Exemplar MCQs

Hint

Given the equation of displacement of the particle
\(
\begin{aligned}
& y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right) \\
& y=3 \cos \left[-\left(2 \omega t-\frac{\pi}{4}\right)\right]
\end{aligned}
\)
We know \(\cos (-\theta)=\cos \theta\)
Hence \(y=3 \cos \left(2 \omega t-\frac{\pi}{4}\right) \dots(i)\)
Comparing with \(y=a \cos \left(\omega t+\phi_0\right)\)
Hence (i) represents simple harmonic motion with angular frequency \(2 \omega\).
Hence its time period, \(T=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\).

Hint

Key concept: There are certain motions that are repeated at equal intervals of time. Let the the interval of time in which motion is repeated. Then \(x(t)=x(t+T)\), where \(T\) is the minimum change in time. The function that repeats itself is known as a periodic function.
Given the equation of displacement of the particle, \(y=\sin ^3 \omega t\)
We know \(\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta\)
\(
\begin{array}{ll}
\text { Hence } & y=\frac{(3 \sin \omega t-4 \sin 3 \omega t)}{4} \\
\Rightarrow & 4 \frac{d y}{d t}=3 \omega \cos \omega t-4 \times[3 \omega \cos 3 \omega t] \\
\Rightarrow & 4 \times \frac{d^2 y}{d t^2}=-3 \omega^2 \sin \omega t+12 \omega \sin 3 \omega t \\
\Rightarrow & \frac{d^2 y}{d t^2}=-\frac{3 \omega^2 \sin \omega t+12 \omega^2 \sin 3 \omega t}{4} \\
\Rightarrow & \frac{d^2 y}{d t^2} \text { is not proportional to } y .
\end{array}
\)
Hence, motion is not SHM.
As the expression is involving sine function, hence it will be periodic.
Also \(\sin ^3 \omega t=(\sin \omega t)^3\)
\(
=[\sin (\omega t+2 \pi)]^3=[\sin \omega(t+2 \pi / \omega)]^3
\)
Hence, \(y=\sin ^3 \omega t\) represents a periodic motion with period \(2 \pi / \omega\).

Hint

Key Concept: In case of simple harmonic motion, the acceleration is always directed towards the mean position and so is always opposite to displacement, i.e. a \(a=-x\) or \(a=-\omega^2 x\). In option (d) \(a_x=-2 x\) or a \(a \propto x\), the acceleration of the particle is proportional to negative of displacement. Hence it represents S.H.M.

Hint

Key Concept: If the liquid in U-tube is filled to a height \(h\) and cross-section of the tube is uniform and the liquid is incompressible and non-viscous. Initially, the level of liquid in the two limbs will be at the same height equal to \(h\). If the liquid is pressed by \(y\) in one limb, it will rise by \(y\) along the length of the tube in the other limb, so the restoring force will be developed by the hydrostatic pressure difference
Restoring force \(F=\) Weight of liquid column of height \(2 y\)
\(
\begin{aligned}
& \Rightarrow \quad F=-(A \times 2 y \times \rho) \times g=-2 A \rho g y \\
& \Rightarrow \quad F \propto-y
\end{aligned}
\)
Motion is SHM with force constant
\(
\begin{aligned}
& k=2 A \rho g \\
& \text { Timeperiod } T=2 \pi \sqrt{\frac{m}{k}} \\
&=2 \pi \sqrt{\frac{A \times 2 h \times \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{h}{g}}
\end{aligned}
\)
which is independent of the density of the liquid.

Hint

\(x=a \cos \omega t\) and \(y=a \sin \omega t\)
\(
\therefore \quad x^2+y^2=a^2\left(\cos ^2 \omega t+\sin ^2 \omega t\right)=a^2
\)
It is an equation of a circle. Thus trajectory of motion of the particle will be a circle.

Hint

Key concept: The sum of two S.H.Ms of same frequencies is a S.H.M.
According to the question, the displacement
\(
y=a \sin \omega t+b \cos \omega t
\)
Given \(\quad x=a \sin \omega t+b \cos \omega t \dots(i)\)
Let \(\quad a=A \cos \phi \dots(ii)\)
and \(\quad b=A \sin \phi \dots(iii)\)
Squaring and adding (ii) and (iii), we get
\(
\begin{aligned}
& a^2+b^2=A^2 \cos ^2 \phi+A^2 \sin ^2 \phi=A^2 \\
&=A^2 \Rightarrow A=\sqrt{a^2+b^2} \\
& y=A \sin \phi \cdot \sin \omega t+A \cos \phi \cdot \cos \omega t \\
&= \sin (\omega t+\phi) \\
& \frac{d y}{d t}=A \omega \cos (\omega t+\phi) \\
& \frac{d^2 y}{d t^2}=-A \omega^2 \sin (\omega t+\phi) \\
&=-A y \omega^2=\left(-A \omega^2\right) y \\
& \Rightarrow \quad \frac{d^2 y}{d t^2} \propto(-y)
\end{aligned}
\)
Hence, it is an equation of SHM with amplitude
\(
A=\sqrt{a^2+b^2}
\)

Hint

Here A is given a transverse displacement. Through the elastic support, the disturbance is transferred to all the pendulums.
\(\mathrm{A}\) and \(\mathrm{C}\) are having the same length, hence they will be in resonance, because of their time period of oscillation. Since the length of pendulums \(\mathrm{A}\) and \(\mathrm{C}\) is the same and \(T=2 \pi \sqrt{L} / \mathrm{g}\), hence their time period is same and they will have the frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.

Hint

Let the angular velocity of the particle executing circular motion is \(\omega\) and when it is at \(P\) makes an angle \(\theta\) as shown in the diagram.
As \(\sin \theta=\frac{x}{O P}=\frac{x}{B}\),
Clearly, \(\omega=\omega t\)
\(
\begin{aligned}
& x=B \sin \theta=B(\sin \omega t)=B \sin \left(\frac{\pi t}{15}\right) \\
& x=B \sin \left(\frac{2 \pi}{30} t\right)
\end{aligned}
\)

Hint

The equation of motion of a particle is
\(
x=\mathrm{a} \cos (\alpha \mathrm{t})^2
\)
is a cosine function and \(x\) varies between \(-a\) and \(+a\), the motion is oscillatory. Now checking for periodic motion, putting \(\mathrm{t}+\mathrm{T}\) in place of \(t\). \(T\) is supposed as period of the function \(\omega(t)\).
\(
\begin{aligned}
x(t+T) & =a \cos [\alpha(t+T)]^2 \\
& =a \cos \left[\alpha^2 t^2+\alpha^2 T^2+2 \alpha^2 t T\right] \neq x(t)
\end{aligned}
\)
Hence, it is not periodic.

Hint

Key concept: Let equation of an SHM is represented by \(y=a \sin \omega t\)
\(
v=\frac{d y}{d t}=a \omega \cos \omega t
\)
Hence \((v)_{\max }=a \omega\)
Acceleration \((A)=\frac{d x^2}{d t^2}=-a \omega^2 \sin \omega t\)
Hence \(A_{\max }=\omega^2 a\)
Maximum speed, \(v_{\max }=\omega A \dots(i)\)
Maximum acceleration, \(a_{\max }=\omega^2 \mathrm{~A} \dots(ii)\)
Divide eqn. (ii) by eqn. (i), we get
\(
\begin{aligned}
& \frac{a_{\max }}{v_{\max }}=\frac{\omega^2 A}{\omega A}=\omega \quad \therefore \frac{a_{\max }}{v_{\max }}=\frac{2 \pi}{T} \\
& T=2 \pi\left(\frac{v_{\max }}{a_{\max }}\right)
\end{aligned}
\)
Here, \(v_{\max }=30 \mathrm{cms}^{-1}, a_{\max }=60 \mathrm{cms}^2\)
\(
\therefore T=2 \pi\left(\frac{30 \mathrm{cms}^{-1}}{60 \mathrm{cms}^{-2}}\right)=\pi \mathrm{s}
\)

Hint

When the mass is connected to the two springs individually,
\(
\begin{aligned}
& v_1=\frac{1}{2 \pi} \sqrt{\frac{k_1}{m}} \dots(i)\\
& v_2=\frac{1}{2 \pi} \sqrt{\frac{k_2}{m}} \dots(ii)
\end{aligned}
\)
Now the block is connected with two springs considered as parallel.
Here equivalent spring constant, \(k_{\mathrm{eq}}=k_1+k_2\)
Time period of oscillation of the spring block-system is
\(
T=2 \pi \sqrt{\frac{m}{k_{\mathrm{eq}}}}=2 \pi \sqrt{\frac{m}{k_1+k_2}}
\)
Hence frequency,
\(
\begin{aligned}
& v=\frac{1}{T}=\frac{1}{2 \pi} \times \sqrt{\frac{k_1+k_2}{m}} \dots(iii)\\
& v=\frac{1}{2 \pi}\left[\frac{k_1}{m}+\frac{k_2}{m}\right]^{1 / 2}
\end{aligned}
\)
From Eq. (i) \(\frac{k_1}{m}=4 \pi^2 v_1^2\) and from Eq. (ii), \(\frac{k_2}{m}=4 \pi^2 v_2^2\)
\(
\begin{array}{ll}
\Rightarrow v=\frac{1}{2 \pi}\left[\frac{4 \pi^2 v_1^2}{1}+\frac{4 \pi^2 v_2^2}{1}\right]^{1 / 2}=\frac{2 \pi}{2 \pi}\left[v_1^2+v_2^2\right]^{1 / 2} \\
\Rightarrow \quad v=\sqrt{v_1^2+v_2^2}
\end{array}
\)

Hint

\(
(a, c) \text { Rotation of earth about its axis repeats its motion after a fixed interval of lime, so its motion is periodic.}
\)

The rotation of the earth is obviously not a to-and-fro type of motion about a fixed point, hence its motion is not an oscillation. Also, this motion does not follow S.H.M equation, a \(\propto-x\).
Hence, this motion is not a S.H.M.

Hint

\((a, c)\) For small angular displacement, the situation is shown in the figure. Only one restoring force creates.
motion in ball inside bowl.
Only one restoring force creates motion in ball inside bowl.
\(
F=-m g \sin \theta
\)
As \(\theta\) is small, \(\sin \theta=\theta\)
So, \(\quad m a=-m g \frac{x}{R}\)
or, \(\quad a=-\left(\frac{g}{R}\right) x \Rightarrow a \propto-x\)
\(
\text { So, the motion of the ball is S.H.M and periodic. }
\)

Hint

Simple harmonic motion: Simple harmonic motion, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.
In option (a): From the figure the state at \(t=0\) and \(t=2\) are not the same as the displacement are not equal. So, the statement is incorrect.

In option (b): From the figure the state at \(\mathrm{t}=2 \mathrm{~s}\) and \(\mathrm{t}=6 \mathrm{~s}\) are in the same state as the displacement and direction are the same. So, the statement is correct.
In option (c): From the figure though the displacement is same at \(t=1 \mathrm{~s}\) and \(\mathrm{t}=7 \mathrm{~s}\) but the direction is opposite. So, the statement is incorrect.

In option (d): From the figure the state at \(\mathrm{t}=1 \mathrm{~s}\) and \(\mathrm{t}=5 \mathrm{~s}\) are in the same state as the displacement and direction are the same. So, the statement is correct.
Final Answer: Option (b), (d)

Hint

Key concept: Simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. When the system is displaced from its equilibrium position, a restoring force that obeys Hooke’s law tends to restore the system to equilibrium. As a result, it accelerates and starts going back to the equilibrium position. An oscillation follows simple harmonic motion if it fulfills the following two rules:
1. Acceleration is always in the opposite direction to the displacement from the equilibrium position.

2. Acceleration is proportional to the displacement from the equilibrium position.
Let us write the equation for the \(\operatorname{SHM} x=a \sin (\omega t+\phi)\) Clearly, it is a periodic motion as it involves sine function.
Let us find the velocity of the particle, \(v=\frac{d x}{d t}=\frac{d}{d t}(a \sin (\omega t+\phi))=a \omega \cos (\omega t+\phi)\) Velocity is also periodic because it is a cosine function.
Now let us find acceleration, \(A=\frac{d v}{d t}=\frac{d^2 x}{d t^2}=-a \omega^2 \sin (\omega t+\phi)\)
The acceleration is a sine function, hence cannot be constant.
\(
\begin{aligned}
& \Rightarrow \quad A=-\left(\omega^2 a\right) \sin (\omega t+\phi)=-\omega^2 x \\
& \text { Force, } F=\text { mass } \times \text { acceleration } \\
& =m A=-m \omega^2 x
\end{aligned}
\)
Hence, the force acting is directly proportional to the displacement from the mean position and opposite to it.

Hint

\((\mathbf{a}, \mathbf{b}, \mathbf{c})\) Displacement of the particle at any time \(t\),
\(
y=A_0 \cos \omega t=A_0 \cos \frac{2 \pi}{T} t
\)
And acceleration \(A=-\omega^2 y=\omega^2 A_0 \cos \frac{2 \pi}{T} t\).
In option (a) Force \(F=m \omega^2 y\).
At
\(
\begin{aligned}
t=\frac{3 T}{4}, y & =A_0 \cos \left(\frac{2 \pi}{T} \times \frac{3 T}{4}\right) \\
& =A_0 \cos \left(\frac{3 \pi}{2}\right)=0
\end{aligned}
\)
Hence \(F=0\), which makes option correct.
In option (b) acceleration \(A=-\omega^2 y=\omega^2 A_0 \cos \frac{2 \pi}{T} t\). At \(t=\frac{4 T}{4}, y=A_0 \cos \left(\frac{2 \pi}{T} \times T\right)=A_0 \cos 2 \pi=+A_0\) \(\therefore a=\omega^2 A_0\), which is maximum value of \(a\). Hence option (b) is correct.
In option (c) The velocity \(v=\omega \sqrt{A_0^2-y^2}\)
At \(t=T / 4\)
\(
y=A_0 \cos \left(\frac{2 \pi}{T} \times \frac{T}{4}\right)=A_0 \cos \left(\frac{\pi}{2}\right) \Rightarrow y=0
\)
Hence, \(v=\omega A_0\), which is maximum value of velocity. Hence option (c) is correct.
In option (d) At \(t=T / 2\),
\(
y=A_0 \cos \left(\frac{2 \pi}{T} \times \frac{T}{2}\right)=-A_0
\)
Displacement is maximum, i.e corresponds to extreme position, it means \(P E\) is maximum and KE is zero.

Hint

(a, b, d) In case of S.H.M, average total energy per cycle
\(
\text { = Maximum kinetic energy }\left(K_0\right)
\)
\(
\text { = Maximum potential energy }\left(U_0\right)
\)
Average KE per cycle \(=\frac{0+K_0}{2}=\frac{K_0}{2}\)
Let us write the equation for the SHM \(x=a \sin \omega t\).
Clearly, it is a periodic motion as it involves sine function.
Let us find velocity of the particle, \(v=\frac{d x}{d t}=\frac{d}{d t}(a \sin \omega t)=a \omega \cos \omega t\)
Mean velocity over a complete cycle,
\(
v_{\text {mean }}=\frac{\int_0^{2 \pi} \omega a \cos \theta d \theta}{2 \pi}=\frac{\omega a[\sin \theta]_0^{2 \pi}}{2 \pi}=0
\)
So, \(\quad v_{\text {mean }} \neq \frac{2}{\pi} v_{\max }\)
Root mean square speed,
\(
\begin{aligned}
& v_{\mathrm{mms}}=\sqrt{\frac{v_{\min }^2+v_{\max }^2}{2}}=\sqrt{\frac{0+v_{\max }^2}{2}} \\
& v_{\mathrm{rms}}=\frac{1}{\sqrt{2}} v_{\max }
\end{aligned}
\)

Hint

In option (a): When the particle is \(3 \mathrm{~cm}\) away from A going towards B. So, velocity is towards AB, i.e. positive. In SHM, acceleration is always towards the mean position \((O)\) it means both force and acceleration act towards \(\mathrm{O}\), have positive sign.
Hence option (a) is correct.
In option (b): When the particle is at C, velocity is towards B hence positive. Hence option (b) is not correct. In option (c): When the particle is \(4 \mathrm{~cm}\) away from \(B\) going towards \(A\) velocity is negative and acceleration is towards mean position \((O)\), hence negative. Hence option (c) is correct.

In option (d): Acceleration is always towards mean position (O). When the particle is at \(B\), acceleration and force are towards BA which is negative. Hence option (d) is correct.

Hint

When mass is displaced from equilibrium position by a distance \(x\) towards right, the right spring gets compressed by \(x\) developing a restoring force \(k x\) towards left on the block. The left spring is stretched by an amount \(x\) developing a restoring force \(k x\) left on the block.
developing a restoring force \(k x\) towards left on the block.
\(
\begin{aligned}
& F_1=-k x \text { (for left spring) and } \\
& F_2=-k x \text { (for right spring) }
\end{aligned}
\)
Restoring force, \(F=F_1+F_2=-2 k x\)
\(\therefore \quad F=2 k x\) towards left.

Hint

Let us write the displacement equation for the \(\operatorname{SHM} x=a \sin (\omega t+\phi)\).
We have to find velocity by differentiating the equation representing displacement and acceleration by differentiating the equation relating velocity and time.
The velocity of the particle, \(v=\frac{d x}{d t}=\frac{d}{d t}(a \sin (\omega t+\phi))=a \omega \cos (\omega t+\phi)\)
Maximum velocity \(|v|_{\max }=a \omega\)
Now let us find acceleration, \(A=\frac{d v}{d t}=\frac{d^2 x}{d t^2}=-a \omega^2 \sin (\omega t+\phi)\)
Maximum acceleration \(|A|_{\max }=\omega^2 a\)
\(
\frac{|v|_{\max }}{|A|_{\max }}=\frac{\omega a}{\omega^2 a}=\frac{1}{\omega} \Rightarrow \frac{a_{\max }}{v_{\max }}=\omega
\)

Hint

In the diagram shown a particle is executing SHM between P and Q. The particle starts from mean position ‘ \(O\) ‘ moves to amplitude position ‘ \(P\) ‘, then particle turn back and moves from ‘ \(P\) ‘ to \( Q\) Finally the particle turns back again and return to mean position ‘ \(O\) ‘. In this way, the particle completes one oscillation in one time period. Total distance travelled while it goes from \(\mathrm{O} \rightarrow \mathrm{P} \rightarrow \mathrm{O} \rightarrow Q \rightarrow \mathrm{O}\) \(=\mathrm{OP}+\mathrm{PO}+\mathrm{OQ}+\mathrm{QO}=\mathrm{A}+\mathrm{A}+\mathrm{A}+\mathrm{A}=4 \mathrm{~A}\)
Amplitude \(=O P=A\)
Hence, ratio of distance and amplitude \(=4 \mathrm{~A} / \mathrm{A}=4\)

Hint

Key concept: A second’s pendulum means a simple pendulum having time period \(T=2 \mathrm{~s}\).
The time period of a simple pendulum, \(T=2 \pi \sqrt{\frac{l}{g}}\)
where, \(\quad l=\) length of the pendulum and
\(g=\) acceleration due to gravity on surface of the earth.
Thus, \(\quad \frac{l_{\text {Moon }}}{g_{\text {Moon }}}=\frac{l_{\text {Earth }}}{g_{\text {Earth }}}\)
or
\(
l_{\text {Moon }}=\left(l_{\text {Earth }}\right) \frac{g_{\text {Moon }}}{g_{\text {Earth }}}=(1 \mathrm{~m})\left(\frac{1}{6}\right)=\frac{1}{6} \mathrm{~m}
\)

Hint

If mass \(m\) moves down by \(h\), then the spring extends by \(2 h\) (because each side expands by \(h\) ). The tension along the string and spring is the same.
In equilibrium
\(
m g=2(k. 2 h)
\)
where \(k\) is the spring constant.
On pulling the mass down by \(x\),
\(
\begin{aligned}
F & =mg-2 k(2 h+2 x) \\
& =-4 k x
\end{aligned}
\)
So. \(T=2 \pi \sqrt{\frac{m}{4 k}}\)

Hint

Let 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,
\(
P E=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2 \dots(i)
\)
Maximum energy of the oscillator = Maximum potential energy \(=\) Total energy
\(
T E=\frac{1}{2} m \omega^2 A^2 \dots(ii)
\)
where, \(A=\) amplitude of motion
We are given, \(P E=\frac{1}{2} T E\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} m \omega^2 x^2=\frac{1}{2}\left[\frac{1}{2} m \omega^2 A^2\right] \\
& \Rightarrow \quad \cdot \quad x^2=\frac{A^2}{2} \text { or } x=\sqrt{\frac{A^2}{2}}= \pm \frac{A}{\sqrt{2}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& U=U_0(1-\cos \alpha x) \\
& F=\frac{-d U}{d x}=\frac{-d}{d x}\left(U_o-U_o \cos a x\right) \\
& =-U_0 \alpha \sin \alpha \mathrm{x}-U_o \alpha \alpha x \text { (for small } \alpha x, \sin \alpha x =\alpha x \text { ) } \\
=-U_0 \alpha^2 x
\end{aligned}
\)
We know that \(F=-k x\)
So, \(k=U_o \alpha^2\)
\(
T=2 \pi \sqrt{\frac{m}{U_o \alpha^2}}
\)

Hint

Given : \(m=2 \mathrm{~kg}, \mathrm{k}=50 \mathrm{Nm}^{-1}, A=5 \mathrm{~cm}\)
\(
\mathrm{w}=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5 \mathrm{~s}^{-1}
\)
At \(t=0, x=0\)
\(\therefore\) Displacement at any time \(t\)
\(x=A \sin w t\) or \(x=5 \sin 5 t\)

Hint

\(
\begin{aligned}
& \theta_1=\theta_o \sin \left(\omega t+\delta_1\right) \\
& \theta_2=\theta_o \sin \left(\omega t+\delta_2\right)
\end{aligned}
\)
For the first, \(\theta=2^{\circ}, \therefore \sin \left(\omega t+\delta_1\right)=1\)
For the 2nd, \(\theta=-1^{\circ}, \therefore \sin \left(\omega t+\delta_2\right)=-1 / 2\)
\(
\begin{aligned}
& \therefore \omega t+\delta_1=90^{\circ}, \omega t+\delta_2=-30^{\circ} \\
& \therefore \delta_1-\delta_2=120^{\circ}
\end{aligned}
\)

Hint

\(
\text { Amplitude }=\frac{\text { distance between } 2 \text { extreme positions }}{2}=\frac{4 \mathrm{~cm}}{2}=2 \mathrm{~cm}
\)

(b) The frequency of oscillation of mass \(m\) suspended from a spring is \(f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}}\) where \(l\) is the extension in spring due to mass \(m\) in equilibrium position.
Неге, \(l=2 \mathrm{~cm}, g=980 \mathrm{~cm} / \mathrm{s}^2 \therefore f=\frac{1}{2 \pi} \sqrt{\frac{980}{2}}=\frac{7 \times 7}{2 \times 22} \sqrt{10}=3.52 \mathrm{~s}\)

Hint

A body is said to be in simple harmonic motion only when,
\(
F=-k x \dots(1)
\)
where \(F\) is force,
\(k\) is force constant, and
\(x\) is displacement of the body from the mean position.
Given:
\(
F=-k \sqrt{x} \dots(2)
\)
On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, \(k\) should be proportional to \(\sqrt{x}\).
Thus, as \(x\) increases \(k\) increases.

Hint

\(
\text { One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started. }
\)

Hint

\(
\text { Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation. }
\)

Hint

Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.

Hint

In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by the particle is,
\(
2 A+2 A=4 A \text {. }
\)

Hint

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
\(
A \omega^2+\left(-A \omega^2\right)=0
\)

Hint

simple harmonic with amplitude \(\sqrt{A^2+B^2}\)
\(x=A \sin \omega t+B \cos \omega t \dots(1)\)
Acceleration,
\(
a=\frac{\mathrm{d}^2 x}{\mathrm{dt}^2}=\frac{\mathrm{d}^2}{\mathrm{dt}^2}(\mathrm{~A} \sin \omega t+\mathrm{B} \cos \omega t)
\)
\(
\begin{aligned}
& =\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{A} \omega \cos \omega t-\mathrm{B} \omega \sin \omega t) \\
& =-\mathrm{A} \omega^2 \sin \omega t-\mathrm{B} \omega^2 \cos \omega t \\
& =-\omega^2(\mathrm{~A} \sin \omega t+\mathrm{B} \cos \omega t) \\
& =-\omega^2 x
\end{aligned}
\)
For a body to undergo simple harmonic motion,
acceleration, \(a=-k x \dots(2)\)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude, A
\(
=\sqrt{A^2+B^2}
\)

Hint

The magnitude of the displacement
\(
r=\sqrt{ }\left(A^2 \cos ^2 \omega t+A^2 \sin ^2 \omega t\right)=\sqrt{A^2}=A
\)
The direction of the displacement from \(X\)-axis is
\(\tan \alpha=A \cdot \sin \omega t / A \cdot \cos \omega t=\tan \omega t\)
\(\rightarrow \alpha=\omega t\)
It means the position vector of the particle can take any angle but its magnitude is constant.
So the motion is on a circle with center at the origin.

Hint

At \(t=0\),
Displacement \(\left(x_0\right)\) is given by, \(x(0)=\mathrm{A}+B\sin \omega(0)=\mathrm{A}\)
Displacement \(x\) will be maximum when \(\sin \omega(t)=1\) or,
\(
x(m)=\mathrm{A}+\mathrm{B}
\)
Amplitude will be:
\(
x(m)-x(0)=A+B-A=B
\)

Hint

\(
\text { Because the direction of motion of particles } A \text { and } B \text { is just opposite to each other. }
\)

Hint

Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by, \(E=\frac{1}{2} m \omega^2 A^2\) where \(m\) is mass of body, and \(\omega\) is angular frequency.
Let \(m_1\) be the mass of the other particle and \(\omega_1\) be its angular frequency.
New angular frequency \(\omega_1\) is given by, \(\omega_1=\sqrt{\frac{k}{m_1}}=\sqrt{\frac{k}{2 m}}\left(m_1=2 m\right)\)
New energy \(E_1\) is given as,
\(
\begin{aligned}
& E_1=\frac{1}{2} m_1 \omega_1^2 A^2 \\
& =\frac{1}{2}(2 m)\left(\sqrt{\frac{k}{2 m}}\right)^2 A^2 \\
& =\frac{1}{2} m \omega^2 A^2=E
\end{aligned}
\)

Hint

Average Energy means the total energy of the Simple Harmonic Motion in One Time Period.

Energy is the Scalar Quantity and not the vector quantity, thus it only depends upon the magnitude and not on the direction.

Therefore, It cannot be zero as in case of the velocity, displacement, etc. It will must be having some numerical terms.
Total energy of the Simple Harmonic Motion = Kinetic Energy + Potential energy.
\(
\begin{aligned}
& =\frac{1}{2} \mathrm{k}\left(\mathrm{A}^2-\mathrm{x}^2\right)+\frac{1}{2} \mathrm{kx}^2 \\
& =\frac{1}{2} \mathrm{kA}^2-\frac{1}{2} \mathrm{kx}^2+\frac{1}{2} \mathrm{kx}^2 \\
& =\frac{1}{2} \mathrm{kA}^2=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y=a \sin w t \\
& d y / d t=a w \cos w t \\
& K . E=1 / 2 \times m \times V^2 \\
& =m a^2 w^2(\cos w t)^2 / 2 \\
& =m a^2 w^2(1+\cos 2 w t) / 4 \\
& =m a^2 w^2 / 4+m a^2 w^2 \cos 2 w t / 4
\end{aligned}
\)
During one complete oscillation, the kinetic energy of the particle executing SHM will be twice. KE will oscillate with frequency double the frequency of SHM, Which is \(2 \mathrm{f}\)

Hint

Time period \((T)\) is given by,
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
where \(m\) is the mass, and
\(k\) is spring constant.
When the spring is divided into two parts, the new spring constant \(k_1\) is given as,
\(
k_1=2 k
\)
New time period \(T_1\) :
\(
T_1=2 \pi \sqrt{\frac{m}{2 k}}=\frac{1}{\sqrt{2}} 2 \pi \sqrt{\frac{m}{k}}=\frac{1}{\sqrt{2}} T
\)

Hint

Maximum velocity, \(v=A \omega\)
where \(A\) is amplitude and \(\omega\) is the angular frequency.
Further, \(\omega=\sqrt{\frac{k}{m}}\)
Let \(A\) and \(B\) be the amplitudes of particles \(A\) and \(B\) respectively. As the maximum velocity of particles are equal,
i.e. \(v_A=v_B\)
or,
\(
\begin{aligned}
& A \omega_A=B \omega_B \\
& \Rightarrow A \sqrt{\frac{k_1}{m_A}}=B \sqrt{\frac{k_2}{m_B}} \\
& \Rightarrow A \sqrt{\frac{k_1}{m}}=B \sqrt{\frac{k_2}{m}}\left(m_A=m_B=m\right) \\
& \Rightarrow \frac{A}{B}=\sqrt{\frac{k_2}{k_1}}
\end{aligned}
\)

Hint

\(
\text { Because the frequency }\left(f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\right) \text { of the system is independent of the acceleration of the system. }
\)

Hint

In general, a spring mass system is a system where an object with mass is hung on a spring, which is then displaced. After displacement, the spring mass system has an oscillatory motion. As in all oscillatory motion, it oscillates with a particular frequency. The frequency for a spring mass system is given by
\(f=\sqrt{\frac{k}{m}}\) where \(k\) is the force constant of the spring, and \(m\) is the mass of the object hung on the spring.
Now, since neither of these quantities is dependent on the acceleration of the system, if the acceleration is increased, neither the spring constant nor the mass will change. Hence the frequency remains constant.

Hint

The acceleration due to gravity at moon is \(g / 6\).
Time period of pendulum is given by,
\(
T=2 \pi \sqrt{\frac{l}{g}}
\)
Therefore, on moon, time period will be:
\(
T_{\text {moon }}=2 \pi \sqrt{\frac{l}{g_{\text {moon }}}}=2 \pi \sqrt{\frac{l}{\left(\frac{g}{6}\right)}}=\sqrt{6}\left(2 \pi \sqrt{\frac{l}{g}}\right)=\sqrt{6} T
\)

Hint

For a spring-mass system, the time period of oscillation \(T=2 \pi \sqrt{\frac{m}{k}}\)
Therefore time period of the clock does not depend upon the value of acceleration due to gravity on the surface of the earth. The clock will give the correct time.

Hint

At high altitudes, the value of g decreases. Since the time period of the pendulum \(T\) \(=2 \pi \sqrt{ }(l / g)\), it will increase at high altitudes.
The clock will not give the correct time. To keep the correct time \(T\) should not change and to keep \(T\) constant the ratio \(l / g\) should not change. Since the value of \(g\) is getting decreased at high altitudes, the length \(l\) should also be proportionately decreased to keep the ratio \(l / g\) constant.

Hint

As the acceleration due to gravity acting on the bob of the pendulum, due to free fall gives a torque to the pendulum, the bob goes in a circular path.

Hint

(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.
A periodic motion need not be necessarily oscillatory. For example, the moon revolving around the earth.
Also, an oscillatory motion need not be necessarily periodic. For example, damped harmonic motion.

Hint

\(
\text { Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position. }
\)

Hint

\(
\text { Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position. }
\)

Hint

\(
\text { As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position. }
\)

Hint

Because it completes one oscillation in a fixed interval of time and the oscillations are in terms of rotation of the body through some angle.

Hint

In S.H.M.,
\(
F=-k x
\)
Therefore,
\(\vec{F} . \vec{r}\). will always be negative. As acceleration has the same direction as the force,
\(\vec{a} . \vec{r}\) Will also be negative, always .

Hint

As the direction of force and acceleration are always same, \(\vec{F} . \vec{a}\) is always positive.

Hint

As \(\vec{F}, \vec{a}, \vec{r}, \vec{v}\) are either parallel or anti-parallel to each other, their cross products will always be zero.

Hint

If the particle were dropped from the surface of the earth, the motion of the particle would be SHM. But when it is dropped from a height \(h\), the motion of the particle is not SHM because there is no horizontal velocity imparted. In that case, the motion of the particle would be periodic and in a straight line.

Hint

displacement from the mean position. Answer is option a.
For S.H.M.,
\(
\begin{aligned}
& F=-k x \\
& m a=-k x \quad(F=m a)
\end{aligned}
\)
or,
\(
a=-\frac{k}{m} x
\)
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.

Hint

\(r=A \cdot \cos \omega t\) is the equation of an SHM, (ì+2j) is the direction of a straight line. So the given equation is the simple harmonic motion of the particle on a straight line, hence also periodic. So (a), (c) and (d) are true. When the motion is on the straight line it can’t be on an ellipse, so (b) is not true.

Hint

(d) with time period \(\frac{\pi}{\omega}\)
Given equation:
\(
\begin{aligned}
& x=x_0 \sin ^2 \omega t \\
& \Rightarrow x=\frac{x_0}{2}(\cos 2 \omega t-1)
\end{aligned}
\)
Now, the amplitude of the particle is \(x_0 / 2\) and the angular frequency of the SHM is \(2 \omega\).
Thus, time period of the \(\mathrm{SHM}=\)
\(
\frac{2 \pi}{\text { angular frequency }}=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}
\)

Hint

(d) the average potential energy in one time period is equal to the average kinetic energy in this period.
The kinetic energy of the motion is given as,
\(
\frac{1}{2} k A^2 \cos ^2 \omega t
\)
The potential energy is calculated as,
\(
\frac{1}{2} k A^2 \sin ^2 \omega t
\)
As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.

Hint

(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy
In SHM,
maximum kinetic energy \(=\frac{1}{2} k A^2\)
maximum potential energy \(=\frac{1}{2} k A^2\)
The minimum value of both kinetic and potential energy is zero.
Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.

Hint

(a) the measured times are same
(b) the measured speeds are same
The effect of gravity on the object as well as on the pendulum clock is same in both cases; the time measured is also same. As the time measured is same, the speeds are same.

Hint

(a) A simple pendulum
(b) A physical pendulum
As the time period of a simple pendulum and a physical pendulum depends on the acceleration due the gravity, the time period of these pendulums changes when they are taken to the moon.