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#### Question 1:

No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.

#### Question 2:

No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.

No. The resultant force on the particle is maximum at the extreme positions.

#### Question 3:

No. The resultant force on the particle is maximum at the extreme positions.

Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.

#### Question 4:

Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.

No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.

#### Question 5:

No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.

Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.

#### Question 6:

Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.

No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.

#### Question 7:

No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.

The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.

#### Question 8:

The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.

Figure (a) shows the graph of the applied force against the position of the particle.

(a)

(b)

(c)

#### Question 9:

Figure (a) shows the graph of the applied force against the position of the particle.

(a)

(b)

(c)

No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.

However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.

#### Question 10:

No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.

However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.

Statement A is more appropriate because the energy of a system in simple harmonic motion is given by

If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A2).
i.e. E A2

Therefore, according to the relation, energy increases as the amplitude increases.

#### Question 11:

Statement A is more appropriate because the energy of a system in simple harmonic motion is given by

If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A2).
i.e. E A2

Therefore, according to the relation, energy increases as the amplitude increases.

According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$

The time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
i.e. $T\propto \sqrt{\frac{1}{g}}$

Also, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.

#### Question 12:

According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$

The time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
i.e. $T\propto \sqrt{\frac{1}{g}}$

Also, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.

No. According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its geffective (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.

#### Question 13:

No. According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its geffective (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.

The time period of a pendulum depends on the length and is given by the relation, $T=2\pi \sqrt{\frac{l}{g}}$.
As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.

#### Question 14:

The time period of a pendulum depends on the length and is given by the relation, $T=2\pi \sqrt{\frac{l}{g}}$.
As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.

Yes.

Time period of a spring mass system is given by,
$T=2\pi \sqrt{\frac{m}{k}}$                        ...(1)              .
where m is mass of the block, and
is the spring constant

Time period is also given by the relation,
$T=2\pi \sqrt{\frac{{x}_{0}}{g}}$                       ...(2)
where, x0 is extension of the spring, and
g is acceleration due to gravity

From the equations (1) and (2), we have:
$mg=k{x}_{0}$

$⇒k=\frac{mg}{{x}_{0}}$

Substituting the value of k in the above equation, we get:
$T=2\pi \sqrt{\frac{m}{\frac{mg}{{x}_{0}}}}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{x}}_{0}}{\mathrm{g}}}$
Thus, we can find the time period if the value of extension x0 is known.

#### Question 15:

Yes.

Time period of a spring mass system is given by,
$T=2\pi \sqrt{\frac{m}{k}}$                        ...(1)              .
where m is mass of the block, and
is the spring constant

Time period is also given by the relation,
$T=2\pi \sqrt{\frac{{x}_{0}}{g}}$                       ...(2)
where, x0 is extension of the spring, and
g is acceleration due to gravity

From the equations (1) and (2), we have:
$mg=k{x}_{0}$

$⇒k=\frac{mg}{{x}_{0}}$

Substituting the value of k in the above equation, we get:
$T=2\pi \sqrt{\frac{m}{\frac{mg}{{x}_{0}}}}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{x}}_{0}}{\mathrm{g}}}$
Thus, we can find the time period if the value of extension x0 is known.

When the frequency of soldiers' feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers' feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers' feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.

#### Question 16:

When the frequency of soldiers' feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers' feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers' feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.

As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.

#### Question 1:

As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.

(a) As x increases k increases.

A body is said to be in simple harmonic motion only when,
F = $-$kx                          ...(1)
where F is force,
k is force constant, and
x is displacement of the body from the mean position.

Given:
F = -k$\sqrt{x}$                         ...(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.

#### Question 2:

(a) As x increases k increases.

A body is said to be in simple harmonic motion only when,
F = $-$kx                          ...(1)
where F is force,
k is force constant, and
x is displacement of the body from the mean position.

Given:
F = -k$\sqrt{x}$                         ...(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.

(b) an extreme position

One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.

#### Question 3:

(b) an extreme position

One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.

(a) vmax

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.

#### Question 4:

(a) vmax

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.

(d) zero

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.

#### Question 5:

(d) zero

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.

(c) 4A

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 particle is,
2A + 2A = 4A.

#### Question 6:

(c) 4A

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 particle is,
2A + 2A = 4A.

(d) zero

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω2+ ( -Aω2) = 0

#### Question 7:

(d) zero

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω2+ ( -Aω2) = 0

(d) simple harmonic with amplitude  $\sqrt{{A}^{2}+{B}^{2}}$

x =
A sin ωt + B cos ωt                                           ...(1)

For a body to undergo simple harmonic motion,
acceleration, a = $-$kx.                                        ...(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude,  A $=\sqrt{{A}^{2}+{B}^{2}}$.2

#### Question 8:

(d) simple harmonic with amplitude  $\sqrt{{A}^{2}+{B}^{2}}$

x =
A sin ωt + B cos ωt                                           ...(1)

For a body to undergo simple harmonic motion,
acceleration, a = $-$kx.                                        ...(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude,  A $=\sqrt{{A}^{2}+{B}^{2}}$.2

(c) on a circle

We know,
But there is a phase difference of 90o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.

#### Question 9:

(c) on a circle

We know,
But there is a phase difference of 90o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.

(b) B

At t = 0,
Displacement $\left({x}_{0}\right)$ is given by,
x0 = A + sin ω(0) = A

Displacement x will be maximum when sinωt is 1
or,
xm = A + B

Amplitude will be:
xm$-$ xo = A + B $-$ A = B

#### Question 10:

(b) B

At t = 0,
Displacement $\left({x}_{0}\right)$ is given by,
x0 = A + sin ω(0) = A

Displacement x will be maximum when sinωt is 1
or,
xm = A + B

Amplitude will be:
xm$-$ xo = A + B $-$ A = B

(c) phase

Because the direction of motion of particles A and B is just opposite to each other.

#### Question 11:

(c) phase

Because the direction of motion of particles A and B is just opposite to each other.

(d) remain E
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 m1 be the mass of the other particle and ω1 be its angular frequency.
New angular frequency ω1 is given by,

New energy E1 is given as,

#### Question 12:

(d) remain E
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 m1 be the mass of the other particle and ω1 be its angular frequency.
New angular frequency ω1 is given by,

New energy E1 is given as,

(a)$\frac{1}{2}m{\omega }^{2}{A}^{2}$

It is the total energy in simple harmonic motion in one time period.

#### Question 13:

(a)$\frac{1}{2}m{\omega }^{2}{A}^{2}$

It is the total energy in simple harmonic motion in one time period.

(c) 2v

Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.

#### Question 14:

(c) 2v

Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.

(d) become T/$\sqrt{2}$T/2

Time period $\left(T\right)$ is given by,

where m is the mass, and
k is spring constant.

When the spring is divided into two parts, the new spring constant k1 is given as,
k1 = $2k$

New time period T1:
T= $2\mathrm{\pi }\sqrt{\frac{m}{2k}}=\frac{1}{\sqrt{2}}2\mathrm{\pi }\sqrt{\frac{m}{k}}=\frac{1}{\sqrt{2}}\mathrm{T}$

#### Question 15:

(d) become T/$\sqrt{2}$T/2

Time period $\left(T\right)$ is given by,

where m is the mass, and
k is spring constant.

When the spring is divided into two parts, the new spring constant k1 is given as,
k1 = $2k$

New time period T1:
T= $2\mathrm{\pi }\sqrt{\frac{m}{2k}}=\frac{1}{\sqrt{2}}2\mathrm{\pi }\sqrt{\frac{m}{k}}=\frac{1}{\sqrt{2}}\mathrm{T}$

(d)$\sqrt{\frac{{k}_{2}}{{k}_{1}}}$

Maximum velocity, v =
where A is amplitude and ω is the angular frequency.

Further, ω = $\sqrt{\frac{k}{m}}$
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,

#### Question 16:

(d)$\sqrt{\frac{{k}_{2}}{{k}_{1}}}$

Maximum velocity, v =
where A is amplitude and ω is the angular frequency.

Further, ω = $\sqrt{\frac{k}{m}}$
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,

(c) remain same

Because the frequency ($\nu =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$) of the system is independent of the acceleration of the system.

#### Question 17:

(c) remain same

Because the frequency ($\nu =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$) of the system is independent of the acceleration of the system.

(c) remain same

As the frequency of the system is independent of the acceleration of the system.

#### Question 18:

(c) remain same

As the frequency of the system is independent of the acceleration of the system.

(d) $\sqrt{6}$ times slower

The acceleration due to gravity at moon is g/6.

Time period of pendulum is given by,

Therefore, on moon, time period will be:

Tmoon = $2\pi \sqrt{\frac{l}{{g}_{moon}}}=2\pi \sqrt{\frac{l}{\left(g}{6}\right)}}=\sqrt{6}\left(2\pi \sqrt{\frac{l}{g}}\right)=\sqrt{6}T$

#### Question 19:

(d) $\sqrt{6}$ times slower

The acceleration due to gravity at moon is g/6.

Time period of pendulum is given by,

Therefore, on moon, time period will be:

Tmoon = $2\pi \sqrt{\frac{l}{{g}_{moon}}}=2\pi \sqrt{\frac{l}{\left(g}{6}\right)}}=\sqrt{6}\left(2\pi \sqrt{\frac{l}{g}}\right)=\sqrt{6}T$

(d) give correct time

Because the time period of a spring-mass system does not depend on the acceleration due to gravity.

#### Question 20:

(d) give correct time

Because the time period of a spring-mass system does not depend on the acceleration due to gravity.

(c) its length should be decreased to keep correct time

Time period of pendulum,
T = $2\pi \sqrt{\frac{l}{g}}$
At higher altitudes, the value of acceleration due to gravity decreases. Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.

#### Question 21:

(c) its length should be decreased to keep correct time

Time period of pendulum,
T = $2\pi \sqrt{\frac{l}{g}}$
At higher altitudes, the value of acceleration due to gravity decreases. Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.

(c) will go in a circular path

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

#### Question 1:

(c) will go in a circular path

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

(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.

#### Question 2:

(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.

(a) periodic

Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.

#### Question 3:

(a) periodic

Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.

(a) periodic

Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.

#### Question 4:

(a) periodic

Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.

(d) none of them

As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.

#### Question 5:

(d) none of them

As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.

(a) periodic
(b) oscillatory
(d) angular simple harmonic

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.

#### Question 6:

(a) periodic
(b) oscillatory
(d) angular simple harmonic

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.

In S.H.M.,
F = -kx
Therefore, $\stackrel{⇀}{F.}\stackrel{⇀}{r}$ will always be negative. As acceleration has the same direction as the force, $\stackrel{⇀}{a}.\stackrel{⇀}{r}$ will also be negative, always.

#### Question 7:

In S.H.M.,
F = -kx
Therefore, $\stackrel{⇀}{F.}\stackrel{⇀}{r}$ will always be negative. As acceleration has the same direction as the force, $\stackrel{⇀}{a}.\stackrel{⇀}{r}$ will also be negative, always.

As the direction of force and acceleration are always same, is always positive.

#### Question 8:

As the direction of force and acceleration are always same, is always positive.

As are either parallel or anti-parallel to each other, their cross products will always be zero.

#### Question 9:

As are either parallel or anti-parallel to each other, their cross products will always be zero.

(c) on a straight line
(d) periodic

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.

#### Question 1:

(c) on a straight line
(d) periodic

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.

It is given,
Amplitude of the simple harmonic motion, A =10 cm

At t = 0 and  x = 5 cm,
Time period of the simple harmonic motion, T  = 6 s

Angular frequency (ω) is given by,

Consider the equation of motion of S.H.M,
Y = Asin $\left(\omega t+\varphi \right)$                                 ...(1)
where Y is displacement of the particle, and
$\varphi$ is phase of the particle.

On substituting the values of A, t and ω in equation (1), we get:
5 = 10sin(ω × 0 + Ï•)
$⇒$5 = 10sin Ï•

Equation of displacement can be written as,

(ii) At t = 4 s,

Acceleration is given by,
a = −ω2x

#### Question 2:

It is given,
Amplitude of the simple harmonic motion, A =10 cm

At t = 0 and  x = 5 cm,
Time period of the simple harmonic motion, T  = 6 s

Angular frequency (ω) is given by,

Consider the equation of motion of S.H.M,
Y = Asin $\left(\omega t+\varphi \right)$                                 ...(1)
where Y is displacement of the particle, and
$\varphi$ is phase of the particle.

On substituting the values of A, t and ω in equation (1), we get:
5 = 10sin(ω × 0 + Ï•)
$⇒$5 = 10sin Ï•

Equation of displacement can be written as,

(ii) At t = 4 s,

Acceleration is given by,
a = −ω2x

It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms−1.
Acceleration of the particle, a = 10 ms−2.

Let $\omega$ be the angular frequency of the particle.
The acceleration of the particle is given by,
a = ω2x

Now, the amplitude A is calculated as,

#### Question 3:

It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms−1.
Acceleration of the particle, a = 10 ms−2.

Let $\omega$ be the angular frequency of the particle.
The acceleration of the particle is given by,
a = ω2x

Now, the amplitude A is calculated as,

It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm

To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
$\omega$ be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.

Equating the mathematical expressions for K.E. and P.E. of the particle, we get:

A2y2 = y2
2y2 = A2

The kinetic energy and potential energy of the particle are equal at a distance of $5\sqrt{2}$ cm from the mean position.

#### Question 4:

It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm

To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
$\omega$ be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.

Equating the mathematical expressions for K.E. and P.E. of the particle, we get:

A2y2 = y2
2y2 = A2

The kinetic energy and potential energy of the particle are equal at a distance of $5\sqrt{2}$ cm from the mean position.

It is given that:
Maximum speed of the particle, ${v}_{Max}$ = 10 cms$-1$
Maximum acceleration of the particle, ${a}_{Max}$ = 50 cms−2â€‹

The maximum velocity of a particle executing simple harmonic motion is given by,
${v}_{Max}=A\omega$
where
A is amplitude of the particle.

Substituting the value of ${v}_{Max}$ in the above expression, we get:
= 10                           $...\left(1\right)$
$⇒{\mathrm{\omega }}^{2}=\frac{100}{{A}^{2}}$

aMax = ω2A = 50 cms−1

To determine the positions where the speed of the particle is 8 ms-1, we may use the following formula:
v2ω2 (A2y2)
where y is distance of particle from the mean position, and
v is velocity of the particle.

On substituting the given values in the above equation, we get:
64 = 25 (4 − y2)
$⇒\frac{64}{25}=4-{y}^{2}$

⇒ 4 − y2 = 2.56
⇒       y2 = 1.44
⇒â€‹       y  = $\sqrt{1.44}$
⇒        y = ± 1.2 cm   (from the mean position)

#### Question 5:

It is given that:
Maximum speed of the particle, ${v}_{Max}$ = 10 cms$-1$
Maximum acceleration of the particle, ${a}_{Max}$ = 50 cms−2â€‹

The maximum velocity of a particle executing simple harmonic motion is given by,
${v}_{Max}=A\omega$
where
A is amplitude of the particle.

Substituting the value of ${v}_{Max}$ in the above expression, we get:
= 10                           $...\left(1\right)$
$⇒{\mathrm{\omega }}^{2}=\frac{100}{{A}^{2}}$

aMax = ω2A = 50 cms−1

To determine the positions where the speed of the particle is 8 ms-1, we may use the following formula:
v2ω2 (A2y2)
where y is distance of particle from the mean position, and
v is velocity of the particle.

On substituting the given values in the above equation, we get:
64 = 25 (4 − y2)
$⇒\frac{64}{25}=4-{y}^{2}$

⇒ 4 − y2 = 2.56
⇒       y2 = 1.44
⇒â€‹       y  = $\sqrt{1.44}$
⇒        y = ± 1.2 cm   (from the mean position)

Given:
Equation of motion of the particle executing S.H.M.,
...(1)
General equation of the particle is given by,
$x=A\mathrm{sin}\left(\omega t+\varphi \right)$                                        ...(2)

On comparing the equations (1) and (2) we get:

(a) Amplitude, A is 2 cm.
Angular frequency, ω is 100 s−1â€‹.

Also, we know -

(b) At t = 0 and x = 2 cm $\mathrm{sin}\frac{\mathrm{\pi }}{6}$

We know:
x = A sin (ωt + Ï•)

Using $v=\frac{dx}{dt},$ we get:
v = Aω cos (ωt + Ï•)

(c) Acceleration of the particle is given by,
a = $-{\omega }^{2}$x
= 1002×1 = 10000 cm/s2

#### Question 6:

Given:
Equation of motion of the particle executing S.H.M.,
...(1)
General equation of the particle is given by,
$x=A\mathrm{sin}\left(\omega t+\varphi \right)$                                        ...(2)

On comparing the equations (1) and (2) we get:

(a) Amplitude, A is 2 cm.
Angular frequency, ω is 100 s−1â€‹.

Also, we know -

(b) At t = 0 and x = 2 cm $\mathrm{sin}\frac{\mathrm{\pi }}{6}$

We know:
x = A sin (ωt + Ï•)

Using $v=\frac{dx}{dt},$ we get:
v = Aω cos (ωt + Ï•)

(c) Acceleration of the particle is given by,
a = $-{\omega }^{2}$x
= 1002×1 = 10000 cm/s2

Given:
The equation of motion of a particle executing S.H.M. is,

The general equation of S..H.M. is give by,

(a) Maximum displacement from the mean position is equal to the amplitude of the particle.
As the velocity of the particle is zero at extreme position, it is at rest.
x = 5, which is also the amplitude of the particle.

The particle will come to rest at

(b)  Acceleration is given as,
a = ω2x

For a = 0,

(c) The maximum speed $\left(v\right)$ is given by,
(using $v=\frac{dx}{dt}$)

For maximum velocity:

#### Question 7:

Given:
The equation of motion of a particle executing S.H.M. is,

The general equation of S..H.M. is give by,

(a) Maximum displacement from the mean position is equal to the amplitude of the particle.
As the velocity of the particle is zero at extreme position, it is at rest.
x = 5, which is also the amplitude of the particle.

The particle will come to rest at

(b)  Acceleration is given as,
a = ω2x

For a = 0,

(c) The maximum speed $\left(v\right)$ is given by,
(using $v=\frac{dx}{dt}$)

For maximum velocity:

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50$\mathrm{\pi }$t + tan−10.75)
= 2.0 cos (50$\mathrm{\pi }$t + 0.643)

(a) Velocity of the particle is given by,
$v=\frac{dx}{dt}$
v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643)

As the particle comes to rest, its velocity becomes be zero.
⇒â€‹ v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643) = 0
⇒                   sin (50$\mathrm{\pi }$t + 0.643) = 0 = sin $\mathrm{\pi }$

When the particle initially comes to rest,
50$\mathrm{\pi }$t + 0.643 = $\pi$
⇒                 t = 1.6 × 10−2s

(b) Acceleration is given by,

For maximum acceleration:
cos (50$\mathrm{\pi }$t + 0.643) = −1 = cos $\mathrm{\pi }$ (max)             (so that a is max)
⇒                             t = 1.6 ×  10−2 s

(c) When the particle comes to rest for the second time, the time is given as,
50$\mathrm{\pi }$t + 0.643 = 2$\mathrm{\pi }$
⇒ â€‹                    t = 3.6 × 10−2 s

#### Question 8:

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50$\mathrm{\pi }$t + tan−10.75)
= 2.0 cos (50$\mathrm{\pi }$t + 0.643)

(a) Velocity of the particle is given by,
$v=\frac{dx}{dt}$
v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643)

As the particle comes to rest, its velocity becomes be zero.
⇒â€‹ v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643) = 0
⇒                   sin (50$\mathrm{\pi }$t + 0.643) = 0 = sin $\mathrm{\pi }$

When the particle initially comes to rest,
50$\mathrm{\pi }$t + 0.643 = $\pi$
⇒                 t = 1.6 × 10−2s

(b) Acceleration is given by,

For maximum acceleration:
cos (50$\mathrm{\pi }$t + 0.643) = −1 = cos $\mathrm{\pi }$ (max)             (so that a is max)
⇒                             t = 1.6 ×  10−2 s

(c) When the particle comes to rest for the second time, the time is given as,
50$\mathrm{\pi }$t + 0.643 = 2$\mathrm{\pi }$
⇒ â€‹                    t = 3.6 × 10−2 s

As per the conditions given in the question,
(for the given two positions)

Let y1and y2 be the displacements at the two positions and A be the amplitude.
Equation of motion for the displacement at the first position is given by,
y1 = Asinωt1
As displacement is equal to the half of the amplitude,

The displacement at second position is given by,
y2 = A sin ωt2
As displacement is equal to the amplitude at this position,
⇒        A = A sin ωt2
⇒ sinωt2 = 1

#### Question 9:

As per the conditions given in the question,
(for the given two positions)

Let y1and y2 be the displacements at the two positions and A be the amplitude.
Equation of motion for the displacement at the first position is given by,
y1 = Asinωt1
As displacement is equal to the half of the amplitude,

The displacement at second position is given by,
y2 = A sin ωt2
As displacement is equal to the amplitude at this position,
⇒        A = A sin ωt2
⇒ sinωt2 = 1

Given:
Spring constant, k =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.

The relation between time period and spring constant is given as,

On substituting the respective values, we get:

#### Question 10:

Given:
Spring constant, k =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.

The relation between time period and spring constant is given as,

On substituting the respective values, we get:

An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,
${T}_{\mathrm{p}}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)}$
where l is the length of the pendulum, and
g is acceleration due to gravity.

Time period of the spring is given by,
${T}_{\mathrm{s}}=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where m is the mass, and
is the spring constant.

Let x be the extension of the spring.
For small frequency, TP â€‹can be taken as equal to TS.

$⇒\sqrt{\left(\frac{l}{g}\right)}=\sqrt{\left(\frac{m}{k}\right)}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{l}{g}\right)=\left(\frac{m}{k}\right)\phantom{\rule{0ex}{0ex}}⇒l=\frac{mg}{k}=\frac{F}{k}=x$
($\because$ restoring force = weight = mg

l = x (proved)

#### Question 11:

An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,
${T}_{\mathrm{p}}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)}$
where l is the length of the pendulum, and
g is acceleration due to gravity.

Time period of the spring is given by,
${T}_{\mathrm{s}}=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where m is the mass, and
is the spring constant.

Let x be the extension of the spring.
For small frequency, TP â€‹can be taken as equal to TS.

$⇒\sqrt{\left(\frac{l}{g}\right)}=\sqrt{\left(\frac{m}{k}\right)}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{l}{g}\right)=\left(\frac{m}{k}\right)\phantom{\rule{0ex}{0ex}}⇒l=\frac{mg}{k}=\frac{F}{k}=x$
($\because$ restoring force = weight = mg

l = x (proved)

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T  = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5$×$10 = 5 kg

Total force exerted on the block = Weight of the block + spring force

Periodic time of spring is given by,

∴ The force exerted by the spring on the block $\left(F\right)$ is,
F = kx = 200.0 × 0.1 = 20 N

Maximum force = F + weight of the block
= 20 + 5 = 25 N

#### Question 12:

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T  = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5$×$10 = 5 kg

Total force exerted on the block = Weight of the block + spring force

Periodic time of spring is given by,

∴ The force exerted by the spring on the block $\left(F\right)$ is,
F = kx = 200.0 × 0.1 = 20 N

Maximum force = F + weight of the block
= 20 + 5 = 25 N

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where k is the spring constant.

On substituting the respective values, we get:

As the restoring force is balanced by the weight, we can write:
F = mg = kx
$⇒x=\frac{mg}{k}=\frac{2×10}{5}=4$

∴ Potential Energy $\left(U\right)$ of the spring is,

#### Question 10:

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where k is the spring constant.

On substituting the respective values, we get:

As the restoring force is balanced by the weight, we can write:
F = mg = kx
$⇒x=\frac{mg}{k}=\frac{2×10}{5}=4$

∴ Potential Energy $\left(U\right)$ of the spring is,

(a) displacement from the mean position

For S.H.M.,
F = -kx
ma = - kx                  (F = ma)
or,
a = $-\frac{k}{m}x$
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.

#### Question 11:

(a) displacement from the mean position

For S.H.M.,
F = -kx
ma = - kx                  (F = ma)
or,
a = $-\frac{k}{m}x$
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.

(a) on a straight line
(c) periodic
(d) simple harmonic

The given equation is a solution to the equation of simple harmonic motion. The amplitude is $\left(\stackrel{⇀}{i}+2\stackrel{⇀}{j}\right)A$, following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.

#### Question 12:

(a) on a straight line
(c) periodic
(d) simple harmonic

The given equation is a solution to the equation of simple harmonic motion. The amplitude is $\left(\stackrel{⇀}{i}+2\stackrel{⇀}{j}\right)A$, following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.

(d) with time period $\frac{\pi }{\omega }$

Given equation:
x = xosin2ωt

⇒â€‹

Now, the amplitude of the particle is xo/2 and the angular frequency of the SHM is 2ω.

Thus, time period of the SHM =

#### Question 13:

(d) with time period $\frac{\pi }{\omega }$

Given equation:
x = xosin2ωt

⇒â€‹

Now, the amplitude of the particle is xo/2 and the angular frequency of the SHM is 2ω.

Thus, time period of the SHM =

(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,

The potential energy is calculated as,

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.

#### Question 14:

(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,

The potential energy is calculated as,

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.

(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.

#### Question 15:

(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.

(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.

#### Question 16:

(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.

(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.

#### Question 16:

(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.

It is given that:
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N

(a) In the equilibrium position,
$F=kx$
where x is the compression of the spring, and
k is the spring constant.

(b) The blow imparts a speed of 2 ms-1 to the block, towards left.
Potential energy of spring, U = $\frac{1}{2}k{x}^{2}$
Kinetic energy, $K$ = $\frac{1}{2}M{v}^{2}$

(c) Time period $\left(T\right)$ is given by,

(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, compression of the spring will be (A + x).

As the total energy in S.H.M.  remains constant, we can write:

$\therefore$ 50(A + 0.1)2 = 2.5 + 10x
$⇒$ 50A2 + 0.5 + 10A = 2.5 + 10A
$⇒$ 50A2 = 2

(e) Potential Energy at the left extreme will be,

(f) Potential Energy at the right extreme is calculated as:

Distance between the two extremes = 2A
$\mathrm{P}.\mathrm{E}.=\frac{1}{2}k{\left(A+x\right)}^{2}-F\left(2A\right)$
= 4.5 − 10 (0.4) = 0.5 J

As the work is done by the external force of 10 N, different values of options (b), (e) and (f) do not violate the law of conservation of energy.

#### Question 17:

It is given that:
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N

(a) In the equilibrium position,
$F=kx$
where x is the compression of the spring, and
k is the spring constant.

(b) The blow imparts a speed of 2 ms-1 to the block, towards left.
Potential energy of spring, U = $\frac{1}{2}k{x}^{2}$
Kinetic energy, $K$ = $\frac{1}{2}M{v}^{2}$

(c) Time period $\left(T\right)$ is given by,

(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, compression of the spring will be (A + x).

As the total energy in S.H.M.  remains constant, we can write:

$\therefore$ 50(A + 0.1)2 = 2.5 + 10x
$⇒$ 50A2 + 0.5 + 10A = 2.5 + 10A
$⇒$ 50A2 = 2

(e) Potential Energy at the left extreme will be,

(f) Potential Energy at the right extreme is calculated as:

Distance between the two extremes = 2A
$\mathrm{P}.\mathrm{E}.=\frac{1}{2}k{\left(A+x\right)}^{2}-F\left(2A\right)$
= 4.5 − 10 (0.4) = 0.5 J

As the work is done by the external force of 10 N, different values of options (b), (e) and (f) do not violate the law of conservation of energy.

(a) Spring constant of a parallel combination of springs is given as,
K = k1 + k2  (parallel)
Using the relation of time period for S.H.M. for the given spring-mass system, we have:
$T=2\mathrm{\pi }\sqrt{\frac{m}{K}}=2\mathrm{\pi }\sqrt{\frac{m}{{k}_{1}+{k}_{2}}}$

(b) Let x be the displacement of the block of mass m, towards left.
Resultant force is calculated as,
F = F1 + F2 = (k1 + k2)x

Acceleration $\left(a\right)$ is given by,
$a=\left(\frac{\mathrm{F}}{m}\right)=\frac{\left({k}_{1}+{k}_{2}\right)}{m}x$

Time period $\left(T\right)$ is given by,

Required spring constant, K = k1 + k2

(c) Let K be the equivalent spring constant of the series combination.

#### Question 18:

(a) Spring constant of a parallel combination of springs is given as,
K = k1 + k2  (parallel)
Using the relation of time period for S.H.M. for the given spring-mass system, we have:
$T=2\mathrm{\pi }\sqrt{\frac{m}{K}}=2\mathrm{\pi }\sqrt{\frac{m}{{k}_{1}+{k}_{2}}}$

(b) Let x be the displacement of the block of mass m, towards left.
Resultant force is calculated as,
F = F1 + F2 = (k1 + k2)x

Acceleration $\left(a\right)$ is given by,
$a=\left(\frac{\mathrm{F}}{m}\right)=\frac{\left({k}_{1}+{k}_{2}\right)}{m}x$

Time period $\left(T\right)$ is given by,

Required spring constant, K = k1 + k2

(c) Let K be the equivalent spring constant of the series combination.

(a) We know-
f = kx
$⇒x=\frac{F}{k}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}=\frac{F}{m}$

Using the relation of time period of S.H.M.,

.

Amplitude = Maximum displacement
$=\frac{F}{k}$

When the block passes through the equilibrium position, the energy contained by the spring is given by,
$E=\frac{1}{2}k{x}^{2}=\frac{1}{2}k{\left(\frac{F}{k}\right)}^{2}=\frac{1}{2}\left(\frac{{F}^{2}}{k}\right)$

(b) At the mean position, potential energy is zero.
Kinetic energy is given by,
$\frac{1}{2}k{x}^{2}=\frac{1}{2}\frac{{\mathrm{F}}^{2}}{k}$

#### Question 19:

(a) We know-
f = kx
$⇒x=\frac{F}{k}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}=\frac{F}{m}$

Using the relation of time period of S.H.M.,

.

Amplitude = Maximum displacement
$=\frac{F}{k}$

When the block passes through the equilibrium position, the energy contained by the spring is given by,
$E=\frac{1}{2}k{x}^{2}=\frac{1}{2}k{\left(\frac{F}{k}\right)}^{2}=\frac{1}{2}\left(\frac{{F}^{2}}{k}\right)$

(b) At the mean position, potential energy is zero.
Kinetic energy is given by,
$\frac{1}{2}k{x}^{2}=\frac{1}{2}\frac{{\mathrm{F}}^{2}}{k}$

(a) Let us push the particle lightly against the spring C through displacement x.

As a result of this movement, the resultant force on the particle is kxâ€‹.
The force on the particle due to springs A and B is $\frac{kx}{\sqrt{2}}$.
Total Resultant force$=kx+\sqrt{{\left(\frac{kx}{\sqrt{2}}\right)}^{2}+{\left(\frac{kx}{\sqrt{2}}\right)}^{2}}$
= kx + kx = 2kx
Acceleration is given by $=\frac{2kx}{m}$

#### Question 20:

(a) Let us push the particle lightly against the spring C through displacement x.

As a result of this movement, the resultant force on the particle is kxâ€‹.
The force on the particle due to springs A and B is $\frac{kx}{\sqrt{2}}$.
Total Resultant force$=kx+\sqrt{{\left(\frac{kx}{\sqrt{2}}\right)}^{2}+{\left(\frac{kx}{\sqrt{2}}\right)}^{2}}$
= kx + kx = 2kx
Acceleration is given by $=\frac{2kx}{m}$

As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.

If the angle between each pair of the springs is 120Ëš then the net force applied by the springs A and B is given as,

Total resultant force$\left(F\right)$ acting on mass m will be,

#### Question 21:

As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.

If the angle between each pair of the springs is 120Ëš then the net force applied by the springs A and B is given as,

Total resultant force$\left(F\right)$ acting on mass m will be,

As the block of mass M is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.

Now, springs k2 and k3 are in connected as a series combination.
Let k4be the equivalent spring constant.
$\therefore \frac{1}{{k}_{4}}=\frac{1}{{k}_{2}}+\frac{1}{{k}_{3}}=\frac{{k}_{2}+{k}_{3}}{{k}_{2}{k}_{3}}\phantom{\rule{0ex}{0ex}}{k}_{4}=\frac{{k}_{2}{k}_{3}}{{k}_{2}+{k}_{3}}$

k4 and k1 form a parallel combination of springs. Hence, equivalent spring constant k = k1 + k4.

(b) Frequency$\left(v\right)$ is given by,
$v=\frac{1}{T}$
$=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{k}_{2}{k}_{3}+{k}_{1}{k}_{2}+{k}_{1}{k}_{3}}{M\left({k}_{2}+{k}_{3}\right)}}$

(c) Amplitude ( x ) is given by,
$x=\frac{F}{k}=\frac{F\left({k}_{2}+{k}_{3}\right)}{{k}_{1}{k}_{2}+{k}_{2}{k}_{3}+{k}_{1}{k}_{3}}$

#### Question 22:

As the block of mass M is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.

Now, springs k2 and k3 are in connected as a series combination.
Let k4be the equivalent spring constant.
$\therefore \frac{1}{{k}_{4}}=\frac{1}{{k}_{2}}+\frac{1}{{k}_{3}}=\frac{{k}_{2}+{k}_{3}}{{k}_{2}{k}_{3}}\phantom{\rule{0ex}{0ex}}{k}_{4}=\frac{{k}_{2}{k}_{3}}{{k}_{2}+{k}_{3}}$

k4 and k1 form a parallel combination of springs. Hence, equivalent spring constant k = k1 + k4.

(b) Frequency$\left(v\right)$ is given by,
$v=\frac{1}{T}$
$=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{k}_{2}{k}_{3}+{k}_{1}{k}_{2}+{k}_{1}{k}_{3}}{M\left({k}_{2}+{k}_{3}\right)}}$

(c) Amplitude ( x ) is given by,
$x=\frac{F}{k}=\frac{F\left({k}_{2}+{k}_{3}\right)}{{k}_{1}{k}_{2}+{k}_{2}{k}_{3}+{k}_{1}{k}_{3}}$

All three spring attached to the mass M are in series.
k1, k2, k3 are the spring constants.
Let k be the resultant spring constant.

As force is equal to the weight of the body,
F = weight = Mg
Let x1, x2, and x3 be the displacements of the springs having spring constants k1, k2 and k3 respectively.
â€‹For spring k1,

#### Question 13:

All three spring attached to the mass M are in series.
k1, k2, k3 are the spring constants.
Let k be the resultant spring constant.

As force is equal to the weight of the body,
F = weight = Mg
Let x1, x2, and x3 be the displacements of the springs having spring constants k1, k2 and k3 respectively.
â€‹For spring k1,

It is given that:
Energy stored in the spring, E = 5 J
Frequency of the mass-spring system, f = 5
Extension in the length of the spring, x = 25 cm = 0.25 m

#### Question 14:

It is given that:
Energy stored in the spring, E = 5 J
Frequency of the mass-spring system, f = 5
Extension in the length of the spring, x = 25 cm = 0.25 m

(a) Consider the free body diagram.
Weight of the body, W = mg
Force, F = ma = mω2x

x is the small displacement of mass m.
As normal reaction R is acting vertically in the upward direction, we can write:
R + mω2xmg = 0                ....(1)

Resultant force = mω2x = mgR

(b) R = mgmω2x
$=mg-m\frac{k}{M+N}x\phantom{\rule{0ex}{0ex}}=mg-\frac{mkx}{M+N}$
It can be seen from the above equations that, for R to be smallest, the value of mω2x should be maximum which is only possible when the particle is at the highest point.

(c) R = mgmω2x
As the two blocks oscillate together becomes greater than zero.
When limiting condition follows,
i.e. R = 0
mg = mω2x
$x=\frac{mg}{m{\mathrm{\omega }}^{2}}=\frac{mg·\left(M+m\right)}{mk}$

Required maximum amplitude$=\frac{g\left(M+m\right)}{k}$

#### Question 15:

(a) Consider the free body diagram.
Weight of the body, W = mg
Force, F = ma = mω2x

x is the small displacement of mass m.
As normal reaction R is acting vertically in the upward direction, we can write:
R + mω2xmg = 0                ....(1)

Resultant force = mω2x = mgR

(b) R = mgmω2x
$=mg-m\frac{k}{M+N}x\phantom{\rule{0ex}{0ex}}=mg-\frac{mkx}{M+N}$
It can be seen from the above equations that, for R to be smallest, the value of mω2x should be maximum which is only possible when the particle is at the highest point.

(c) R = mgmω2x
As the two blocks oscillate together becomes greater than zero.
When limiting condition follows,
i.e. R = 0
mg = mω2x
$x=\frac{mg}{m{\mathrm{\omega }}^{2}}=\frac{mg·\left(M+m\right)}{mk}$

Required maximum amplitude$=\frac{g\left(M+m\right)}{k}$

(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m1 + m2) g sin θ

At equilibrium,
kx = (m1 + m2) g sin θ

(b) It is given that:
Distance at which the spring is pushed,

As the system is released, it executes S.H.M.
where $\mathrm{\omega }=\sqrt{\frac{k}{{m}_{1}+{m}_{2}}}$

When the blocks lose contact, becomes zero.                   (is the force exerted by mass m1on mass m2)

Therefore, the blocks lose contact with each other when the spring attains its natural length.

(c) Let v be the common speed attained by both the blocks.

#### Question 23:

(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m1 + m2) g sin θ

At equilibrium,
kx = (m1 + m2) g sin θ

(b) It is given that:
Distance at which the spring is pushed,

As the system is released, it executes S.H.M.
where $\mathrm{\omega }=\sqrt{\frac{k}{{m}_{1}+{m}_{2}}}$

When the blocks lose contact, becomes zero.                   (is the force exerted by mass m1on mass m2)

Therefore, the blocks lose contact with each other when the spring attains its natural length.

(c) Let v be the common speed attained by both the blocks.

Let l be the extension in the spring when massis hung.

Let T1 be the tension in the string; its value is given by,
T1 =kl = mg
Let x be the extension in the string on applying a force F.
Then, the new value of tension T2 is given by,
T2 = k(x + l)
Driving force is the difference between tensions T1 and T2.
∴ Driving force = T2T1 = k(x + l) − kl
= kx

#### Question 24:

Let l be the extension in the spring when massis hung.

Let T1 be the tension in the string; its value is given by,
T1 =kl = mg
Let x be the extension in the string on applying a force F.
Then, the new value of tension T2 is given by,
T2 = k(x + l)
Driving force is the difference between tensions T1 and T2.
∴ Driving force = T2T1 = k(x + l) − kl
= kx

Let us try to solve the problem using energy method.
If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by,
$\frac{mg}{k}$
Let x be any position below the equilibrium during oscillation.
Let v be the velocity of mass m and ω be the angular velocity of the pulley.
If r is the radius of the pulley then
v = rω
As total energy remains constant for simple harmonic motion, we can write:

By taking derivatives with respect to t, on both sides, we have:

#### Question 25:

Let us try to solve the problem using energy method.
If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by,
$\frac{mg}{k}$
Let x be any position below the equilibrium during oscillation.
Let v be the velocity of mass m and ω be the angular velocity of the pulley.
If r is the radius of the pulley then
v = rω
As total energy remains constant for simple harmonic motion, we can write:

By taking derivatives with respect to t, on both sides, we have:

The centre of mass of the system should not change during simple harmonic motion.
Therefore, if the block m on the left hand side moves towards right by distance x, the block on the right hand side should also move towards left by distance x. The total compression of the spring is 2x.
If v is the velocity of the block. Then
Using energy method, we can write:
$\frac{1}{2}k{\left(2x\right)}^{2}+\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}=\mathrm{C}$
mv2 + 2kx2 = C
By taking the derivative of both sides with respect to t, we get:

#### Question 26:

The centre of mass of the system should not change during simple harmonic motion.
Therefore, if the block m on the left hand side moves towards right by distance x, the block on the right hand side should also move towards left by distance x. The total compression of the spring is 2x.
If v is the velocity of the block. Then
Using energy method, we can write:
$\frac{1}{2}k{\left(2x\right)}^{2}+\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}=\mathrm{C}$
mv2 + 2kx2 = C
By taking the derivative of both sides with respect to t, we get:

Let m is the mass of rectangular plate and x is the displacement of the rectangular plate.
During the oscillation, the centre of mass does not change.
Driving force$\left(F\right)$ is given as,
F = mgsin θ
Comparing the above equation with F = ma, we get:

For small values of θ, sinθ can be taken as equal to θ.
Thus, the above equation reduces to:

It can be seen from the above equation that, a α x.
Hence, the motion is simple harmonic.
Time period of simple harmonic motion $\left(T\right)$ is given by,

#### Question 27:

Let m is the mass of rectangular plate and x is the displacement of the rectangular plate.
During the oscillation, the centre of mass does not change.
Driving force$\left(F\right)$ is given as,
F = mgsin θ
Comparing the above equation with F = ma, we get:

For small values of θ, sinθ can be taken as equal to θ.
Thus, the above equation reduces to:

It can be seen from the above equation that, a α x.
Hence, the motion is simple harmonic.
Time period of simple harmonic motion $\left(T\right)$ is given by,

It is given that:
Amplitude of simple harmonic motion, x  = 0.1 m
Total mass of the system, M = 3 + 1 = 4 kg          (when both the blocks move together)
Spring constant, k = 100 N/m
â€‹Time period of SHM $\left(T\right)$ is given by,

Let v be the velocity of the 1 kg block, at mean position.

where x = amplitude = 0.1 m

When the 3 kg block is gently placed on the 1 kg block, the 4 kg mass and the spring become one system. As a spring-mass system experiences external force, momentum should be conserved.
Let V be the velocity of 4 kg block.
Now,
Initial momentum = Final momentum
∴ 1 × v = 4 × V

Thus, at the mean position, two blocks have a velocity of $\frac{1}{4}{\mathrm{ms}}^{-1}$.

At the extreme position, the spring-mass system has only potential energy.
$\mathrm{PE}=\frac{1}{2}k{\mathrm{\delta }}^{2}=\frac{1}{2}×\frac{1}{4}$
where δ is the new amplitude.

#### Question 28:

It is given that:
Amplitude of simple harmonic motion, x  = 0.1 m
Total mass of the system, M = 3 + 1 = 4 kg          (when both the blocks move together)
Spring constant, k = 100 N/m
â€‹Time period of SHM $\left(T\right)$ is given by,

Let v be the velocity of the 1 kg block, at mean position.

where x = amplitude = 0.1 m

When the 3 kg block is gently placed on the 1 kg block, the 4 kg mass and the spring become one system. As a spring-mass system experiences external force, momentum should be conserved.
Let V be the velocity of 4 kg block.
Now,
Initial momentum = Final momentum
∴ 1 × v = 4 × V

Thus, at the mean position, two blocks have a velocity of $\frac{1}{4}{\mathrm{ms}}^{-1}$.

At the extreme position, the spring-mass system has only potential energy.
$\mathrm{PE}=\frac{1}{2}k{\mathrm{\delta }}^{2}=\frac{1}{2}×\frac{1}{4}$
where δ is the new amplitude.

According to the question, the collision is elastic and the surface is frictionless, therefore, when the left block A moves with speed v and collides with the right block B, it transfers all the energy to the right block B.
The left block A moves a distance x against the spring; the right block returns to the original position and completes half of the oscillation.

Therefore, the period of right block B will be,
$T=\frac{2\pi \sqrt{\left(\frac{m}{k}\right)}}{2}=\pi \sqrt{\left(\frac{m}{k}\right)}$
Right block B collides with left block A and comes to rest.
Let L be the distance moved by the block to return to its original position.
The time taken is given by,
$\frac{\mathit{L}}{\mathit{V}}+\frac{\mathit{L}}{\mathit{V}}=2\left(\frac{\mathit{L}}{\mathit{V}}\right)$
Hence, time period of the periodic motion is, $2\frac{\mathit{L}}{\mathit{V}}+\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$.

#### Question 29:

According to the question, the collision is elastic and the surface is frictionless, therefore, when the left block A moves with speed v and collides with the right block B, it transfers all the energy to the right block B.
The left block A moves a distance x against the spring; the right block returns to the original position and completes half of the oscillation.

Therefore, the period of right block B will be,
$T=\frac{2\pi \sqrt{\left(\frac{m}{k}\right)}}{2}=\pi \sqrt{\left(\frac{m}{k}\right)}$
Right block B collides with left block A and comes to rest.
Let L be the distance moved by the block to return to its original position.
The time taken is given by,
$\frac{\mathit{L}}{\mathit{V}}+\frac{\mathit{L}}{\mathit{V}}=2\left(\frac{\mathit{L}}{\mathit{V}}\right)$
Hence, time period of the periodic motion is, $2\frac{\mathit{L}}{\mathit{V}}+\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$.

Let t1 and t2  be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a1 =g sin 45°
The distance travelled along AB is s1.

Let v be the velocity at point B, and
u be the initial velocity.

Using the third equation of motion, we have:
v2u2 = 2a1s1

For the distance BC,
Acceleration, a2 = $-$gsin 60°

Thus, the total time period, t = 2(t1 + t2) = 2 (0.2 + 0.163) = 0.73 s

#### Question 30:

Let t1 and t2  be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a1 =g sin 45°
The distance travelled along AB is s1.

Let v be the velocity at point B, and
u be the initial velocity.

Using the third equation of motion, we have:
v2u2 = 2a1s1

For the distance BC,
Acceleration, a2 = $-$gsin 60°

Thus, the total time period, t = 2(t1 + t2) = 2 (0.2 + 0.163) = 0.73 s

Let x1 and x2 be the amplitudes of oscillation of masses m and M respectively.

(a) As the centre of mass should not change during the motion, we can write:
mx1 = Mx2                            $\dots \left(1\right)$

Let k be the spring constant. By conservation of energy, we have:

where x0 is the length to which spring is stretched.

From equation (2) we have,

On substituting the value of x2 from equation (1) in equation (2), we get:
${x}_{0}={x}_{1}+\frac{m{x}_{1}}{M}\phantom{\rule{0ex}{0ex}}⇒{x}_{0}=\left(1+\frac{m}{M}\right){x}_{1}\phantom{\rule{0ex}{0ex}}⇒{x}_{1}=\left(\frac{M}{M+m}\right){x}_{0}$

On substituting the value of x1 from above equation, we get:

Thus, the amplitude of the simple harmonic motion of a car, as seen from the road is $\frac{m{x}_{0}}{M+m}$.

(b) At any position,
Let v1 and v2 be the velocities.

Using law of conservation of energy we have,

Here, (v1v2) is the absolute velocity of mass m as seen from the road.

Now, from the principle of conservation of momentum, we have:
Mx2 = mx1

Putting the above values in equation (3), we get:
$\frac{1}{2}\mathrm{M}{v}_{2}^{2}+\frac{1}{2}m\frac{{\mathrm{M}}^{2}}{{m}^{2}}{v}_{2}^{2}+\frac{1}{2}k{x}_{2}^{2}{\left(1+\frac{\mathrm{M}}{m}\right)}^{2}=\mathrm{constant}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{M}\left(1+\frac{\mathrm{M}}{m}\right){v}_{2}^{2}+k\left(1+\frac{\mathrm{M}}{m}\right){x}_{2}^{2}=\mathrm{constant}\phantom{\rule{0ex}{0ex}}⇒\mathrm{M}{v}_{2}^{2}+k\left(1+\frac{\mathrm{M}}{m}\right){x}_{2}^{2}=\mathrm{constant}$

Taking derivative of both the sides, we get:

#### Question 31:

Let x1 and x2 be the amplitudes of oscillation of masses m and M respectively.

(a) As the centre of mass should not change during the motion, we can write:
mx1 = Mx2                            $\dots \left(1\right)$

Let k be the spring constant. By conservation of energy, we have:

where x0 is the length to which spring is stretched.

From equation (2) we have,

On substituting the value of x2 from equation (1) in equation (2), we get:
${x}_{0}={x}_{1}+\frac{m{x}_{1}}{M}\phantom{\rule{0ex}{0ex}}⇒{x}_{0}=\left(1+\frac{m}{M}\right){x}_{1}\phantom{\rule{0ex}{0ex}}⇒{x}_{1}=\left(\frac{M}{M+m}\right){x}_{0}$

On substituting the value of x1 from above equation, we get:

Thus, the amplitude of the simple harmonic motion of a car, as seen from the road is $\frac{m{x}_{0}}{M+m}$.

(b) At any position,
Let v1 and v2 be the velocities.

Using law of conservation of energy we have,

Here, (v1v2) is the absolute velocity of mass m as seen from the road.

Now, from the principle of conservation of momentum, we have:
Mx2 = mx1

Putting the above values in equation (3), we get:
$\frac{1}{2}\mathrm{M}{v}_{2}^{2}+\frac{1}{2}m\frac{{\mathrm{M}}^{2}}{{m}^{2}}{v}_{2}^{2}+\frac{1}{2}k{x}_{2}^{2}{\left(1+\frac{\mathrm{M}}{m}\right)}^{2}=\mathrm{constant}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{M}\left(1+\frac{\mathrm{M}}{m}\right){v}_{2}^{2}+k\left(1+\frac{\mathrm{M}}{m}\right){x}_{2}^{2}=\mathrm{constant}\phantom{\rule{0ex}{0ex}}⇒\mathrm{M}{v}_{2}^{2}+k\left(1+\frac{\mathrm{M}}{m}\right){x}_{2}^{2}=\mathrm{constant}$

Taking derivative of both the sides, we get:

Let x be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement x.

At the displaced position,
R1 + R2 = mg

Taking moment about g, we get:

#### Question 32:

Let x be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement x.

At the displaced position,
R1 + R2 = mg

Taking moment about g, we get:

It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g = $\mathrm{\pi }$2 ms−2
The relation between time period and acceleration due to gravity is given by,
$T=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)}$
where l is the length of the second pendulum.

Substituting the values of T and g, we get:

Hence, the length of the pendulum is 1 m.

#### Question 33:

It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g = $\mathrm{\pi }$2 ms−2
The relation between time period and acceleration due to gravity is given by,
$T=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)}$
where l is the length of the second pendulum.

Substituting the values of T and g, we get:

Hence, the length of the pendulum is 1 m.

#### Question 46:

It is given that a car is moving with speed v on a circular horizontal road of radius r.
(a) Let T be the tension in the string.

According to the free body diagram, the value of T is given as,
$T=\sqrt{{\left(mg\right)}^{2}+{\left(\frac{m{v}^{2}}{r}\right)}^{2}}$
$=m\sqrt{{g}^{2}+\frac{{v}^{4}}{{r}^{2}}}=ma,$
where acceleration, a $=\sqrt{{g}^{2}+\frac{{v}^{4}}{{r}^{2}}}$
The time period $\left(T\right)$ is given by,

#### Question 34:

It is given that a car is moving with speed v on a circular horizontal road of radius r.
(a) Let T be the tension in the string.

According to the free body diagram, the value of T is given as,
$T=\sqrt{{\left(mg\right)}^{2}+{\left(\frac{m{v}^{2}}{r}\right)}^{2}}$
$=m\sqrt{{g}^{2}+\frac{{v}^{4}}{{r}^{2}}}=ma,$
where acceleration, a $=\sqrt{{g}^{2}+\frac{{v}^{4}}{{r}^{2}}}$
The time period $\left(T\right)$ is given by,

Given,
Time period of the clock pendulum = 2.04 s

The number of oscillations made by the pendulum in one day is calculated as
= 43200

In each oscillation, the clock gets slower by (2.04 − 2.00) s, i.e., 0.04 s.
In one day, it is slowed by = 43200 × (0.04)
= 28.8 min
Thus, the clock runs 28.8 minutes slow during 24 hours.

#### Question 35:

Given,
Time period of the clock pendulum = 2.04 s

The number of oscillations made by the pendulum in one day is calculated as
= 43200

In each oscillation, the clock gets slower by (2.04 − 2.00) s, i.e., 0.04 s.
In one day, it is slowed by = 43200 × (0.04)
= 28.8 min
Thus, the clock runs 28.8 minutes slow during 24 hours.

Let T1 be the time period of pendulum clock at a place where acceleration due to gravity $\left({g}_{1}\right)$ is 9.8 ms−2.
Let T1 = 2 s
g1 = 9.8 ms$-2$

Let T2 be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is $\left({g}_{2}\right)$.

As $T\propto \frac{1}{\sqrt{g}}$
$\therefore \frac{{T}_{1}}{{T}_{2}}=\sqrt{\left(\frac{{g}_{2}}{{g}_{1}}\right)}$

#### Question 36:

Let T1 be the time period of pendulum clock at a place where acceleration due to gravity $\left({g}_{1}\right)$ is 9.8 ms−2.
Let T1 = 2 s
g1 = 9.8 ms$-2$

Let T2 be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is $\left({g}_{2}\right)$.

As $T\propto \frac{1}{\sqrt{g}}$
$\therefore \frac{{T}_{1}}{{T}_{2}}=\sqrt{\left(\frac{{g}_{2}}{{g}_{1}}\right)}$

It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms-2
Acceleration due to gravity at the moon, g' = 1.67 ms-2

(a) Time period $\left(T\right)$ is given by,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$

i.e. the body will take  2$\mathrm{\pi }$(0.7) seconds to complete an oscillation.

Now, frequency $\left(f\right)$ is given by,
$f=\frac{1}{T}$

(b) Let $g\text{'}$ be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon $\left(T\text{'}\right)$, is given as:
$T\text{'}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g\text{'}}\right)}$
On substituting the respective values in the above formula, we get:
$T\text{'}=2\mathrm{\pi }\sqrt{\frac{5}{1.67}}$
Therefore, frequency $\left(f\text{'}\right)$ will be,

#### Question 37:

It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms-2
Acceleration due to gravity at the moon, g' = 1.67 ms-2

(a) Time period $\left(T\right)$ is given by,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$

i.e. the body will take  2$\mathrm{\pi }$(0.7) seconds to complete an oscillation.

Now, frequency $\left(f\right)$ is given by,
$f=\frac{1}{T}$

(b) Let $g\text{'}$ be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon $\left(T\text{'}\right)$, is given as:
$T\text{'}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g\text{'}}\right)}$
On substituting the respective values in the above formula, we get:
$T\text{'}=2\mathrm{\pi }\sqrt{\frac{5}{1.67}}$
Therefore, frequency $\left(f\text{'}\right)$ will be,

Let the speed of bob of the pendulum at an angle $\theta$ be v.
Using the principle of conservation of energy between the mean and extreme positions, we get:
$\frac{1}{2}$mv2 − 0 = mgl(1 − cos θ)
v2 = 2gl(1 − cos θ)                  ...(1)

In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
Tmax = mg + 2mg(1 − cos θ)
Minimum tension at the extreme position is given by
Tmin = m g cosθ
According to the question,
Tmax = 2Tmin
mg + 2mg − 2m g cosθ = 2m g cosθ
⇒ 3mg = 4 mg cosθ

#### Question 38:

Let the speed of bob of the pendulum at an angle $\theta$ be v.
Using the principle of conservation of energy between the mean and extreme positions, we get:
$\frac{1}{2}$mv2 − 0 = mgl(1 − cos θ)
v2 = 2gl(1 − cos θ)                  ...(1)

In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
Tmax = mg + 2mg(1 − cos θ)
Minimum tension at the extreme position is given by
Tmin = m g cosθ
According to the question,
Tmax = 2Tmin
mg + 2mg − 2m g cosθ = 2m g cosθ
⇒ 3mg = 4 mg cosθ

It is given that R is the radius of the concave surface.
â€‹Let N be the normal reaction force.

Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
$\therefore \mathrm{\theta }=\frac{x}{R}\phantom{\rule{0ex}{0ex}}⇒a=g\mathrm{\theta }=g\left(\frac{x}{R}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{a}{x}\right)=\left(\frac{g}{R}\right)$
$⇒a=x\frac{g}{R}$
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.

Time period$\left(T\right)$ is given by,
$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{displacement}}{\mathrm{Acceleration}}}$
$=2\mathrm{\pi }\sqrt{\frac{x}{gx\mathit{/}R}}\mathit{=}2\mathrm{\pi }\sqrt{\frac{R}{g}}$

#### Question 39:

It is given that R is the radius of the concave surface.
â€‹Let N be the normal reaction force.

Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
$\therefore \mathrm{\theta }=\frac{x}{R}\phantom{\rule{0ex}{0ex}}⇒a=g\mathrm{\theta }=g\left(\frac{x}{R}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{a}{x}\right)=\left(\frac{g}{R}\right)$
$⇒a=x\frac{g}{R}$
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.

Time period$\left(T\right)$ is given by,
$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{displacement}}{\mathrm{Acceleration}}}$
$=2\mathrm{\pi }\sqrt{\frac{x}{gx\mathit{/}R}}\mathit{=}2\mathrm{\pi }\sqrt{\frac{R}{g}}$

Let ω be the angular velocity of the system about the point of suspension at any time.
Velocity of the ball rolling on a rough concave surface $\left({v}_{C}\right)$ is given by,
vc = (Rr)ω
Also, vc = 1
where ω1 is the rotational velocity of the sphere.

As total energy of a particle in S.H.M. remains constant,

Taking derivative on both sides, we get:

Therefore, the motion is S.H.M.

#### Question 40:

Let ω be the angular velocity of the system about the point of suspension at any time.
Velocity of the ball rolling on a rough concave surface $\left({v}_{C}\right)$ is given by,
vc = (Rr)ω
Also, vc = 1
where ω1 is the rotational velocity of the sphere.

As total energy of a particle in S.H.M. remains constant,

Taking derivative on both sides, we get:

Therefore, the motion is S.H.M.

It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth's surface, g = 9.8 ms$-2$

Let $g\text{'}$ be the acceleration due to gravity at a depth of 1600 km from the surface of the earth.
Its value is given by,

Time period is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{l}{g\text{'}}\right)}$

#### Question 41:

It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth's surface, g = 9.8 ms$-2$

Let $g\text{'}$ be the acceleration due to gravity at a depth of 1600 km from the surface of the earth.
Its value is given by,

Time period is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{l}{g\text{'}}\right)}$

Given:
Radius of the earth is R.
Let M be the total mass of the earth and $\rho$ be the density.
Let mass of the part of earth having radius x be M'.
$\therefore \frac{M\text{'}}{M}=\frac{\rho ×\frac{4}{3}\mathrm{\pi }{x}^{3}}{\rho ×\frac{4}{3}\mathrm{\pi }{R}^{3}}=\frac{{x}^{3}}{{R}^{3}}\phantom{\rule{0ex}{0ex}}⇒M\text{'}=\frac{M{x}^{3}}{{R}^{3}}$

Force on the particle is calculated as,

Now, acceleration $\left({a}_{x}\right)$ of mass M' at that position is given by,

(a) Velocity-displacement equation in S.H.M is written as,

When the particle is at y = R,
The velocity of the particle is $\sqrt{gR}$ and .
On substituting these values in the velocity-displacement equation, we get:

Let t1 and t2be the time taken by the particle to reach the positions X and Y.
Now, phase of the particle at point X will be greater than $\frac{\mathrm{\pi }}{2}$ but less than $\mathrm{\pi }$.
Also, the phase of the particle on reaching Y will be greater than $\mathrm{\pi }$ but less than $\frac{3\mathrm{\pi }}{2}$.

Displacement-time relation is given by,
y = A sin ωt

Substituting y = R and A =$\sqrt{2R}$ , in the above relation, we get:

$⇒\mathrm{\omega }{t}_{1}=\frac{3\mathrm{\pi }}{4}$

Also,

Time taken by the particle to travel from X to Y:
${t}_{2}-{t}_{1}=\frac{\mathrm{\pi }}{2\omega }=\frac{\mathrm{\pi }}{2}\sqrt{\frac{R}{g}}$ s

(b) When the body is dropped from a height R

Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
$⇒\frac{\mathrm{GM}m}{\mathrm{R}}-\frac{\mathrm{GM}m}{2\mathrm{R}}=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}⇒v=\sqrt{\left(g\mathrm{R}\right)}$

As the velocity is same as that at X, the body will take the same time to travel XY.

(c) The body is projected vertically upwards from the point X with a velocity $\sqrt{g\mathrm{R}}$. Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches X again will be,
$v=\sqrt{\left(g\mathrm{R}\right)}$
Hence, the body will take same time i.e. $\frac{\mathrm{\pi }}{2}\sqrt{\left(\frac{\mathrm{R}}{g}\right)}$ s to travel XY.

#### Question 42:

Given:
Radius of the earth is R.
Let M be the total mass of the earth and $\rho$ be the density.
Let mass of the part of earth having radius x be M'.
$\therefore \frac{M\text{'}}{M}=\frac{\rho ×\frac{4}{3}\mathrm{\pi }{x}^{3}}{\rho ×\frac{4}{3}\mathrm{\pi }{R}^{3}}=\frac{{x}^{3}}{{R}^{3}}\phantom{\rule{0ex}{0ex}}⇒M\text{'}=\frac{M{x}^{3}}{{R}^{3}}$

Force on the particle is calculated as,

Now, acceleration $\left({a}_{x}\right)$ of mass M' at that position is given by,

(a) Velocity-displacement equation in S.H.M is written as,

When the particle is at y = R,
The velocity of the particle is $\sqrt{gR}$ and .
On substituting these values in the velocity-displacement equation, we get:

Let t1 and t2be the time taken by the particle to reach the positions X and Y.
Now, phase of the particle at point X will be greater than $\frac{\mathrm{\pi }}{2}$ but less than $\mathrm{\pi }$.
Also, the phase of the particle on reaching Y will be greater than $\mathrm{\pi }$ but less than $\frac{3\mathrm{\pi }}{2}$.

Displacement-time relation is given by,
y = A sin ωt

Substituting y = R and A =$\sqrt{2R}$ , in the above relation, we get:

$⇒\mathrm{\omega }{t}_{1}=\frac{3\mathrm{\pi }}{4}$

Also,

Time taken by the particle to travel from X to Y:
${t}_{2}-{t}_{1}=\frac{\mathrm{\pi }}{2\omega }=\frac{\mathrm{\pi }}{2}\sqrt{\frac{R}{g}}$ s

(b) When the body is dropped from a height R

Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
$⇒\frac{\mathrm{GM}m}{\mathrm{R}}-\frac{\mathrm{GM}m}{2\mathrm{R}}=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}⇒v=\sqrt{\left(g\mathrm{R}\right)}$

As the velocity is same as that at X, the body will take the same time to travel XY.

(c) The body is projected vertically upwards from the point X with a velocity $\sqrt{g\mathrm{R}}$. Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches X again will be,
$v=\sqrt{\left(g\mathrm{R}\right)}$
Hence, the body will take same time i.e. $\frac{\mathrm{\pi }}{2}\sqrt{\left(\frac{\mathrm{R}}{g}\right)}$ s to travel XY.

If $\rho$ is the density of the earth, then mass of the earth $\left(M\right)$ is given by,

(a) Let F be the gravitational force exerted by the earth on the particle of mass m. Then, its value is given by,

(b)

(c)

${F}_{x}=\frac{GMm}{{R}^{2}}$
$\because$ Normal force exerted by the wall N = Fx
(d)The resultant force is $\frac{\mathrm{G}Mmx}{{R}^{3}}$
(e) Acceleration = Driving force/mass
$=\frac{GMmx}{{R}^{3}m}\phantom{\rule{0ex}{0ex}}=\frac{GMx}{{R}^{3}}$

$⇒$a $\propto$ x      (the body executes S.H.M.)
$\frac{a}{x}={\omega }^{2}=\frac{\mathrm{G}M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}⇒\omega =\sqrt{\frac{\mathrm{G}m}{{R}^{3}}}\phantom{\rule{0ex}{0ex}}⇒T=2\mathrm{\pi }\sqrt{\frac{{R}^{3}}{\mathrm{G}M}}$

#### Question 43:

If $\rho$ is the density of the earth, then mass of the earth $\left(M\right)$ is given by,

(a) Let F be the gravitational force exerted by the earth on the particle of mass m. Then, its value is given by,

(b)

(c)

${F}_{x}=\frac{GMm}{{R}^{2}}$
$\because$ Normal force exerted by the wall N = Fx
(d)The resultant force is $\frac{\mathrm{G}Mmx}{{R}^{3}}$
(e) Acceleration = Driving force/mass
$=\frac{GMmx}{{R}^{3}m}\phantom{\rule{0ex}{0ex}}=\frac{GMx}{{R}^{3}}$

$⇒$a $\propto$ x      (the body executes S.H.M.)
$\frac{a}{x}={\omega }^{2}=\frac{\mathrm{G}M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}⇒\omega =\sqrt{\frac{\mathrm{G}m}{{R}^{3}}}\phantom{\rule{0ex}{0ex}}⇒T=2\mathrm{\pi }\sqrt{\frac{{R}^{3}}{\mathrm{G}M}}$

The length of the simple pendulum is l.
â€‹Let x be the displacement of the simple pendulum..
(a)

From the diagram, the driving forces f is given by,
f = m(g + a0)sinθ                 ...(1)
Acceleration (a) of the elevator is given by,

[ when θ is very small, sin θθ = x/l]

$\therefore a=\left(\frac{g+{a}_{0}}{l}\right)x$                 ...(2)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =${\omega }^{2}x$, we get:
${\omega }^{2}=\frac{g+{a}_{0}}{l}$

Thus, time period of small oscillations when elevator is going upward(T) will be:
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+{a}_{0}}}$

(b)

When the elevator moves downwards with acceleration a0,
Driving force (F) is given by,
F = m(ga0)sinθ
On comparing the above equation with the expression, F = ma,

(c) When the elevator moves with uniform velocity, i.e. a0 = 0,
For a simple pendulum, the driving force $\left(F\right)$ is given by,

#### Question 44:

The length of the simple pendulum is l.
â€‹Let x be the displacement of the simple pendulum..
(a)

From the diagram, the driving forces f is given by,
f = m(g + a0)sinθ                 ...(1)
Acceleration (a) of the elevator is given by,

[ when θ is very small, sin θθ = x/l]

$\therefore a=\left(\frac{g+{a}_{0}}{l}\right)x$                 ...(2)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =${\omega }^{2}x$, we get:
${\omega }^{2}=\frac{g+{a}_{0}}{l}$

Thus, time period of small oscillations when elevator is going upward(T) will be:
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+{a}_{0}}}$

(b)

When the elevator moves downwards with acceleration a0,
Driving force (F) is given by,
F = m(ga0)sinθ
On comparing the above equation with the expression, F = ma,

(c) When the elevator moves with uniform velocity, i.e. a0 = 0,
For a simple pendulum, the driving force $\left(F\right)$ is given by,

It is given that:
Length of the simple pendulum, l = 1  feet
Time period of simple pendulum, T =
Acceleration due to gravity, g = 32 ft/s2
â€‹
â€‹Let a be the acceleration of the elevator while moving upwards.

Driving force$\left(f\right)$ is given by,
f = m(g + a)sinθ

Comparing the above equation with the expression, f = ma, we get:
Acceleration, a  =  (g + a)sinθ = (g +a)θ              (For small angle θ, sin θ → θ)
$=\frac{\left(g+a\right)x}{l}={\omega }^{2}x$                       (From the diagram $\theta =\frac{x}{l}$)

Time period $\left(T\right)$ is given as,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+a}}$
On substituting the respective values in the above formula, we get:

#### Question 45:

It is given that:
Length of the simple pendulum, l = 1  feet
Time period of simple pendulum, T =
Acceleration due to gravity, g = 32 ft/s2
â€‹
â€‹Let a be the acceleration of the elevator while moving upwards.

Driving force$\left(f\right)$ is given by,
f = m(g + a)sinθ

Comparing the above equation with the expression, f = ma, we get:
Acceleration, a  =  (g + a)sinθ = (g +a)θ              (For small angle θ, sin θ → θ)
$=\frac{\left(g+a\right)x}{l}={\omega }^{2}x$                       (From the diagram $\theta =\frac{x}{l}$)

Time period $\left(T\right)$ is given as,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+a}}$
On substituting the respective values in the above formula, we get:

It is given that:
When the car is moving uniformly, time period of simple pendulum, T = 4.0 s
As the accelerator is pressed, new time period of the pendulum, T' = 3.99 s
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,

Let the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:

On solving the above equation for a, we get:

#### Question 47:

It is given that:
When the car is moving uniformly, time period of simple pendulum, T = 4.0 s
As the accelerator is pressed, new time period of the pendulum, T' = 3.99 s
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,

Let the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:

On solving the above equation for a, we get:

Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
â€‹Acceleration due to gravity,= 9.8 ms$-2$
(a) Time period $\left(T\right)$ is given by

(b)  Velocity of merry-go-round, v = 4 ms$-1$
Radius of circle, r = 2 m
As the lady sits on the merry-go-round, her earring experiences centripetal acceleration.
Centripetal acceleration $\left(a\right)$ is given by,

Resultant acceleration $\left(A\right)$ is given by,

Time period, $T=2\mathrm{\pi }\sqrt{\left(\frac{l}{A}\right)}$

#### Question 48:

Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
â€‹Acceleration due to gravity,= 9.8 ms$-2$
(a) Time period $\left(T\right)$ is given by

(b)  Velocity of merry-go-round, v = 4 ms$-1$
Radius of circle, r = 2 m
As the lady sits on the merry-go-round, her earring experiences centripetal acceleration.
Centripetal acceleration $\left(a\right)$ is given by,

Resultant acceleration $\left(A\right)$ is given by,

Time period, $T=2\mathrm{\pi }\sqrt{\left(\frac{l}{A}\right)}$

(a) Moment of inertia $\left(I\right)$ about the point X is given by,

I = IC.G + mh2
$=\frac{m{l}^{2}}{12}+m{h}^{2}\phantom{\rule{0ex}{0ex}}=\frac{m{l}^{2}}{12}+m{\left(0.3\right)}^{2}\phantom{\rule{0ex}{0ex}}=m\left(\frac{1}{12}+0.09\right)\phantom{\rule{0ex}{0ex}}=m\left(\frac{1+1.08}{12}\right)\phantom{\rule{0ex}{0ex}}=m\left(\frac{2.08}{12}\right)$

The time period $\left(T\right)$ is given by,

(b) Moment of inertia $\left(I\right)$ about A is given as,
I = IC.G. + mr2 = mr2 + mr2 = 2mr2

The time period (T) will be,

(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.
$I=m\left(\frac{{a}^{2}+{a}^{2}}{3}\right)=\frac{2m}{3}{a}^{2}$

In the $△$ABC, l2 + l2 = a2

(d)

Moment of inertia about A will be:
l = IC.G. + mh2
$=\frac{m{r}^{2}}{2}+m{\left(\frac{r}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=m{r}^{2}\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{3}{4}m{r}^{2}$

Time period (T) will be,

#### Question 49:

(a) Moment of inertia $\left(I\right)$ about the point X is given by,

I = IC.G + mh2
$=\frac{m{l}^{2}}{12}+m{h}^{2}\phantom{\rule{0ex}{0ex}}=\frac{m{l}^{2}}{12}+m{\left(0.3\right)}^{2}\phantom{\rule{0ex}{0ex}}=m\left(\frac{1}{12}+0.09\right)\phantom{\rule{0ex}{0ex}}=m\left(\frac{1+1.08}{12}\right)\phantom{\rule{0ex}{0ex}}=m\left(\frac{2.08}{12}\right)$

The time period $\left(T\right)$ is given by,

(b) Moment of inertia $\left(I\right)$ about A is given as,
I = IC.G. + mr2 = mr2 + mr2 = 2mr2

The time period (T) will be,

(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.
$I=m\left(\frac{{a}^{2}+{a}^{2}}{3}\right)=\frac{2m}{3}{a}^{2}$

In the $△$ABC, l2 + l2 = a2

(d)

Moment of inertia about A will be:
l = IC.G. + mh2
$=\frac{m{r}^{2}}{2}+m{\left(\frac{r}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=m{r}^{2}\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{3}{4}m{r}^{2}$

Time period (T) will be,

It is given that the length of the rod is l.

Let point A be the suspension point and point B be the centre of gravity.

Separation between the point of suspension and the centre of mass, l' = $\frac{l}{2}$

Also, h = $\frac{l}{2}$

Using parallel axis theorem, the moment of inertia about A is given as,

Let T' be the time period of simple pendulum of length x.

Time period $\left(T\text{'}\right)$ is given by,

#### Question 50:

It is given that the length of the rod is l.

Let point A be the suspension point and point B be the centre of gravity.

Separation between the point of suspension and the centre of mass, l' = $\frac{l}{2}$

Also, h = $\frac{l}{2}$

Using parallel axis theorem, the moment of inertia about A is given as,

Let T' be the time period of simple pendulum of length x.

Time period $\left(T\text{'}\right)$ is given by,

Let m be the mass of the disc andbe its radius.

Consider a point at a distance x from the centre of gravity.
Thus, l = x

Moment of intertia $\left(I\right)$ about the point x will be,
I = IC.G +mx2
$=\frac{m{r}^{2}}{2}+m{x}^{2}\phantom{\rule{0ex}{0ex}}=m\left(\frac{{r}^{2}}{2}+{x}^{2}\right)\phantom{\rule{0ex}{0ex}}$

Time period(T) is given as,

To determine the minimum value of T,
$\frac{{\mathrm{d}}^{2}T}{\mathrm{d}{x}^{2}}=0$

$Now,\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}T}{\mathrm{d}{x}^{2}}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{4{\pi }^{2}{r}^{2}}{2gx}+\frac{4{\pi }^{2}2{x}^{2}}{2gx}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2{\pi }^{2}{r}^{2}}{g}\left(-\frac{1}{{x}^{2}}\right)+\frac{4{\pi }^{2}}{g}=0\phantom{\rule{0ex}{0ex}}⇒-\frac{{\pi }^{2}{r}^{2}}{g{x}^{2}}+\frac{2{\pi }^{2}}{g}=0\phantom{\rule{0ex}{0ex}}⇒\frac{{\pi }^{2}{r}^{2}}{g{x}^{2}}=\frac{2{\pi }^{2}}{g}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{r}{\sqrt{2}}$

Substituting this value of x in equation (1), we get:

#### Question 51:

Let m be the mass of the disc andbe its radius.

Consider a point at a distance x from the centre of gravity.
Thus, l = x

Moment of intertia $\left(I\right)$ about the point x will be,
I = IC.G +mx2
$=\frac{m{r}^{2}}{2}+m{x}^{2}\phantom{\rule{0ex}{0ex}}=m\left(\frac{{r}^{2}}{2}+{x}^{2}\right)\phantom{\rule{0ex}{0ex}}$

Time period(T) is given as,

To determine the minimum value of T,
$\frac{{\mathrm{d}}^{2}T}{\mathrm{d}{x}^{2}}=0$

$Now,\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}T}{\mathrm{d}{x}^{2}}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{4{\pi }^{2}{r}^{2}}{2gx}+\frac{4{\pi }^{2}2{x}^{2}}{2gx}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2{\pi }^{2}{r}^{2}}{g}\left(-\frac{1}{{x}^{2}}\right)+\frac{4{\pi }^{2}}{g}=0\phantom{\rule{0ex}{0ex}}⇒-\frac{{\pi }^{2}{r}^{2}}{g{x}^{2}}+\frac{2{\pi }^{2}}{g}=0\phantom{\rule{0ex}{0ex}}⇒\frac{{\pi }^{2}{r}^{2}}{g{x}^{2}}=\frac{2{\pi }^{2}}{g}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{r}{\sqrt{2}}$

Substituting this value of x in equation (1), we get:

It is given that:
Radius of the hollow sphere, r = 2 cm
Length of the long thread, l = 18 cm = .

Let I be the moment of inertia and $\omega$ be the angular speed.
Using the energy equation, we can write:



On substituting the value of I in equation (1) and differentiating it, we get:

For a simple pendulum, time period (T) is given by,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$
= 0.86 s

∴ It is about 0.3% greater than the calculated value.

#### Question 52:

It is given that:
Radius of the hollow sphere, r = 2 cm
Length of the long thread, l = 18 cm = .

Let I be the moment of inertia and $\omega$ be the angular speed.
Using the energy equation, we can write:

On substituting the value of I in equation (1) and differentiating it, we get:

For a simple pendulum, time period (T) is given by,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$
= 0.86 s

∴ It is about 0.3% greater than the calculated value.

It is given that:
Time period of oscillation, T = 2 s
Acceleration due to gravity, g = ${\mathrm{\pi }}^{2}$ ms−2
Let I be the moment of inertia of the circular wire having mass m and radius r.

(a) Time period of compound pendulum $\left(T\right)$ is given by,
$T=2\sqrt{\frac{I}{mgl}}=2\sqrt{\frac{I}{mgr}}$   $\left(\because l=r\right)$             ...(1)
Moment of inertia about the point of suspension  is calculated as,
I = mr2 + mr2 = 2mr2
On substituting the value of moment of inertia I in equation (1), we get:

(b) From the energy equation, we have:

(c) The acceleration is found to be centripetal at the extreme position.
Centripetal acceleration at the extreme position $\left({a}_{n}\right)$ is given by,
an = ω2(2r) = (0.11) × 100 = 12 cm/s2
The direction of an is towards the point of suspension.

(d) The particle has zero centripetal acceleration at the extreme position. However, the particle will still have acceleration due to the S.H.M.
Angular frequency$\left(\omega \right)$ is given by,

$\therefore$ Angular acceleration $\left(a\right)$ at the extreme position is given as,

Thus, tangential acceleration$=\mathrm{\alpha }\left(2r\right)=\left(\frac{2{\mathrm{\pi }}^{3}}{180}\right)×100$
= 34 cm/s2

#### Question 53:

It is given that:
Time period of oscillation, T = 2 s
Acceleration due to gravity, g = ${\mathrm{\pi }}^{2}$ ms−2
Let I be the moment of inertia of the circular wire having mass m and radius r.

(a) Time period of compound pendulum $\left(T\right)$ is given by,
$T=2\sqrt{\frac{I}{mgl}}=2\sqrt{\frac{I}{mgr}}$   $\left(\because l=r\right)$             ...(1)
Moment of inertia about the point of suspension  is calculated as,
I = mr2 + mr2 = 2mr2
On substituting the value of moment of inertia I in equation (1), we get:

(b) From the energy equation, we have:

(c) The acceleration is found to be centripetal at the extreme position.
Centripetal acceleration at the extreme position $\left({a}_{n}\right)$ is given by,
an = ω2(2r) = (0.11) × 100 = 12 cm/s2
The direction of an is towards the point of suspension.

(d) The particle has zero centripetal acceleration at the extreme position. However, the particle will still have acceleration due to the S.H.M.
Angular frequency$\left(\omega \right)$ is given by,

$\therefore$ Angular acceleration $\left(a\right)$ at the extreme position is given as,

Thus, tangential acceleration$=\mathrm{\alpha }\left(2r\right)=\left(\frac{2{\mathrm{\pi }}^{3}}{180}\right)×100$
= 34 cm/s2

It is given that:
Mass of disc = m
The time period of torsional oscillations is T.
Moment of inertia of the disc at the centre, I$=\frac{m{r}^{2}}{2}$

Time period of torsional pendulum$\left(T\right)$ is given by,

where I is the moment of inertia, and
k is the torsional constant.

On substituting the value of moment of inertia in the expression for time period T, we have:

#### Question 54:

It is given that:
Mass of disc = m
The time period of torsional oscillations is T.
Moment of inertia of the disc at the centre, I$=\frac{m{r}^{2}}{2}$

Time period of torsional pendulum$\left(T\right)$ is given by,

where I is the moment of inertia, and
k is the torsional constant.

On substituting the value of moment of inertia in the expression for time period T, we have:

It is given that the mass of both the balls is m and they are connected to each other with the help of a light rod of length L.

Moment of inertia of the two-ball system $\left(I\right)$ is given by,
$I=2m{\left(\frac{L}{2}\right)}^{2}=\frac{m{L}^{2}}{2}$

Torque $\left(\tau \right)$, produced at any given position θ is given as:
$\tau$ =
$⇒$ Work done during the displacement of system from 0 to θ0 will be,

On applying work-energy theorem, we get:

From the free body diagram of the rod, we can write:

#### Question 55:

It is given that the mass of both the balls is m and they are connected to each other with the help of a light rod of length L.

Moment of inertia of the two-ball system $\left(I\right)$ is given by,
$I=2m{\left(\frac{L}{2}\right)}^{2}=\frac{m{L}^{2}}{2}$

Torque $\left(\tau \right)$, produced at any given position θ is given as:
$\tau$ =
$⇒$ Work done during the displacement of system from 0 to θ0 will be,

On applying work-energy theorem, we get:

From the free body diagram of the rod, we can write:

It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.

Amplitude of first motion, A1 = 3 cm
Amplitude of second motion, A2 = 4 cm

Let Ï• be the phase difference.

The resultant amplitude $\left(R\right)$ is given by,

(a) When Ï• = 0°

(b) When Ï• = 60°

(c) When Ï• = 90°

#### Question 56:

It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.

Amplitude of first motion, A1 = 3 cm
Amplitude of second motion, A2 = 4 cm

Let Ï• be the phase difference.

The resultant amplitude $\left(R\right)$ is given by,

(a) When Ï• = 0°

(b) When Ï• = 60°

(c) When Ï• = 90°

It is given that three S.H.M.s of equal amplitudes A and equal time periods are combined in the same direction.

Let be the three vectors representing the motions, as shown in the figure given below.

According to the question:
.

By using the vector method, we can find the resultant vector.
Resultant amplitude = Vector sum of the three vectors
= A + A cos 60° + A cos 60°
$=\mathrm{A}+\frac{\mathrm{A}}{2}+\frac{\mathrm{A}}{2}=2A$

#### Question 57:

It is given that three S.H.M.s of equal amplitudes A and equal time periods are combined in the same direction.

Let be the three vectors representing the motions, as shown in the figure given below.

According to the question:
.

By using the vector method, we can find the resultant vector.
Resultant amplitude = Vector sum of the three vectors
= A + A cos 60° + A cos 60°
$=\mathrm{A}+\frac{\mathrm{A}}{2}+\frac{\mathrm{A}}{2}=2A$

Given are the equations of motion of a particle:
x1 = 2.0sin100$\mathrm{\pi }$t
${x}_{2}=2.0\mathrm{sin}\left(120\mathrm{\pi }t+\frac{\mathrm{\pi }}{3}\right)$

The resultant displacement $\left(x\right)$ will be,
x = x1 + x2
$=2\left[\mathrm{sin}\left(100\mathrm{\pi }t\right)+\mathrm{sin}\left(120\mathrm{\pi }t+\frac{\mathrm{\pi }}{3}\right)\right]$

(a) At t = 0.0125 s

(b) At t = 0.025 s

#### Question 58:

Given are the equations of motion of a particle:
x1 = 2.0sin100$\mathrm{\pi }$t
${x}_{2}=2.0\mathrm{sin}\left(120\mathrm{\pi }t+\frac{\mathrm{\pi }}{3}\right)$

The resultant displacement $\left(x\right)$ will be,
x = x1 + x2
$=2\left[\mathrm{sin}\left(100\mathrm{\pi }t\right)+\mathrm{sin}\left(120\mathrm{\pi }t+\frac{\mathrm{\pi }}{3}\right)\right]$

(a) At t = 0.0125 s

(b) At t = 0.025 s

Given:
Equation of motion along X axis, x = x0sinωt
Equation of motion along Y axis, s = s0sinωt
Angle between the two motions, $\theta$ = 45â‚€
Resultant motion (R) will be,

Hence, the resultant amplitude is ${\left[{x}_{0}^{2}+{s}_{0}^{2}+\sqrt{2{x}_{0}{s}_{0}}\right]}^{1/2}$.

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