-x^2\right)$
Potential energy $\mathrm{U}=\frac{1}{2} m \omega^2 x^2 \Rightarrow \frac{\mathrm{K}}{\mathrm{U}}=\frac{a^2-x^2}{x^2}$
Question 28.
A particle is oscillating according to the equation $x=5 \cos (0.5 \pi t)$ where $t$ is in seconds. The particle moves from the position of equilibrium to the position of maximum displacement in time
(a) $1 \mathrm{~s}$
(b) $2 \mathrm{~s}$
(c) $0.5 \mathrm{~s}$
(d) $4 \mathrm{~s}$
Answer:
(a) $1 \mathrm{~s}$
Solution:
$
\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{0.5 \pi}=4 \mathrm{~s}
$
Time taken to move from the position of equilibrium to the position of maximum displacement is $t=\frac{T}{4}=$ Is
Question 29 .
A seconds pendulum is placed in a space laboratory orbiting around the Earth at a height $3 R$ from the Earth's surface where $\mathrm{R}$ is the radius of the Earth. The time period of the pendulum will be
(a) zero
(b) $2 / 3 \mathrm{~s}$
(c) $4 \mathrm{~s}$
(d) infinite
Answer:
(d) infinite

Solution:
In a space laboratory $g=0$. Therefore, $T=2 \pi \sqrt{\frac{l}{g}}=\infty$
Question 30 .
A mass $\mathrm{m}$ is vertically suspended from a spring of negligible mass; the system oscillates with a frequency $\mathrm{n}$. What will be the frequency of the system, if a mass $4 \mathrm{~m}$ is suspended from the same spring?
(a) $\frac{n}{2}$
(b) $2 n$
(c) $\frac{n}{4}$
(d) $4 n$
Answer:
(a) $\frac{n}{2}$
Solution:
$
n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \quad ; \quad n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k}{4 m}}=\frac{n}{2}
$
Question 31 .
Two simple pendulums of lengths $0.5 \mathrm{~m}$ and $2.0 \mathrm{~m}$ respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has completed oscillations.
(a) 5
(b) 3
(c) 1
(d) 2
Answer:
(d) 2
Solution:
The time period of the shorter pendulum is half that of the longer pendulum. Therefore, the pendulums will again be in phase ( at the mean position). When the shorter pendulum has completed 2 oscillations.
Question 32.
A body is executing simple harmonic motion with an angular frequency $2 \mathrm{rod} / \mathrm{sec}$. The velocity of the body at $20 \mathrm{~mm}$ displacement, when the amplitude of motion is $60 \mathrm{~mm}$, is
(a) $90 \mathrm{~mm} / \mathrm{s}$
(b) $113 \mathrm{~mm} / \mathrm{s}$
(c) $118 \mathrm{~mm} / \mathrm{s}$
(d) $131 \mathrm{~mm} / \mathrm{s}$
Answer:
(b) $113 \mathrm{~mm} / \mathrm{s}$
Solution:
$
v=\omega \sqrt{\mathrm{A}^2-y^2}=2 \times \sqrt{(60)^2-(20)^2}=30 \sqrt{2}=113 \mathrm{~mm} / \mathrm{s}
$

Question 33.
If the displacement of a particle executing SHM, is given by $y=0.30 \sin (220 t+0.64)$ in metre, then the frequency and the maximum velocity of the particle are ( $\mathrm{t}$ is in seconds)
(a) $35 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$
(b) $45 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$
(c) $58 \mathrm{~Hz}, 113 \mathrm{~m} / \mathrm{s}$
(d) $35 \mathrm{~Hz}, 132 \mathrm{~m} / \mathrm{s}$
Answer:
(a) $35 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$
Solution:
Frequency $f=\frac{\omega}{2 \pi}=\frac{220}{6.28}=35 \mathrm{~Hz}$
Question 34.
The kinetic energy of a particle, executing SHM, is $16 \mathrm{~J}$ when it is at its mean position. If the amplitude of oscillations is $25 \mathrm{~cm}$, and the mass of the particle is $5.12 \mathrm{~kg}$, the time period of its oscillation is
(a) $\pi / 5 \mathrm{~s}$
(b) $2 \pi \mathrm{s}$
(c) $20 \pi \mathrm{s}$
(d) $5 \pi \mathrm{s}$
Answer:
(a) $\pi / 5 \mathrm{~s}$
Solution:
$
\begin{aligned}
& \mathrm{E}=\frac{1}{2} m \omega^2 \mathrm{~A}^2 \text { or } \omega=\sqrt{\frac{2 \mathrm{E}}{m \mathrm{~A}^2}} \\
& \mathrm{~T}=\frac{2 \pi}{\omega}=\pi \mathrm{A} \sqrt{\frac{2 m}{\mathrm{E}}}=\pi \times 0.25 \times \sqrt{\frac{2 \times 5.12}{16}}=\frac{\pi}{4} \times \sqrt{0.64}=\frac{\pi}{5} \mathrm{~s}
\end{aligned}
$
Question 35 .
A particle of mass $m$ is executing oscillations about the origin on the $x$-axis. Its potential energy is $V(x)=$ $\mathrm{kx}^2$. Where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then its time period $\mathrm{T}$ is

(a) Proportional to $\frac{1}{\sqrt{a}}$
(b) independent of $a$
(c) proportional to $\sqrt{a}$
(d) proportional to $a^{3 / 2}$
Answer:
(b) independent of a
Solution:
Since $V(x)=\mathrm{Kx}^2$, the motion is simple harmonic. In SHM, the time period is independent of the amplitude of oscillation.
Question 36.
The amplitude of a damped oscillation reduces to one third of its original value $\mathrm{a}_0$ in 20 s. The amplitude of such oscillation after a period of 40 s will be
(a) $\mathrm{a}_0 / 9$
(b) $\mathrm{a}_0 / 6$
(c) $\mathrm{a}_0 / 2$
(d) $\mathrm{a}_0 / 27$
Answer:
(a) $\mathrm{a}_0 / 9$
Solution:
In the first 20 s, the amplitude reduces to one-third of the original value, i.e., to $\mathrm{a}_0 / 3$, In the next 20 s, it will reduce to one-third of the reduced value, i.e., to $\mathrm{a}_0 / 9$.
Question 37.
Masses $m_A$ and $m_B$ hanging from the ends of strings of lengths $1_A$ and $l_B$ are executing. Simple harmonic motions. If their frequencies are related as $\mathrm{f}_{\mathrm{A}}=2 \mathrm{f}_{\mathrm{B}}$, then
(a) $l_{\mathrm{A}}=2 l_{\mathrm{B}}$ and $m_{\mathrm{A}}=m_{\mathrm{B}} / 2$
(b) $l_{\mathrm{A}}=4 l_{\mathrm{B}}$ regardless of masses.
(c) $l_{\mathrm{A}}=l_{\mathrm{B}} / 4$ regardless of masses.
(d) $l_{\mathrm{A}}=2 l_{\mathrm{B}}$ and $m_{\mathrm{A}}=2 m_{\mathrm{B}}$
Answer:
(c) $1_A=1_B / 4$ regardless of masses.
Solution:

$f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}}$ Thus, the correct choice is $(c)$
Question 38 .
Two simple harmonic motions act on a particle. These harmonic motions are $\mathrm{x}=\mathrm{A} \cos (\omega t+\delta) ; \mathrm{y}=\mathrm{A} \cos$
$(\omega \mathrm{t}+\alpha)$ When $\delta=\alpha+\frac{\pi}{2}$ the resulting motion is
(a) A circle and the actual motion is clockwise
(b) an ellipse and the actual motion is counter clockwise
(c) a ellipse and the actual motion is clockwise
(d) a circle and the actual motion is counter clockwise
Answer:
(d) a circle and the actual motion is counter clockwise
$
\begin{aligned}
& \left.x=\mathrm{A} \cos \left(\omega t+\alpha+\frac{\pi}{2}\right)=-\mathrm{A} \sin \quad+\alpha\right) \\
& y=\mathrm{A} \cos (\omega t+\alpha)
\end{aligned}
$
Solution:
Squaring and adding the two equations, We get $x^2+y^2=A^2$. This is an equation of a circle. Hence the resultant motion is circular. The motion is counter clockwise.
Question 39.
A metal bob is suspended from a coiled spring. When set into vertical vibrations on the earth. It oscillates up and down with frequency $f$ If the same experiment is carried out in a satellite circling the Earth the frequency of vibration will be
(a) $\mathrm{f}$
(b) zero
(c) infinite
(d) depend on the distance of the satellite from the earth.
Answer:
(a) $\mathrm{f}$
Solution:
The frequency of oscillation of a mass spring system depends only on the mass and the spring constant.

Question 40 .
In forced oscillations of a particle, the amplitude is maximum for a frequency $\omega_1$ of the force, while the energy is maximum for a frequency $\omega_2$ of the force. Then ......
(a) $\omega_1<\omega_2$
(b) $\omega_1>\omega_2$ when damping is small and $\omega_1>\omega_2$ when damping is large.
(c) $\omega_1>\omega_2$
(d) $\omega_1=\omega_2$
Answer:
(d) $\omega_1=\omega_2$
Question 41.
Which one of the following statements is true for the speed $\mathrm{v}$ and the acceleration a of a particle executing simple harmonic motion?
(a) When $v$ is maximum, a is maximum
(b) Value of a is zero, whatever may be the value of $\mathrm{V}$
(c) When $v$ is zero, $a$ is zero
(d) When $v$ is maximum, a is zero
Answer:
(d) When $\mathrm{v}$ is maximum, a is zero
Solution:
At the mean position, $v$ is maximum and a is zero.
Question 42 .
The function $\sin ^2(\omega \mathrm{t})$ represents
(a) a simple harmonic motion with a period $\pi / \omega$
(b) a simple harmonic motion with a period $2 \pi / \omega$
(c) a periodic, but not simple harmonic motion with a period $\pi / \omega$
(d) a periodic, but not simple harmonic motion with a period $2 \pi / \omega$
Answer:
(a) a simple harmonic motion with a period $\pi / \omega$
Solution: $\sin ^2 \omega t=\frac{1}{2}(1-\cos 2 \omega t)$
This is an SHM with period $\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$. Its equilibrium position is at $\frac{1}{2}$ instead of zero.
Question 43.
A particle executing simple harmonic motion has a kinetic energy $\mathrm{K}_0 \cos ^2 \omega \mathrm{t}$. The maximum values of the potential energy and the total energy are, respectively ....
(a) $\mathrm{k}_0 / 2$ and $\mathrm{k}_0$
(b) $\mathrm{k}_0$ and $2 \mathrm{k}_0$
(c) $\mathrm{k}_0$ and $\mathrm{k}_0$
(d) 0 and $2 \mathrm{k}_0$
Answer:

(c) $\mathrm{k}_0$ and $\mathrm{k}_0$
Question 44.
A particle executing simple harmonic motion of amplitude $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is
(a) $3 \mathrm{~Hz}$
(b) $4 \mathrm{~Hz}$
(c) $2 \mathrm{~Hz}$
(d) $1 \mathrm{~Hz}$
Answer:
(d) $1 \mathrm{~Hz}$
Solution: $f_0=\omega \mathrm{A}=2 \pi v \mathrm{~A}$
or
$
f=\frac{v}{2 \pi \mathrm{A}}=\frac{31.4}{2 \times 3.14 \times 5}=1 \mathrm{~Hz}
$
Question 45 .
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is ........
(a) $0.5 \pi$
(b) $\pi$
(c) $0.707 \pi$
(d) zero
Answer:
(a) $0.5 \pi$
Question 46.
Which one of the following equations of motion represents simple harmonic motion?
(a) Acceleration $=-\mathrm{k}_0 \mathrm{x}+\mathrm{k}_1 \mathrm{x}^2$
(b) Acceleration $=-\mathrm{k}(\mathrm{x}+\mathrm{a})$
(c) Acceleration $=\mathrm{k}(\mathrm{x}+\mathrm{a})$
(d) Acceleration $=\mathrm{kx}$
Answer:
(b) Acceleration $=\mathrm{k}(\mathrm{x}+\mathrm{a})$

Question 47.
Which of the following functions represent SHM?
I. $y=\sin \omega t-\cos \omega t$
II. $y=\sin ^3 t$
III. $y=5 \cos \left(\frac{3 \pi}{4}-3 \omega t\right)$
IV. $y=1+\omega t+\omega^2 t^2$
(a) I and III
(b) I and II
(c) only I
(d) I, II and III
Answer:
(a) I and III
Question 48 .
Two simple harmonic motions of angular frequencies 100 and $1000 \mathrm{rad} / \mathrm{s}$ have the same displacement amplitude. The ratio of their maximum accelerations is
(a) $1: 10$
(b) $1: 10^2$
(c) $1: 10^3$
(d) $1: 10^4$
Answer:
(b) $1: 10$
Solution:
The magnitude of the maximum acceleration is given by
$
a_{\max }=\omega^2 \mathrm{~A} \Rightarrow \frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{\omega_1^2}{\omega_2^2}=\left(\frac{100}{1000}\right)^2=\frac{1}{10^2}
$
Question 49.
The period of oscillation of a simple pendulum is $\mathrm{T}$ in a stationary lift. If the lift moves upwards with an acceleration of $8 \mathrm{~g}$, the period will
(a) remain the same
(b) decrease by $T / 2$
(c) increase by $\mathrm{T} / 3$

(d) none of these
Answer:
(c) increase by $\mathrm{T} / 3$
Solution:
Thus, the new time period is $T / 3$. Hence the correct option is (d).
Question 50 .
A simple harmonic oscillator consist of a particle of mass $\mathrm{m}$ and an ideal spring with spring constant $\mathrm{k}$ The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be
(a) $\mathrm{T} / 2$
(b) $\frac{\mathrm{T}}{\sqrt{2}}$
(c) $\sqrt{2} \mathrm{~T}$
(d) $2 \mathrm{~T}$
Answer:
(b) $\frac{T}{\sqrt{2}}$
Solution:
$
\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}
$
If the spring is cut into two equal parts, the force constant of each part becomes $2 \mathrm{k}$. Therefore,
$
\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{m}{2 k}}=\frac{T}{\sqrt{2}}
$
Question 51.
A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is
(a) $3 \mathrm{~Hz}$
(b) $4 \mathrm{~Hz}$
(c) $2 \mathrm{~Hz}$
(d) $1 \mathrm{~Hz}$
Answer:
(d) $1 \mathrm{~Hz}$
Solution:

$
v_{\max }=\mathrm{A} \omega=2 \pi \mathrm{A} f \Rightarrow f=\frac{v_{\max }}{2 \pi \mathrm{A}}=\frac{31.4}{2 \times 3.14 \times 5}=1 \mathrm{~Hz}
$
2 Marks Questions
Question 1.
What is meant by Oscillatory motion?
Answer:
When an object or a particle moves back and forth repeatedly for some duration of time, its - motion is said to be oscillatory (or vibratory).
Question 2.
What is meant by simple harmonic motion (SHM)?
Answer:
A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.
Question 3.
What is meant by displacement in SHM?
Answer:
The distance travelled by the ribrating particle at any instant of time $t$ from its mean position is known as displacement.
$y=A \sin \omega \mathrm{t}$
The maximum displacement from the mean position is known as amplitude (A) of the vibrating particle.
Question 4 .
Define velocity in SHM.
Answer:
Velocity: The rate of change of displacement is velocity.
$
\begin{aligned}
& v=\frac{d y}{d t}=\frac{d}{d t}(\mathrm{~A} \sin \omega t) \\
& v=\frac{d y}{d t}=\mathrm{A} \omega \cos \omega t
\end{aligned}
$
Question 5.
Define acceleration in SHM.
Answer:

Acceleration: The rate of change of velocity is acceleration.
$
\begin{aligned}
a & =\frac{d v}{d t}=\frac{d}{d t}(\mathrm{~A} \omega \cos \omega t) \\
a & =-\omega^2 \mathrm{~A} \sin \omega t=-\omega^2 y \\
\therefore \quad a & =\frac{d^2 y}{d t^2}=-\omega^2 y
\end{aligned}
$
Question 6.
What is meant by phase in SHM?
Answer:
Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position.
Question 7.
What is meant by angular oscillation?
Answer:
When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation.
Question 8.
Define Amplitude.
Answer:
The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude.
3 Marks Questions
Question 1.
Derive the expression for resultant spring constant when two springs having constant $k_1$ and $k_2$ are connected in series.

Answer:
Let $x_1$ and $x_2$ be the elongation of springs from their equilibrium position (un-stretched position) due to applied force $F$. Then, the net displacement of the mass point is

$
x=x_1+x_2
$
From Hooke's law, the net force
$
\mathrm{F}=-k_s\left(x_1+x_2\right) \Rightarrow x_1+x_2=-\frac{\mathrm{F}}{k_s}
$

For springs in series connection
$
\begin{aligned}
\Rightarrow \quad-k_1 x_1 & =-k_2 x_2=\mathrm{F} \\
x_1 & =-\frac{\mathrm{F}}{k_1} \text { and } x_2=-\frac{\mathrm{F}}{k_2}
\end{aligned}
$
Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as
$
\begin{aligned}
-\frac{\mathrm{F}}{k_1}-\frac{\mathrm{F}}{k_2} & =-\frac{\mathrm{F}}{k_s} \\
\frac{1}{k_s} & =\frac{1}{k_1}+\frac{1}{k_2} \text { (or) } k_s=\frac{k_1 k_2}{k_1+k_2} \mathrm{Nm}^{-1}
\end{aligned}
$
Suppose we have $\mathrm{n}$ springs connected in series, the effective spring constant in series is
$
\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots+\frac{1}{k_n}=\sum_{i=1}^n \frac{1}{k_i}
$
If all spring constants are identical i.e., $\mathrm{k}_1=\mathrm{k}_2=\ldots=\mathrm{k}_{\mathrm{n}}=\mathrm{k}$ then
$
\frac{1}{k_s}=\frac{n}{k} \Rightarrow k_s=\frac{k}{n}
$
This means that the effective spring constant reduces by the factor $n$. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,
$
k_1 x_1=k_2 x_2
$
Then the ratio of compressed distance or elongated distance $\mathrm{x}_1$ and $\mathrm{x}_2$ is
$
\frac{x_2}{x_1}=\frac{k_1}{k_2}
$The elasticity potential energy stored in first and second springs are
$
\mathrm{V}_1=\frac{1}{2} k_1 x_1^2 \text { and } \mathrm{V}_2=\frac{1}{2} k_2 x_2^2
$
respectively. Then their ratio is
$
\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{1}{2} k_1 x_1^2}{\frac{1}{2} k_2 x_2^2}=\frac{k_1}{k_2}\left(\frac{x_1}{x_2}\right)^2=\frac{k_2}{k_1}
$
Question 2.
Derive the expression for resultant spring constant when two springs having constant $\mathrm{k}_1$ and $\mathrm{k}_2$ are connected in parallel.
Answer:
$\mathrm{k}_1$ and $\mathrm{k}_2$ attached to a mass $\mathrm{m}$ as shown in figure. The results can be generalized to any number of springs in parallel.

Let the force $\mathrm{F}$ be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass $\mathrm{m}$ is $\mathrm{F}=-\mathrm{k}_{\mathrm{p}} \mathrm{X}$
where $\mathrm{k}_{\mathrm{p}}$ is called effective spring constant.
Let the first spring be elongated by a displacement $x$ due to force $F_1$ and second spring be elongated by the same displacement $x$ due to force $F_2$, then the net force
$\mathrm{F}=-\mathrm{k}_1 \mathrm{x}-\mathrm{k}_2 \mathrm{x} \ldots .(2)$
Equating equations (2) and (1), we get $\mathrm{k}_{\mathrm{p}}=\mathrm{k}_1+\mathrm{k}_2 \ldots .(2)$
Generalizing, for $\mathrm{n}$ springs connected in parallel,
$
k_{\mathrm{P}}=\sum_{i=1}^n k_i
$
If all spring constants are identical i.e., $\mathrm{k}_1=\mathrm{kk}_2=\ldots=\mathrm{k}_{\mathrm{n}}=\mathrm{k}$ then
$
k_{\mathrm{P}}=n k
$
This implies that the effective spring constant increases by a factor $n$. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.