Answer

(b) and (c)

The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.

An $\to$ released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

Answer

(b) and (c) are SHMs

and (d) are periodic, but not SHMs

During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

Answer

(b) and (d) are periodic

It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.

In this case, the motion of the particle repeats itself after $2 s$. Hence, it is a periodic motion, having a period of $2 s$.

It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time.

In this case, the motion of the particle repeats itself after $2 s$. Hence, it is a periodic motion, having a period of $2 s$.

Answer

(a) SHM

The given function is:

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

This function represents SHM as it can be written in the form: ${ }^{a} \sin (\omega t+\phi)$

Its period is: $\frac{2 \pi}{\omega}$

(b) Periodic, but not SHM

The given function is:

$\sin ^{3} \omega t$

$=\frac{1}{2}[3 \sin \omega t-\sin 3 \omega t]$

The terms $\sin \omega t$ and $\sin \omega t$ individually represent simple harmonic motion (SHM).

However, the superposition of two SHM is periodic and not simple harmonic.

(c) SHM

The given function is:

$3 \cos [\frac{\pi}{4}-2 \omega t]$

$=3 \cos [2 \omega t-\frac{\pi}{4}]$

This function represents simple harmonic motion because it can be written in the form: $a \cos (\omega t+\phi)$

Its period is: $\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$

(d) Periodic, but not SHM

The given function is $\cos \omega t+\cos 3 \omega t+\cos 5 \omega t$. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

(e) Non-periodic motion

The given function $\exp (-\omega^{2} t^{2})$ is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) The given function $1+\omega t+\omega^{2} t^{2}$ is non-periodic.

Answer

(a) Zero, Positive, Positive

(b) Zero, Negative, Negative

(c) Negative, Zero, Zero

(d) Negative, Negative, Negative

(e) Zero, Positive, Positive

(f) Negative, Negative, Negative

Explanation:

(a) The given situation is shown in the following figure. Points $A$ and $B$ are the two end points, with $AB=10 cm$. $O$ is the midpoint of the path.

A particle is in linear simple harmonic motion between the end points

At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is positive as it is directed along AO.

Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is negative as it is directed along B.

Force is also negative in this case as the particle is directed leftward.

(c)

The particle is executing a simple harmonic motion. $O$ is the mean position of the particle. Its velocity at the mean position $O$ is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)

The particle is moving toward point $O$ from the end $B$. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving toward point $O$ from the end $A$. This direction of motion is from $A$ to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.

(f)

This case is similar to the one given in (d).

Answer

A motion represents simple harmonic motion if it is governed by the force law:

$F=-k x$

$m a=-k$

$\therefore a=-\frac{k}{m} x$

Where,

$F$ is the force

$m$ is the mass (a constant for a body)

$x$ is the displacement

$a$ is the acceleration

$k$ is a constant

Among the given equations, only equation $a=-10 x$ is written in the above form with $\frac{k}{m}=10$.

Hence, this relation represents SHM.

Answer

Initially, at $t=0$ :

Displacement, $x=1 cm$

Initial velocity, $v=\omega cm / sec$.

Angular frequency, $\omega=\pi rad / s^{-1}$

It is given that:

$ \begin{aligned} & x(t)=A \cos (\omega t+\phi) \\ & 1=A \cos (\omega \times 0+\phi)=A \cos \phi \end{aligned} $

$A \cos \phi=1$

Velocity, $v=\frac{d x}{d t}$

$$ \begin{align*} & \omega=-A \omega \sin (\omega t+\phi) \\ & 1=-A \sin (\omega \times 0+\phi)=-A \sin \phi \\ & A \sin \phi=-1 \tag{ii} \end{align*} $$

Squaring and adding equations ( $i$ ) and (ii), we get:

$ \begin{aligned} & A^{2}(\sin ^{2} \phi+\cos ^{2} \phi)=1+1 \\ & A^{2}=2 \\ & \therefore A=\sqrt{2} cm \end{aligned} $

Dividing equation (ii) by equation $(i)$, we get:

$ \begin{aligned} & \tan \phi=-1 \\ & \therefore \phi=\frac{3 \pi}{4}, \frac{7 \pi}{4}, \ldots \ldots \end{aligned} $

SHM is given as:

$ x=B \sin (\omega t+\alpha) $

Putting the given values in this equation, we get:

$$ \begin{equation*} 1=B \sin [\omega \times 0+\alpha] \tag{iii} \end{equation*} $$

$B \sin \alpha=1$

Velocity, $v=\omega B \cos (\omega t+\alpha)$

Substituting the given values, we get:

$$ \begin{align*} & \pi=\pi B \sin \alpha \\ & B \sin \alpha=1 \tag{iv} \end{align*} $$

Squaring and adding equations (iii) and (iv), we get:

$ \begin{aligned} & B^{2}[\sin ^{2} \alpha+\cos ^{2} \alpha]=1+1 \\ & B^{2}=2 \\ & \therefore B=\sqrt{2} cm \end{aligned} $

Dividing equation (iii) by equation (iv), we get: $\frac{B \sin \alpha}{B \cos \alpha}=\frac{1}{1}$

$\tan \alpha=1=\tan \frac{\pi}{4}$

$\therefore \alpha=\frac{\pi}{4}, \frac{5 \pi}{4}, \ldots \ldots$.

Answer

Maximum mass that the scale can read, $M=50 kg$

Maximum displacement of the spring $=$ Length of the scale, $l=20 cm=0.2 m$

Time period, $T=0.6 s$

Maximum force exerted on the spring, $F=M g$

Where,

$g=$ acceleration due to gravity $=9.8 m / s^{2}$

$F=50 \times 9.8=490$

$\therefore$ Spring constant,

$ k=\frac{F}{l}=\frac{490}{0.2}=2450 Nm^{-1} $

Mass $m$, is suspended from the balance.

Time period, $T=2 \pi \sqrt{\frac{m}{k}}$ $\therefore m=(\frac{T}{2 \pi})^{2} \times k=(\frac{0.6}{2 \times 3.14})^{2} \times 2450=22.36 kg$

$\therefore$ Weight of the body $=m g=22.36 \times 9.8=219.167 N$

Hence, the weight of the body is about $219 N$.

Answer

Spring constant, $k=1200 N m^{-1}$

Mass, $m=3 kg$

Displacement, $A=2.0 cm=0.02 cm$

Frequency of oscillation $v$, is given by the relation:

$v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

Where, $T$ is the time period

$ \therefore v=\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}=3.18 m / s $

Hence, the frequency of oscillations is $3.18 m / s$.

Maximum acceleration $(a)$ is given by the relation:

$a=\omega^{2} A$

Where,

$\omega=$ Angular frequency $=\sqrt{\frac{k}{m}}$

$A=$ Maximum displacement

$\therefore a=\frac{k}{m} A=\frac{1200 \times 0.02}{3}=8 m s^{-2}$

Hence, the maximum acceleration of the mass is $8.0 m / s^{2}$.

Maximum velocity, $v _{\max }=A \omega$

$ =A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}=0.4 m / s $

Hence, the maximum velocity of the mass is $0.4 m / s$.

Answer

(a) $x=2 \sin 20 t$

(b) $x=2 \cos 20 t$

(c) $x=-2 \cos 20 t$

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, $A=2.0 cm$

Force constant of the spring, $k=1200 N m^{-1}$

Mass, $m=3 kg$

Angular frequency of oscillation:

$ \begin{aligned} & \omega=\sqrt{\frac{k}{m}} \\ & =\sqrt{\frac{1200}{3}}=\sqrt{400}=20 rad s^{-1} \end{aligned} $

When the mass is at the mean position, initial phase is 0 .

Displacement, $x=A \sin \omega t$

$=2 \sin 20 t$

At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is $\frac{\pi}{2}$.

Displacement, $\quad x=A \sin (\omega t+\frac{\pi}{2})$

$ =2 \sin (20 t+\frac{\pi}{2}) $

$=2 \cos 20 t$

At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is $\frac{3 \pi}{2}$.

Displacement, $x=A \sin (\omega t+\frac{3 \pi}{2})$

$=2 \sin (20 t+\frac{3 \pi}{2})=-2 \cos 20 t$

The functions have the same frequency $(\frac{20}{2 \pi} Hz)$ and amplitude $(2 cm)$, but different initial phases $(0, \frac{\pi}{2}, \frac{3 \pi}{2})$.

Answer

Time period, $T=2 s$

Amplitude, $A=3 cm$

At time, $t=0$, the radius vector OP makes an angle $\frac{\pi}{2}$ with the positive $x$-axis, i.e., phase angle $\phi=+\frac{\pi}{2}$

Therefore, the equation of simple harmonic motion for the $x$-projection of $OP$, at time $t$, is given by the displacement equation:

$ \begin{aligned} & x=A \cos [\frac{2 \pi t}{T}+\phi] \\ & =3 \cos (\frac{2 \pi t}{2}+\frac{\pi}{2})=-3 \sin (\frac{2 \pi t}{2}) \end{aligned} $

$\therefore x=-3 \sin \pi t cm$

Time period, $T=4 s$

Amplitude, $a=2 m$

At time $t=0$, OP makes an angle $\pi$ with the $x$-axis, in the anticlockwise direction. Hence, phase angle, $\Phi=+\pi$

Therefore, the equation of simple harmonic motion for the $x$-projection of $OP$, at time $t$, is given as:

$ \begin{aligned} & x=a \cos (\frac{2 \pi t}{T}+\phi)=2 \cos (\frac{2 \pi t}{4}+\pi) \\ & \therefore x=-2 \cos (\frac{\pi}{2} t) m \end{aligned} $

Answer

$ \begin{aligned} & x=-2 \sin (3 t+\frac{\pi}{3})=+2 \cos (3 t+\frac{\pi}{3}+\frac{\pi}{2}) \\ & =2 \cos (3 t+\frac{5 \pi}{6}) \end{aligned} $

If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get:

Amplitude, $A=2 cm$

Phase angle, $\phi=\frac{5 \pi}{6}=150^{\circ}$

Angular velocity, $\omega=\frac{2 \pi}{T}=3 rad / sec$.

The motion of the particle can be plotted as shown in the following figure.

$ x=\cos (\frac{\pi}{6}-t)=\cos (t-\frac{\pi}{6}) $

If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get:

Amplitude, $A=1$

Phase angle, $\phi=-\frac{\pi}{6}=-30^{\circ}$

Angular velocity, $\omega=\frac{2 \pi}{T}=1 rad / s$

The motion of the particle can be plotted as shown in the following figure.

$ \begin{aligned} & x=3 \sin (2 \pi t+\frac{\pi}{4}) \\ & =-3 \cos [(2 \pi t+\frac{\pi}{4})+\frac{\pi}{2}]=-3 \cos (2 \pi t+\frac{3 \pi}{4}) \end{aligned} $

If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get:

Amplitude, $A=3 cm$

Phase angle, $\phi=\frac{3 \pi}{4}=135^{\circ}$

Angular velocity, $\omega=\frac{2 \pi}{T}=2 \pi rad / s$

The motion of the particle can be plotted as shown in the following figure.

$x=2 \cos \pi t$

If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get:

Amplitude, $A=2 cm$

Phase angle, $\Phi=0$

Angular velocity, $\omega=\pi rad / s$

The motion of the particle can be plotted as shown in the following figure.

Answer

For the one block system:

When a force $F$, is applied to the free end of the spring, an extension $l$, is produced. For the maximum extension, it can be written as:

$F=k l$

Where, $k$ is the spring constant

Hence, the maximum extension produced in the spring, $l=\frac{F}{k}$

For the two block system:

The displacement $(x)$ produced in this case is:

$x=\frac{l}{2}$

Net force, $F=+2 k x=2 k \frac{l}{2}$

$\therefore l=\frac{F}{k}$

For the one block system:

For mass $(m)$ of the block, force is written as:

$F=m a=m \frac{d^{2} x}{d t^{2}}$

Where, $x$ is the displacement of the block in time $t$

$\therefore m \frac{d^{2} x}{d t^{2}}=-k x$

It is negative because the direction of elastic force is opposite to the direction of displacement.

$\frac{d^{2} x}{d t^{2}}=-(\frac{k}{m}) x=-\omega^{2} x$

Where, $\omega^{2}=\frac{k}{m}$

$\omega=\sqrt{\frac{k}{m}}$

Where,

$\omega$ is angular frequency of the oscillation

$\therefore$ Time period of the oscillation, $T=\frac{2 \pi}{\omega}$

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

For the two block system:

$F=m \frac{d^{2} x}{d t^{2}}$

$m \frac{d^{2} x}{d t^{2}}=-2 k x$

It is negative because the direction of elastic force is opposite to the direction of displacement.

$ \frac{d^{2} x}{d t^{2}}=-[\frac{2 k}{m}] x=-\omega^{2} x $

Where,

Angular frequency, $\omega=\sqrt{\frac{2 k}{m}}$

$\therefore$ Time period, $T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}}$

Answer

Angular frequency of the piston, $\omega=200 rad / min$.

Stroke $=1.0 m$

Amplitude, $A=\frac{1.0}{2}=0.5 m$

The maximum speed $(v _{\max })$ of the piston is give by the relation:

$ \begin{aligned} v_{\max } & =A \omega \\ & =200 \times 0.5=100 m / min \end{aligned} $

Answer

Acceleration due to gravity on the surface of moon, $g^{\prime}=1.7 m s^{-2}$

Acceleration due to gravity on the surface of earth, $g=9.8 m s^{-2}$

Time period of a simple pendulum on earth, $T=3.5 s$

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

Where,

$l$ is the length of the pendulum

$ \begin{aligned} \therefore l & =\frac{T^{2}}{(2 \pi)^{2}} \times g \\ & =\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 m \end{aligned} $

The length of the pendulum remains constant.

On moon’s surface, time period, $T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$

$ =2 \pi \sqrt{\frac{\frac{(3.5)^{2}}{\frac{4 \times(3.14)^{2}}{1.7}} \times 9.8}{1.7}}=8.4 s $

Hence, the time period of the simple pendulum on the surface of moon is $8.4 s$.

Answer

The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity $=g$

Centripetal acceleration $=\frac{v^{2}}{R}$

Where,

$v$ is the uniform speed of the car

$R$ is the radius of the track

Effective acceleration $(a _{\text{eff }})$ is given as:

$a _{\text{eff }}=\sqrt{g^{2}+(\frac{v^{2}}{R})^{2}}$

Time period, $T=2 \pi \sqrt{\frac{l}{a _{\text{eff }}}}$

Where, $l$ is the length of the pendulum

$\therefore$ Time period, $T=2 \pi \sqrt{\frac{l}{g^{2}+\frac{v^{4}}{R^{2}}}}$

Answer

Base area of the cork $=A$

Height of the cork $=h$

Density of the liquid $=\rho_l$

Density of the cork $=\rho$

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by $x$. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust $=$ Restoring force, $F=$ Weight of the extra water displaced

$F=-($ Volume $\times$ Density $\times g)$

Volume $=$ Area $\times$ Distance through which the cork is depressed

Volume $=A x$

$\therefore F=-A x^{\rho_l} g \ldots(i)$

According to the force law:

$F=k x$

$k=\frac{F}{x}$

Where, $k$ is a constant

$$ \begin{equation*} k=\frac{F}{x}=-A \rho_1 g \tag{ii} \end{equation*} $$

The time period of the oscillations of the cork:

$$ \begin{equation*} T=2 \pi \sqrt{\frac{m}{k}} \tag{iii} \end{equation*} $$

Where, $m=$ Mass of the cork

$=$ Volume of the cork $\times$ Density

$=$ Base area of the cork $\times$ Height of the cork $\times$ Density of the cork

$=A h \rho$

Hence, the expression for the time period becomes:

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

Answer

Area of cross-section of the U-tube $=A$

Density of the mercury column $=\rho$

Acceleration due to gravity $=g$

Restoring force, $F=$ Weight of the mercury column of a certain height

$F=-($ Volume $\times$ Density $\times g)$

$F=-(A \times 2 h \times \rho \times g)=-2 A \rho g h=-k \times$ Displacement in one of the arms $(h)$

Where,

$2 h$ is the height of the mercury column in the two arms

$k$ is a constant, given by $k=-\frac{F}{h}=2 A \rho g$

Time period, $T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}$

Where,

$m$ is the mass of the mercury column

Let $l$ be the length of the total mercury in the U-tube.

Mass of mercury, $m=$ Volume of mercury $\times$ Density of mercury

$=A l \rho$

$\therefore \quad T=2 \pi \sqrt{\frac{A l \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{l}{2 g}}$

Hence, the mercury column executes simple harmonic motion with time period $2 \pi \sqrt{\frac{l}{2 g}}$.