Answer (b) Given, $y=3 \cos (\frac{\pi}{4}-2 \omega t)$

Velocity of the particle

$$ \begin{aligned} v & =\frac{d y}{d t}=\frac{d}{d t}[3 \cos (\frac{\pi}{4}-2 \omega t)] \\ & =3(-2 \omega)[-\sin (\frac{\pi}{4}-2 \omega t)] \\ & =6 \omega \sin (\frac{\pi}{4}-2 \omega t) \\ \text { Acceleration, } a & =\frac{d v}{d t}=\frac{d}{d t}[6 \omega \sin (\frac{\pi}{4}-2 \omega t)] \\ & =6 \omega \times(-2 \omega) \cos (\frac{\pi}{4}-2 \omega t)=-12 \omega^{2} \cos (\frac{\pi}{4}-2 \omega t) \\ & =-4 \omega^{2}[3 \cos (\frac{\pi}{4}-2 \omega t)] \\ \Rightarrow \quad a & =-4 \omega^{2} y \\ \Rightarrow \text { As acceleration, } a \propto & -y \end{aligned} $$

Hence, due to negative sign motion is SHM.

Clearly, from the equation

$$ \begin{aligned} & \omega^{\prime}=2 \omega \\ & \Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \omega \Rightarrow T^{\prime}=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega} \quad \text { [and given equation } y=3 \cos (-2 \omega t+\frac{\pi}{4}) \text { ] } \end{aligned} $$

So, motion is SHM with period $\frac{\pi}{\omega}$.

Answer (b) Given equation of motion is

$$ \begin{aligned} & \qquad=\sin ^{3} \omega t \\ &=(3 \sin \omega t-4 \sin 3 \omega t) / 4 \quad[\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta] \\ & \Rightarrow \quad \frac{d y}{d t}=[\frac{d}{d t}(3 \sin \omega t)-\frac{d}{d t}(4 \sin 3 \omega t)] / 4 \\ & \Rightarrow \quad 4 \frac{d y}{d t}=3 \omega \cos \omega t-4 \times[3 \omega \cos 3 \omega t] \\ & \Rightarrow \quad 4 \times \frac{d^{2} y}{d t^{2}}=-3 \omega^{2} \sin \omega t+12 \omega \sin 3 \omega t \\ & \Rightarrow \quad \frac{d^{2} y}{d t^{2}}=-\frac{3 \omega^{2} \sin \omega t+12 \omega^{2} \sin 3 \omega t}{4} \\ & \Rightarrow \frac{d^{2} y}{d t^{2}} \text { is not proportional to } y . \\ & \text { Hence, motion is not SHM. } \end{aligned} $$

As the expression is involving sine function, hence it will be periodic.

Answer (d) For motion to be SHM acceleration of the particle must be proportional to negative of displacement.

i.e., $\quad a \propto-(y$ or $x)$

We should be clear that $y$ has to be linear.

Answer (c) Consider the diagram in which a liquid column oscillates. In this case, restoring force acts on the liquid due to gravity. Acceleration of the liquid column, can be calculated in terms of restoring force.

Restoring force

$f=$ Weight of liquid column of height $2 y$

$\Rightarrow \quad f=-(A \times 2 y \times \rho) \times g=-2 A \rho g y$

$\Rightarrow f \propto-y \Rightarrow$ Motion is SHM with force constant $k=2 A \rho g$.

$\Rightarrow$ Time period

$$ \begin{aligned} & T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{A \times 2 h \times \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{h}{g}} \\ & T=2 \pi \sqrt{\frac{l}{g}}, \text { where } l=h \end{aligned} $$

Which is independent of the density of the liquid.

Thinking Process

We have to find resultant-displacement by adding $x$ and $y$-components. According to variation of $x$ and $y$, trajectory will be predicted.

Answer (c) Given,

$$ \begin{gathered} x=a \cos \omega t \\ y=a \sin \omega t \end{gathered} $$

Squaring and adding Eqs. (i) and (ii),

$$ \begin{matrix} a^{2} \Rightarrow y^{2} & =a^{2} & (\cos ^{2} \omega t+\sin ^{2} \omega t) \\ x^{2}+y^{2} & =a^{2} & {[\because \cos ^{2} \omega t+\sin ^{2} \omega t=1]} \end{matrix} $$

This is the equation. of a circle

Clearly, the locus is a circle of constant radius a.

Answer (d) According to the question, the displacement

$$ \begin{aligned} & \text { Let } \\ & y=a \sin \omega t+b \cos \omega t \\ & \text { Let } \\ & a=A \sin \phi \text { and } b=A \cos \phi \\ & \quad \text { Now, } \quad a^{2}+b^{2}=A^{2} \sin ^{2} \phi+A^{2} \cos ^{2} \phi \\ & =A^{2} \Rightarrow A=\sqrt{a^{2}+b^{2}} \\ & y=A \sin \phi \cdot \sin \omega t+A \cos \phi \cdot \cos \omega t \\ & =A \sin (\omega t+\phi) \\ & \frac{d y}{d t}=A \omega \cos (\omega t+\phi) \\ & \frac{d^{2} y}{d t^{2}}=-A \omega^{2} \sin (\omega t+\phi)=-A y \omega^{2}=(-A \omega^{2}) y \\ & \Rightarrow \quad \frac{d^{2} y}{d t^{2}} \propto(-y) \end{aligned} $$

Hence, it is an equation of SHM with amplitude $A=\sqrt{a^{2}+b^{2}}$.

Answer (b) According to the question, $A$ is given a transverse displacement.

Through the elastic support the disturbance is transferred to all the pendulums.

$A$ and $C$ are having same length, hence they will be in resonance, because their time period of oscillation.

$$ T=2 \pi \sqrt{\frac{l}{g}} \text { and hence, frequency is same. } $$

So, amplitude of $A$ and $C$ will be maximum.

Note In this problem, we have assumed that the support is perfectly elastic and there is no damping. Hence, oscillation is considered as undamped.

Answer (a) Let angular velocity of the particle executing circular motion is $\omega$ and when it is at $Q$ makes and angle $\theta$ as shown in the diagram.

Clearly,

$\theta=\omega t$

Now, we can write

$$ \begin{aligned} & O R=O Q \cos (90-\theta) \\ & =O Q \sin \theta=O Q \sin \omega t \\ & =r \sin \omega t \quad[\because O Q=r] \\ & {[\because r=B]} \\ & =B \sin \frac{2 \pi}{T} t=B \sin (\frac{2 \pi}{30} t) \end{aligned} $$

$$ \Rightarrow \quad x=r \sin \omega t=B \sin \omega t $$

Clearly, this equation represents SHM.

Answer (c) As the given equation is

$$ x=a \cos (\alpha t)^{2} $$

is a cosine function. Hence, it is an oscillatory motion.

Now, putting $t+T$ in place of $t$

$$ \begin{aligned} x(t+T) & =a \cos [\alpha(t+T)]^{2} \quad[\because x(t)=acos(\alpha t)^{2}] \\ & =a \cos [\alpha t^{2}+\alpha T^{2}+2 \alpha t T] \neq x(t) \end{aligned} $$

where, $T$ is supposed as period of the function $\omega(t)$.

Hence, it is not periodic.

Answer (a) Let equation of an SHM is represented by $y=a \sin \omega t$

v =$\frac{d y}{d t}=a \omega \cos \omega t $

$\Rightarrow \quad(v)_{\max } =a \omega=30 $

$\text { Acceleration }(A) =\frac{d x^{2}}{d t^{2}}=-a \omega^{2} \sin \omega t $

$A_{\max } =\omega^{2} a=60 $

$\text { Eqs. (i) and (ii), we get } \omega(\omega a) =60 \Rightarrow \omega(30)=60 $

$\Rightarrow \quad \omega =2 rad / s $

$\Rightarrow \quad \frac{2 \pi}{T} =2 rad / s \Rightarrow T=\pi sec$

Thinking Process

To solve this question, we have to find equivalent spring constant of the system when mass is connected in between.

Answer (b)

Consider the diagram, two springs can be considered as parallel.

Hence,

$$ \begin{aligned} k_{eq} & =\text { Equivalent spring constant } \\ & =k_1+k_2 \end{aligned} $$

Time period of oscillation of the spring block- system

$$ \begin{aligned} T & =2 \pi \sqrt{\frac{m}{k_{\text {eq }}}}=2 \pi \sqrt{\frac{m}{k_1+k_2}} \\ \Rightarrow \quad v & =\frac{1}{T}=\frac{1}{2 \pi} \times \sqrt{\frac{k_1+k_2}{m}} \\ & =\text { Equivalent oscillation frequency. } \end{aligned} $$

When the mass is connected to the two springs individually

From Eqs. (i), (ii) and (iii),

$$ \begin{aligned} & v=\frac{1}{2 \pi}[\frac{k_1}{m}+\frac{k_2}{m}]^{1 / 2} \\ & =\frac{1}{2 \pi}[\frac{4 \pi^{2} v_1^{2}}{1}+\frac{4 \pi^{2} v_2^{2}}{1}]^{1 / 2} \\ & {[\because \text { from Eq. (ii) } \frac{k_1}{m}=4 \pi^{2} v_1^{2} \text { and from Eq. (iii), } \frac{k_2}{m}=4 \pi^{2} v_2^{2}]} \\ & =\frac{2 \pi}{2 \pi}[v_1^{2}+v_2^{2}]^{1 / 2} \Rightarrow v=\sqrt{v_1^{2}+v_2^{2}} \end{aligned} $$

Note Do not confuse with parallel and series combinations of springs.

Answer $(a, c)$

The rotation of earth about its axis is periodic because it repeats after a regular interval of time.

The rotation of earth is obviously not a to and fro type of motion about a fixed point, hence its motion is not SHM.

Answer $(a, c)$

Consider the motion of the ball inside a smooth curved bowl.

For small angular displacement or slightly released motion, it can be considered as angular SHM.

When the ball is at an angle of $\theta$ the restoring force $(g \sin \theta) m$ acts as shown.

$$ \begin{aligned} \because & m a & =m g \sin \theta \\ \Rightarrow & a & =g \sin \theta \end{aligned} $$

$$ \Rightarrow \quad \frac{d^{2} x}{d t^{2}}=-g \sin \theta=-g \times \frac{x}{R} \quad[\therefore \sin \theta \simeq \theta=x / R] $$

Hence ,motion is SHM

$$ \begin{matrix} \Rightarrow & d 2 x / d t^{2} \propto(-x) \\ \Rightarrow & \omega & =\sqrt{g / R} \\ & & \text { Time period } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{g} .} . \end{matrix} $$

As motion is SHM, hence it must be periodic.

Answer $(b, d)$

It is clear from the curve that points corresponding to $t=2 s$ and $t=6 s$ are separated by a distance belongs to one time period. Hence, these points must be in same phase.

Similarly, points belong to $t=1 s$ and $t=5 s$ are at separation of one time period, hence must be in phase.

Thinking Process

We have to write equation for the SHM and then find out velocity as well as acceleration corresponding to the displacement.

Answer $(a, b, d)$

Let the equation for the SHM is $x=a \sin \omega t$.

Clearly, it is a periodic motion because it involves sine function.

$$ \text { Velocity } v=\frac{d x}{d t}=\frac{d}{d t}(a \sin \omega t)=a \omega \cos \omega t $$

Velocity is also periodic because of cosine function.

Acceleration,

$$ A=\frac{d v}{d t}=\frac{d^{2} x}{d t^{2}}=-a \omega^{2} \sin \omega t $$

[ $\because$ Acceleration is a sine function, hence cannot be constant]

$$ =-(\omega^{2} a) \sin \omega t=-\omega^{2} x $$

Force,

$$ \begin{aligned} F & =\text { mass } \times \text { acceleration } \\ & =m A=-m \omega^{2} x \end{aligned} $$

Hence, force acting is directly proportional to displacement from the mean position and opposite to it.

Answer $(a, b, c)$

Consider the diagram

From the given diagram; it is clear that

(a) A $t=\frac{3 T}{4}$, the displacement of the particle is zero. Hence, the particle executing SHM will be at mean position i.e., $x=0$. So, acceleration is zero and force is also zero.

(b) At $t=\frac{4 T}{4}$, displacement is maximum i.e., extreme position, so acceleration is maximum.

(c) At $t=\frac{T}{4}$, corresponds to mean position, so velocity will be maximum at this position.

(d) At $t=\frac{2 T}{4}=\frac{T}{2}$, corresponds to extreme position, $so KE=0$ and $PE=$ maximum.

Answer $(a, b, d)$

Let the equation of a SHM is represented as $x=a \sin \omega t$

Assume mass of the body is $m$.

(a) Total mechanical energy of the body at any time $t$ is

$$ E=\frac{1}{2} m \omega^{2} a^{2} $$

Kinetic energy at any instant $t$ is

$$ \begin{matrix} K & =\frac{1}{2} m v^{2}=\frac{1}{2} m[\frac{d x}{d t}]^{2} & {[\because v=\frac{d x}{d t}]} \\ & =\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t & {[\because x=a \sin \omega t]} \\ \Rightarrow \quad K_{\max } & =\frac{1}{2} m \omega^{2} a^{2}=E & {[\because \text { for } k_{\max }, \cos \omega t=1] \ldots \text { (ii) }} \end{matrix} $$

(b) $KE$ at any instant $t$ is

K = $\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t $

$(K_{av}) \text { for a cycle } =\frac{1}{2} m \omega^{2} a^{2}[(\cos ^{2} \omega t)_{av}] \text { for a cycle } $

$ =\frac{1}{2} m \omega^{2} a^{2}[\frac{0+1}{2}] $

$ =\frac{1}{4} m \omega^{2} a^{2}=\frac{K_{\max }}{2}$

[from Eq. (ii)]

(c) Velocity $=v=\frac{d x}{d t}=a \omega \cos \omega t$

$$ \begin{aligned} v_{\text {mean }} & =\frac{v_{\text {max }}+v_{\text {min }}}{2} \\ & =\frac{a \omega+(-a \omega)}{2}=0 \end{aligned} $$

[For a complete cycle]

(d) $v_{\text {rms }}=\sqrt{\frac{v_{\text {max }}^{2}+v_2^{2}}{2}}=\sqrt{\frac{0+a^{2} \omega^{2}}{2}}=\frac{a \omega}{\sqrt{2}}$

$$ \Rightarrow \quad V_{rms}=\frac{V_{\max }}{\sqrt{2}} $$

Answer $(a, c, d)$

Consider the diagram.

(a) When the particle is $3 cm$ away from $A$ going towards $B$, velocity is towards $A B$ i.e., positive.

In SHM, acceleration is always towards mean position $(O)$ in this case. Hence, it is positive.

(b) When the particle is at $C$, velocity is towards $B$ hence positive.

(c) When the particle is $4 cm$ away from $B$ going towards $A$ velocity is negative and acceleration is towards mean position $(O)$ hence negative.

(d) Acceleration is always towards mean position $(O)$. When the particle is at $B$ acceleration and force are towards $B A$ that is negative.

Answer In SHM y-t graph, zero displacement values correspond to mean position; where velocity of the oscillator is maximum.

Whereas the crest and troughs represent extreme positions, where displacement is maximum and velocity of the oscillator is minimum and is zero. Hence,

(a) $A, C, E, G$ are either crest or trough having zero velocity.

(b) speed is maximum at mean positions represented by $B, D, F, H$ points.

Answer Consider the diagram in which the block is displaced right through $x$.

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 $k x$ towards left on the block as gven in the free body diagram of the block.

Hence, total force (restoring) $=(k x+k x) \quad[\because$ Both forces are in same direction $]$ $=2 k x$ towards left

Answer The two basic characteristics of a simple harmonic motion

(i) Acceleration is directly proportional to displacement.

(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.

Answer Consider the diagram of a simple pendulum.

The bob is displaced through an angle $\theta$ shown.

The restoring torque about the fixed point $O$ is

If $\theta$ is small angle in radians, then $\sin \theta \approx \theta$

$$ \tau=-m g \sin \theta $$

$\Rightarrow \quad \tau \approx-m g \theta \Rightarrow \tau \propto(-\theta)$

Hence, motion of a simple pendulum is SHM for small angle of oscillations.

Thinking Process

We have to find velocity by differentiating the equation representing displacement and acceleration by differentiating the equation relating velocity and time.

Answer Let $x=A \sin \omega t$ is the displacement function of SHM.

Velocity,

$$ \begin{aligned} v & =\frac{d x}{d t}=A \omega \cos \omega t \\ v_{\max } & =A \omega|\cos \omega t|_{\max } \\ & =A \omega \times 1=\omega A \end{aligned} $$

$[\because|\cos \omega t|_{\max }=1] \ldots(i)$

Acceleration, $a=\frac{d v}{d t}=-\omega A \cdot \omega \sin \omega t$

$$ \begin{aligned} & =-\omega^{2} A \sin \omega t \\ |a_{\max }| & =|(-\omega^{2} A)(+1)| \\ |a_{\max }| & =\omega^{2} A \end{aligned} $$

From Eqs. (i) and (ii), we get

$ \frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^{2} A}=\frac{1}{\omega} $

$\Rightarrow \frac{a_{\max }}{v_{\max }}=\omega$

Answer The diagram represents

the motion of a particle executing SHM between $A$ and $B$.

Total distance travelled while it goes from $A$ to $B$ and returns to $A$ is

$$ \begin{aligned} & =A O+O B+B O+O A \\ & =A+A+A+A=4 A \quad[\because O A=A] \end{aligned} $$

Amplitude $=OA=A$

Hence, ratio of distance and amplitude $=\frac{4 A}{A}=4$

Answer As the particle on reference circle moves in anti-iclockwise direction. The projection will move from $P$ ’to $O$ towards left.

Hence, in the position shown the velocity is directed from $P^{\prime} arrow P^{\prime \prime}$ i.e., from right to left, hence sign is negative.

Answer Let us assume the displacement function of SHM

where,

$$ x=acos \omega t $$

$$ \text { where, } \quad \begin{aligned} a & =\text { amplitude of motion } \\ \text { velocity } v & =\frac{d x}{d t} \end{aligned} $$

or

$$ \begin{aligned} \frac{d x}{d t} & =a(-\sin \omega t) \omega=-\omega asin \omega t \\ v & =-\omega asin \omega t \\ & =\omega acos(\frac{\pi}{2}+\omega t) \quad[\because \sin \omega t=-\cos (\frac{\pi}{2}+\omega t)] \end{aligned} $$

Now, phase of displacement $=\omega t$

Phase of velocity $=\frac{\pi}{2}+\omega t$

$\therefore$ Difference in phase of velocity to that of phase of displacement

$$ =\frac{\pi}{2}+\omega t-\omega t=\frac{\pi}{2} $$

Answer Potential energy (PE) of a simple harmonic oscillator is

where,

$$ \begin{aligned} & =\frac{1}{2} k x^{2}=\frac{1}{2} m \omega^{2} x^{2} \\ k & =\text { force constant }=m \omega^{2} \end{aligned} $$

When, PE is plotted against displacement $x$, we will obtain a parabola.

When $x=0, PE=0$

When $x= \pm A, PE=$ maximum

$$ \begin{aligned} & =\frac{1}{2} m \omega^{2} A^{2} \\ \text { KE of a simple harmonic oscillator } & =\frac{1}{2} m v^{2} \\ & =\frac{1}{2} m[\omega \sqrt{A^{2}-x^{2}}]^{2} \quad[\because v=\omega \sqrt{A^{2}-x^{2}}] \\ & =\frac{1}{2} m \omega^{2}(A^{2}-x^{2}) \end{aligned} $$

This is also parabola, if plot KE against displacement $x$.

$$ \begin{matrix} \text { i.e., } & KE=0 \text { at } x= \pm A \\ \text { and } & KE=\frac{1}{2} m \omega^{2} A^{2} \text { at } x=0 \end{matrix} $$

$$ \begin{aligned} & \begin{matrix} =\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2}(A^{2}-x^{2}) \end{matrix} \\ & =\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2} \\ & TE=\frac{1}{2} m \omega^{2} A^{2} \end{aligned} $$

Which is constant and does not depend on $x$.

Plotting under the above guidelines KE, PE and TE versus displacement $x$-graph as follows

Answer A second’s pendulum means a simple pendulum having time period $T=2 s$.

For a simple pendulum, $\quad T=2 \pi \sqrt{\frac{l}{g}}$

where, $l=$ length of the pendulum and

$g=$ acceleration due to gravity on surface of the earth.

$$ T_{e}=2 \pi \sqrt{\frac{l_{e}}{g_{e}}} $$

On the surface of the moon,

$$ \begin{aligned} & T_{m}=2 \pi \sqrt{\frac{l_{m}}{g_{m}}} \\ & \frac{T_{e}}{T_{m}}=\frac{2 \pi}{2 \pi} \sqrt{\frac{l_{e}}{g_{e}}} \times \sqrt{\frac{g_{m}}{l_{m}}} \end{aligned} $$

$T_{e}=T_{m}$ to maintain the second’s pendulum time period.

$$ \therefore \quad 1=\sqrt{\frac{l_{e}}{l_{m}} \times \frac{g_{m}}{g_{e}}} $$

But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth, i.e.,

$$ g_{m}=\frac{g_{e}}{6} $$

Squaring Eq. (iii) and putting this value,

1 = $\frac{l_{e}}{l_{m}} \times \frac{g_{e} / 6}{g_{e}}=\frac{l_{e}}{l_{m}} \times \frac{1}{6} $

$\Rightarrow \frac{l_{e}}{6 l_{m}} =1 \Rightarrow 6 l_{m}=l_{e} $

$\Rightarrow l_{m} =\frac{1}{6} l_{e}=\frac{1}{6} \times 1=\frac{1}{6} m$

Thinking Process

To predict the nature of the motion, we have to displace the mass slightly and find the acceleration due to the restoring force.

Answer For the calculation purpose, in this situation we will neglect gravity because it is constant throughout and will not effect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount $x_0$.

Now, if the mass is given a further displacement downwards by an amount $x$. The string and spring both should increase in length by $x$.

But, string is inextensible, hence the spring alone will contribute the total extension $x+x=2 x$, to lower the mass down by $x$ from initial equilibrium mean position $x_0$. So, net extension in the spring $(=2 x+x_0)$.

Now force on the mass before bulling (in the $x_0$ extension case)

But

$$ \begin{aligned} F & =2 T \\ T & =k x_0 \\ F & =2 k x_0 \end{aligned} $$

[where $k$ is spring constant]

When the mass is lowered down further by $x$,

$$ F^{\prime}=2 T^{\prime} $$

But new spring length $=(2 x+x_0)$

$\therefore \quad F^{\prime}=2 k(2 x+x_0)$

Restoring force on the system.

Using Eqs. (i) and (ii), we get

$$ F_{\text {restoring }}=-[F^{\prime}-F] $$

$$ \begin{aligned} F_{\text {restoring }} & =-[2 k(2 x+x_0)-2 k x_0] \\ & =-[2 \times 2 k x+2 k x_0-2 k x_0] \\ & =-4 k x \end{aligned} $$

or

$$ M a=-4 k x $$

where,

$$ a=\text { acceleration } $$

$\Rightarrow$

$$ a=-(\frac{4 k}{M}) x $$

(As, $F=m a)$

k, $M$ being constant.

$$ \therefore \quad a \propto-x $$

Hence, motion is SHM.

Comparing the above acceleration expression with standard SHM equation $a=-\omega^{2} x$, we get

$$ \therefore \quad \text { Time period } T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{4 k}{M}}}=2 \pi \sqrt{\frac{M}{4 k}} $$

Answer We have to convert the given combination of two harmonic functions to a single harmonic (sine or cosine) function.

Given, displacement function

$$ \begin{aligned} y & =\sin \omega t-\cos \omega t \\ & =\sqrt{2}(\frac{1}{\sqrt{2}} \cdot \sin \omega t-\frac{1}{\sqrt{2}} \cdot \cos \omega t) \\ & =\sqrt{2}[\cos (\frac{\pi}{4}) \cdot \sin \omega t-\sin (\frac{\pi}{4}) \cdot \cos \omega t] \\ & =\sqrt{2}[\sin (\omega t-\frac{\pi}{4})]=\sqrt{2}[\sin (\omega t-\frac{\pi}{4})] \end{aligned} $$

Comparing with standard equation

$$ y=a \sin (\omega t+\phi), \text { we get }=\omega=\frac{2 \pi}{T} \Rightarrow T=\frac{2 \pi}{\omega} $$

Clearly, the function represents SHM with a period $T=\frac{2 \pi}{\omega}$.

Answer Let us assume that the required displacement be $x$.

$\therefore$ Potential energy of the simple harmonic oscillator $=\frac{1}{2} k x^{2}$

where,

$$ k=\text { force constant }=m \omega^{2} $$

$$ PE=\frac{1}{2} m \omega^{2} x^{2} $$

Maximum energy of the oscillator

$$ TE=\frac{1}{2} m \omega^{2} A^{2} $$

where,

$$ A=\text { amplitude of motion } $$

Given,

$PE=\frac{1}{2} TE$

$\Rightarrow \quad \frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2}[\frac{1}{2} m \omega^{2} A^{2}]$

$\Rightarrow \quad x^{2}=\frac{A^{2}}{2}$

or

$$ x=\sqrt{\frac{A^{2}}{2}}= \pm \frac{A}{\sqrt{2}} $$

Sign $\pm$ indicates either side of mean position.

Answer Given potential energy associated with the field

$$ U(x)=U_0(1-\cos \alpha x) $$

Now,

$$ \text { force } F=-\frac{d U(x)}{d x} $$

$$ [\because \text { for conservatine force } f, \text { we can write } f=\frac{-d u}{d x}] $$

[We have assumed the field to be conservative]

$$ \begin{aligned} & F=-U_0 \alpha^{2} x \\ & m \omega^{2}=U_0 \alpha^{2} \\ & \omega^{2}=\frac{U_0 \alpha^{2}}{m} \text { or } \omega=\sqrt{\frac{U_0 \alpha^{2}}{m}} \\ & \text { Time period } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{U_0 \alpha^{2}}} \end{aligned} $$

$$ F=-\frac{d}{d x}(U_0-U_0 \cos \alpha x)=-U_0 \alpha \sin \alpha x $$

$[\because$ for small oscillations $\alpha x$ is small, $\sin \alpha x \approx \alpha x]$

Note The motion is SHM only for small oscillations and hence, the time period is valid only in case of small oscillations.

Thinking Process

The spring-block system will perform SHM about the mean position with an amplitude of $5 cm$.

Answer Consider the diagram of the spring block system. It is a SHM with amplitude of $5 cm$ about the mean position shown.

Given, $\quad \quad \quad$ spring constant $k=50 N / m$

$$ m=\text { mass attached }=2 kg $$

$\therefore \quad$ Angular frequency $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5 rad / s$

Assuming the displacement function

$$ y(t)=A \sin (\omega t+\phi) $$

where, $\phi=$ initial phase

But given at $t=0, y(t)=+A$

or

$$ \begin{aligned} & y(0)=+A=A \sin (\omega \times 0+\phi) \\ & \sin \phi=1 \Rightarrow \phi=\frac{\pi}{2} \end{aligned} $$

$\therefore$ The desired equation is $y(t)=A \sin (\omega t+\frac{\pi}{2})=A \cos \omega t$

Putting $A=5 cm, \omega=5 rad / s$

we get, $y(t)=5 \sin 5 t$

where, $t$ is in second and $y$ is in centimetre.

Answer Consider the situations shown in the diagram (i) and (ii)

(i)

(ii)

Assuming the two pendulums follow the following functions of their angular displacements

and

$$ \begin{aligned} & \theta_1=\theta_0 \sin (\omega t+\phi_1) \\ & \theta_2=\theta_0 \sin (\omega t+\phi_2) \end{aligned} $$

As it is given that amplitude and time period being equal but phases being different.

Now, for first pendulum at any time $t$

From Eq. (i), we get

$$ \theta_1=+\theta_0 $$

[Right extreme]

$\Rightarrow$

$$ \begin{aligned} \theta_0 & =\theta_0 \sin (\omega t+\phi_1) \text { or } 1=\sin (\omega t+\phi_1) \\ \sin \frac{\pi}{2} & =\sin (\omega t+\phi_1) \\ (\omega t+\phi_1) & =\frac{\pi}{2} \end{aligned} $$

$$ \Rightarrow \quad \sin \frac{\pi}{2}=\sin (\omega t+\phi_1) $$

or

Similarly, at the same instant $t$ for pendulum second, we have

$$ \theta_2=-\frac{\theta_0}{2} $$

where $\theta_0=2^{\circ}$ is the angular amplitude of first pendulum. For the second pendulum, the angular displacement is one degree, therefore $\theta_2=\frac{\theta_0}{2}$ and negative sign is taken to show for being left to mean position.

From Eq. (ii), then

$$ \begin{aligned} -\frac{\theta_0}{2} & =\theta_0 \sin (\omega t+\phi_2) \\ \sin (\omega t+\phi_2) & =-\frac{1}{2} \Rightarrow(\omega t+\phi_2)=-\frac{\pi}{6} \text { or } \frac{7 \pi}{6} \\ (\omega t+\phi_2) & =-\frac{\pi}{6} \text { or } \frac{7 \pi}{6} \end{aligned} $$

$$ \Rightarrow $$

From Eqs. (iv) and (iii), the difference in phases

or

$$ \begin{gathered} (\omega t+\phi_2)-(\omega t+\phi_1)=\frac{7 \pi}{6}-\frac{\pi}{2}=\frac{7 \pi-3 \pi}{6}=\frac{4 \pi}{6} \\ (\phi_2-\phi_1)=\frac{4 \pi}{6}=\frac{2 \pi}{3}=120^{\circ} \end{gathered} $$

Long Answer Type Questions

Answer In this case acceleration is variable. In accelerated motion, weight of body depends on the magnitude and direction of acceleration for upward or downward motion.

(a) Hence, the weight of the body changes during oscillations

(b) Considering the situation in two extreme positions, as their acceleration is maximum in magnitude.

we have,

$$ m g-N=m a $$

$\because$ At the highest point, the platform is accelerating downward.

$$ \begin{matrix} \Rightarrow & N=m g-m a \\ \text { But } & a=\omega^{2} A \\ \therefore & N=m g-m \omega^{2} A \\ \text { where, } & A=\text { amplitude of motion. } \\ \text { Given, } & m=50 kg, \text { frequency } v=2 s^{-1} \\ \therefore & \omega=2 \pi v=4 \pi rad / s \\ \therefore & A=5 cm=5 \times 10^{-2} m \\ & N=50 \times 9.8-50 \times(4 \pi)^{2} \times 5 \times 10^{-2} \\ & =50[9.8-16 \pi^{2} \times 5 \times 10^{-2}] \\ & =50[9.8-7.89] \\ & =50 \times 1.91 \\ & =95.5 N \end{matrix} $$

$$ \begin{matrix} \text { But } & a=\omega^{2} A & \text { [in magnitude] } \end{matrix} $$

When the platform is at the lowest position of its oscillation,

It is accelerating towards mean position that is vertically upwards.

Writing equation of motion

or

$$ N-m g=m a=m \omega^{2} A $$

$$ \begin{aligned} N & =m g+m \omega^{2} A \\ & =m[g+\omega^{2} A] \end{aligned} $$

Putting the data

$$ \begin{aligned} N & =50[9.8+(4 \pi)^{2} \times 5 \times 10^{-2}] \\ & =50[9.8+(12.56)^{2} \times 5 \times 10^{-2}] \\ & =50[9.8+7.88] \\ & =50 \times 17.68=884 N \end{aligned} $$

Now, the machine reads the normal reaction. It is clear that

maximum weight $=884 N$

(at lowest point)

minimum weight $=95.5 N$

(at top point)

Thinking Process

When we have given the support of hand, net force $=0$ and when it is released, all of its PE will convert into $KE$.

Answer (a) When the support of the hand is removed, the body oscillates about a mean position.

Suppose $x$ is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block $=m g x$

where,

$m=$ mass of the block

Gain in elastic potential energy of the spring

$$ =\frac{1}{2} k x^{2} $$

As the two are equal, conserving the mechanical energy,

we get,

$$ m g x=\frac{1}{2} k x^{2} \text { or } x=\frac{2 m g}{k} $$

Now, the mean position of oscillation will be, when the block is balanced by the spring.

If $x^{\prime}$ is the extension in that case, then

$$ \begin{aligned} & F & =+k x^{\prime} \\ \text { But } & F & =m g \\ \Rightarrow & m g & =+k x^{\prime} \\ \text { or } & x^{\prime} & =\frac{m g}{k} \end{aligned} $$

Dividing Eq. (iii) by Eq. (iv),

$$ \begin{aligned} & & \frac{x}{x^{\prime}} & =\frac{2 m g}{k} / \frac{m g}{k}=2 \\ \Rightarrow & & x & =2 x^{\prime} \end{aligned} $$

But given $x=4 cm$ (maximum extension from the unstretched position)

$$ \begin{aligned} \therefore & 2 x^{\prime} & =4 \\ \therefore & x^{\prime} & =\frac{4}{2}=2 cm \end{aligned} $$

But the displacement of mass from the mean position to the position when spring attains its natural length is equal to amplitude of the oscillation.

$$ \therefore \quad A=x^{\prime}=2 cm $$

where, $A=$ amplitude of the motion.

(b) Time period of the oscillating system depends on mass spring constant given by

$$ T=2 \pi \sqrt{\frac{m}{k}} $$

It does not depend on the amplitude.

But from Eq. (iii),

$ \frac{2 m g}{k} =x $

$\frac{2 m g}{k} =4 cm=4 \times 10^{-2} m $

$\therefore \frac{m}{k} =\frac{4 \times 10^{-2}}{2 g}=\frac{2 \times 10^{-2}}{g} $

$\therefore \frac{k}{m} =\frac{g}{2 \times 10^{-2}} $

$ \text { and } \text { (maximum extension) } $

$\therefore \quad \text { frequency }=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} $

$ =\frac{1}{2 \times 3.14} \sqrt{\frac{g}{2 \times 10^{-2}}} $

$ =\frac{10}{6.28} \times 2.14 \sqrt{\frac{4.9}{10^{-2}}}=\frac{1}{6.28} \times \sqrt{4.9 \times 100}$

Note The block during oscillation cannot go above the position it was released, as it is given that there is no velocity in the upward direction either by the system or external agent.

Thinking Process

To find the time period for the SHM displace the log from equilibrium position by a small displacement and then find the restoring force.

Answer Consider the diagram.

Let the log be pressed and let the vertical displacement at the equilibrium position be $x_0$. At equilibrium,

$$ m g=\text { buoyant force }=(\rho A x_0) g \quad[\because m=V \rho=(A x_0) \rho] $$

When it is displaced by a further displacement $x$, the buoyant force is $A(x_0+x) \rho g$

$\therefore \quad$ Net restoring force $=$ Buoyant force - Weight

$$ \begin{aligned} & =A(x_0+x) \rho g-m g \\ & =(A \rho g) x \end{aligned} $$

$$ [\because m g=\rho A x_0 g] $$

As displacement $x$ is downward and restoring force is upward, we can write

where

$$ \begin{aligned} F_{\text {restoring }} & =-(A \rho g) x \\ & =-k x \\ k & =\text { constant }=A \rho g \end{aligned} $$

So, the motion is SHM

Now,

$$ \text { Acceleration } a=\frac{F_{\text {restoring }}}{m}=-\frac{k}{m} x $$

Comparing with

$$ a=-\omega^{2} x $$

$\Rightarrow$

$$ \omega^{2}=\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}} $$

$\Rightarrow$

$$ \frac{2 \pi}{T}=\sqrt{\frac{k}{m}} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}} $$

$\Rightarrow$

$$ T=2 \pi \sqrt{\frac{m}{A \rho g}} $$

Answer Consider the diagram shown below

Let us consider an infinitesimal liquid column of length $d x$ at a height $x$ from the horizontal line.

If $\rho=$ density of the liquid

$$ A=\text { cross-sectional area of } V \text {-tube } $$

PE of element $d x$ will be given as

$$ PE=d m g x=(A \rho d x) g x \quad[\because d m=p V=p A d x] $$

where

$A \rho d x=d m=$ mass of element $d x$

$\therefore$ Total PE of the left column

$$ \begin{aligned} & =\int_0^{h_1} A \rho g x d x \\ & =A \rho g \int_0^{h_1} x d x \\ & =A \rho g|\frac{x^{2}}{2}|_0^{h_1}=A \rho g \frac{h_1^{2}}{2} \end{aligned} $$

But,

$$ \begin{aligned} h_1 & =l \sin 45^{\circ} \\ PE & =\frac{A \rho g}{2} l^{2} \sin ^{2} 45^{\circ} \end{aligned} $$

In a similar way,

$$ \begin{aligned} \text { PE of right column } & =\frac{A \rho g}{2} l^{2} \sin ^{2} 45^{\circ} \\ \text { Total PE } & =\frac{A \rho g}{2} l^{2} \sin ^{2} 45^{\circ}+\frac{A \rho g}{2} l^{2} \sin ^{2} 45^{\circ} \\ & =2 \times \frac{1}{2} A \rho g l^{2}(\frac{1}{\sqrt{2}})^{2}=\frac{A \rho g l^{2}}{2} \end{aligned} $$

If due to pressure difference created y element of left side moves on the right side, then liquid present in the left arm $=l-y$

liquid present in the right arm $=l+y$

$\therefore \quad$ Total PE $=A \rho g(l-y)^{2} \sin ^{2} 45^{\circ}+A \rho g(l+y)^{2} \sin ^{2} 45^{\circ}$

or

Changes in P E = $PE _{final }-P E _{initial }$

$$ \begin{aligned} \Delta PE & =\frac{A \rho g}{2}[(l-y)^{2}+(l+y)^{2}-l^{2}] \\ & =\frac{A \rho g}{2}[l^{2}+y^{2}-2 l y+l^{2}+y^{2}+2 l y-l^{2}] \\ & =\frac{A \rho g}{2}[2(l^{2}+y^{2})] \\ & =A \rho g(l^{2}+y^{2}) \end{aligned} $$

If $v$ is the change in velocity of the total liquid column, then change in KE

$$ \begin{aligned} \Delta KE & =\frac{1}{2} m v^{2} \\ m & =A \rho(2 l) \end{aligned} $$

But

$\therefore \quad \Delta KE=\frac{1}{2} A \rho 2 l v^{2}=A \rho l v^{2}$

From Eqs. (iv) and (v),

$$ \Delta PE+\Delta KE=A \rho g(l^{2}+y^{2})+A \rho l v^{2} $$

System being conservative.

$\therefore$ Change in total energy $=0$

From Eq. (vi),

$$ A \rho g(l^{2}+y^{2})+A \rho l v^{2}=0 $$

Differentiating both sides with respect to time $(t)$, we get

But, $\quad \frac{d y}{d t}=v$ and $\frac{d v}{d t}=a$

[acceleration]

$\Rightarrow \quad A \rho g(2 y v)+A \rho l(2 v) a=0$

$\Rightarrow \quad(g y+l a) 2 A \rho v=0$

$2 A \rho v=$ constant and $2 A \rho v \neq 0$

$\therefore \quad l a+g y=0$

$$ \begin{matrix} a+(\frac{g}{l}) y=0 \\ \frac{d^{2} y}{d t^{2}}+(\frac{g}{l}) y=0 \end{matrix} $$

This is the standard differential equation for SHM of the form

$ \frac{d^{2} y}{d t^{2}}+\omega^{2} y $ =0

$\therefore \quad =\sqrt{\frac{g}{l}} $

$T=\frac{2 \pi}{\omega} =2 \pi \sqrt{\frac{l}{g}}$

Answer Consider the situation shown in the diagram.

The gravitational force on the particle at a distance $r$ from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the position of the particle. The external shells exert no force on the particle.

More clearly,

let $g^{\prime}$ be the acceleration at $P$.

So,

$g^{\prime}=g(1-\frac{d}{R})=g(\frac{R-d}{R})$

From figure,

$R-d=y$

$\Rightarrow$

$g^{\prime}=g \frac{y}{R}$

Force on body at $p$,

$F=-m g^{\prime}=\frac{-m g}{R} y$

$\Rightarrow$

$F \propto-y \quad$ [where, $y$ is distance from the centre]

So, motion is SHM.

For time period, we can write Eq. (i)

As

$$ m a=-\frac{M g}{R} y \Rightarrow a=-\frac{g}{R} y $$

Comparing with $a=-\omega^{2} y$

$$ \begin{aligned} \omega^{2} & =\frac{g}{R} \\ \Rightarrow \quad(\frac{2 \pi}{T}) & =\frac{g}{R} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}} \end{aligned} $$

Thinking Process

We have to consider $\theta=\theta_0$ as reference point that is when $\theta=\theta_0$ the time will be considered as $t=0$. Then, equation of rectilinear motion will be applied.

Answer Consider the diagram.

Let us assume $t=0$ when $\theta=\theta_0$, then $\theta=\theta_0 \cos \omega t$

Given a seconds pendulum $\omega=2 \pi \Rightarrow \theta=\theta_0 \cos 2 \pi t$

At time $t_1$ let $\theta=\theta_0 / 2$

$$ \begin{matrix} \therefore & \cos 2 \pi t_1 & =1 / 2 \Rightarrow t_1=\frac{1}{6} \\ \frac{d \theta}{d t} & =-(\theta_0 2 \pi) \sin 2 \pi t & {[\because \cos 2 \pi t_1=\cos \frac{\pi}{3} \Rightarrow 2 \pi t_1=\frac{\pi}{3}]} \\ & \text { [from Eq. (i)] } \end{matrix} $$

At

$$ \begin{aligned} t & =t_1=\frac{1}{6} \\ \frac{d \theta}{d t} & =-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0 \end{aligned} $$

Negative sign shows that it is going left.

Thus, the linear velocity is

$$ u=-\sqrt{3} \pi \theta_0 l \text { perpendicular to the string. } $$

The vertical component is

$$ u_{y}=-\sqrt{3} \pi \theta_0 l \sin (\theta_0 / 2) $$

and the horizontal component is

$$ u_{x}=-\sqrt{3} \pi \theta_0 l \cos (\theta_0 / 2) $$

At the time it snaps, the vertical height is

$$ H^{\prime}=H+l(1-\cos (\theta_0 / 2)) $$

Let the time required for fall be $t$, then

$$ \begin{aligned} & H^{\prime}=u_{y} t+(1 / 2) g t^{2} \quad \text { (notice } g \text { also in the negative direction) } \\ & \text { or } \quad \frac{1}{2} g t^{2}+\sqrt{3} \pi \theta_0 l \sin \frac{\theta_{o}}{2} t-H^{\prime}=0 \\ & \therefore \quad t=\frac{-\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} \pm \sqrt{3 \pi^{2} \theta_0^{2} l^{2} \sin ^{2} \frac{\theta_0}{2}+2 g H^{\prime}}}{g} \\ & =\frac{-\sqrt{3} \pi l \frac{\theta_o^{2}}{2} \pm \sqrt{3 \pi^{2}(\frac{\theta_o^{4}}{4}) l^{2}+2 g H^{\prime}}}{g}[\because \sin \frac{\theta_0}{2} \simeq \frac{\theta_0}{2} \text { for small angle }] \end{aligned} $$

Given that $\theta_0$ is small, hence neglecting terms of order $\theta_0^{2}$ and higher

$$ t=\sqrt{\frac{2 H^{\prime}}{g}} $$

$$ \begin{aligned} & \text { Now, } \quad H^{\prime} \approx H+l(1-1) \quad[\therefore \cos \theta_0 / 2 \approx 1] \\ & \begin{aligned} & =H \\ \Rightarrow \quad t & =\sqrt{\frac{2 H}{g}} \end{aligned} \end{aligned} $$

The distance travelled in the $x$-direction is $u_{x} t$ to the left of where the bob is snapped

$$ \begin{aligned} & X=U x t=\sqrt{3} \pi \theta_0 l Cos(\frac{\theta_0}{2}) \sqrt{\frac{2 H}{g}} s \\ & \text { As } \theta_0 \text { is small } \Rightarrow \cos (\frac{\theta_0}{2}) \approx 1 \\ & X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l \pi \end{aligned} $$

At the time of snapping, the bob was at a horizontal distance of $l \sin (\theta_0 / 2) \approx l \frac{\theta_0}{2}$ from $A$.

Thus, the distance of bob from $A$ where it meets the ground is

$$ \begin{aligned} & =\frac{l \theta_0}{2}-X=\frac{l \theta_0}{2}-\sqrt{\frac{6 H}{g}} \theta_0 l \pi \\ & =\theta_0 l(\frac{1}{2}-\pi \sqrt{\frac{6 H}{g}}) \end{aligned} $$