Simple Harmonic Motion MCQs with Answers and Explanations

Which condition is essential for a particle to execute simple harmonic motion?

1. The particle must move in a circular path
2. The restoring force must always have constant magnitude
3. The restoring force must be proportional to displacement and directed toward equilibrium
4. The particle must move with constant speed

Explanation

For simple harmonic motion, the restoring force must satisfy

\[
F=-kx
\]

The negative sign shows that the force always points toward the equilibrium position.

The magnitude increases linearly with displacement. A force that is only opposite to motion is not enough for SHM.

A particle performs SHM about x = 0. At which position is its acceleration zero?

1. At \(x=A\)
2. At \(x=0\)
3. At \(x=-A\)
4. At every position

Explanation

In SHM,

\[
a=-\omega^2x
\]

Therefore, acceleration becomes zero only when displacement is zero.

At \(x=0\), the particle is at the mean or equilibrium position.

At an extreme position of SHM, which combination is correct?

1. Velocity is maximum and acceleration is zero
2. Velocity is zero and acceleration is zero
3. Velocity is zero and acceleration has maximum magnitude
4. Velocity and acceleration are both maximum

Explanation

At an extreme position,

\[
x=\pm A
\]

The particle changes direction there, so its velocity is zero.

Since

\[
|a|=\omega^2|x|
\]

the acceleration magnitude is maximum at the extremes.

A particle in SHM has maximum speed \(v_{\max}\). What is its speed when its displacement is \(\dfrac{A}{2}\)?

1. \(\dfrac{v_{\max}}{2}\)
2. \(\dfrac{v_{\max}}{\sqrt{2}}\)
3. \(\dfrac{\sqrt{3}v_{\max}}{4}\)
4. \(\dfrac{\sqrt{3}v_{\max}}{2}\)

Explanation

For SHM,

\[
v^2=\omega^2(A^2-x^2)
\]

At

\[
x=\dfrac{A}{2}
\]

\[
v^2=\omega^2\left(A^2-\dfrac{A^2}{4}\right)
\]

\[
v^2=\dfrac{3}{4}\omega^2A^2
\]

Since

\[
v_{\max}=\omega A
\]

\[
v=\dfrac{\sqrt{3}}{2}v_{\max}
\]

Which statement correctly describes the energy of an ideal spring-mass system in SHM?

1. Kinetic energy remains constant throughout the motion
2. Potential energy remains constant throughout the motion
3. Total mechanical energy remains constant while kinetic and potential energy change continuously
4. Total mechanical energy becomes zero at the extreme positions

Explanation

In ideal SHM, no energy is lost to friction or air resistance.

The total energy is

\[
E=\dfrac{1}{2}kA^2
\]

Kinetic energy and spring potential energy continuously transform into each other.

A particle starts from the positive extreme position in SHM. After what fraction of a period does it first reach the mean position?

1. \(\dfrac{T}{8}\)
2. \(\dfrac{T}{4}\)
3. \(\dfrac{T}{2}\)
4. \(\dfrac{3T}{4}\)

Explanation

A complete SHM cycle has four equal parts:

\[
+A \rightarrow 0 \rightarrow -A \rightarrow 0 \rightarrow +A
\]

Moving from an extreme position to the mean position takes one-fourth of the time period.

Therefore,

\[
t=\dfrac{T}{4}
\]

In SHM, the velocity is always:

1. In the same direction as displacement
2. Opposite to displacement
3. Perpendicular to acceleration in phase space, but not necessarily in physical space
4. Maximum when acceleration is maximum

Explanation

This question checks a common trap.

Velocity is not always in one fixed direction relative to displacement because the particle moves back and forth.

However, velocity is maximum when acceleration is zero, at the mean position.

The phrase “perpendicular in phase space” is mathematically true but can be misleading in basic physical motion. For a standard one-dimensional SHM question, the clearer physical result is that velocity and acceleration are generally not fixed in the same or opposite direction.

Which graph represents the relationship between acceleration and displacement for SHM?

1. A horizontal straight line
2. A positive-slope straight line through the origin
3. A parabola opening upward
4. A negative-slope straight line through the origin

Explanation

The acceleration equation is

\[
a=-\omega^2x
\]

Therefore, acceleration is directly proportional to displacement but opposite in direction.

The \(a\)-versus-\(x\) graph is a straight line with negative slope.

A particle moves in SHM. At the mean position, which statement is correct?

1. Potential energy is maximum
2. Acceleration is maximum
3. Kinetic energy and speed are maximum
4. Total energy is zero

Explanation

At the mean position,

\[
x=0
\]

Therefore,

\[
U=\dfrac{1}{2}kx^2=0
\]

All mechanical energy is kinetic energy.

Hence, speed and kinetic energy are maximum.

A particle is displaced from equilibrium and released. Its restoring force is proportional to \(x^3\), where x is displacement. Which statement is correct?

1. It must execute SHM because the force is restoring
2. It does not execute ideal SHM because the restoring force is not proportional to x
3. Its acceleration remains constant
4. Its time period is independent of amplitude for all amplitudes

Explanation

For ideal SHM,

\[
F=-kx
\]

If

\[
F\propto-x^3
\]

the force is restoring, but it is not linearly proportional to displacement.

Therefore, the motion is oscillatory but not ideal SHM.

For small displacement, some nonlinear systems can approximate SHM, but not exactly.

A 0.50 kg block is attached to a spring of force constant 200 N/m on a frictionless horizontal surface. What is the time period of its SHM?

1. \(\dfrac{\pi}{10}\) s
2. \(\dfrac{\pi}{5}\) s
3. \(\dfrac{\pi}{8}\) s
4. \(\dfrac{\pi}{2}\) s

Explanation

For a spring-mass system,

\[
T=2\pi\sqrt{\dfrac{m}{k}}
\]

\[
T=2\pi\sqrt{\dfrac{0.50}{200}}
\]

\[
T=2\pi\sqrt{\dfrac{1}{400}}
\]

\[
T=\dfrac{\pi}{10}\text{ s}
\]

A particle executes SHM with amplitude 0.20 m and angular frequency 5 rad/s. What is its maximum speed?

1. 0.20 m/s
2. 1.0 m/s
3. 2.5 m/s
4. 5.0 m/s

Explanation

Maximum speed in SHM is

\[
v_{\max}=\omega A
\]

\[
v_{\max}=5\times0.20
\]

\[
v_{\max}=1.0\text{ m/s}
\]

A particle has displacement \(x=A\cos(\omega t)\). What is its velocity at \(t=\dfrac{\pi}{2\omega}\)?

1. 0
2. \(\omega A\)
3. -\(\omega A\)
4. \(-A\)

Explanation

Differentiate displacement:

\[
v=\dfrac{dx}{dt}
\]

\[
v=-\omega A\sin(\omega t)
\]

At

\[
t=\dfrac{\pi}{2\omega}
\]

\[
\omega t=\dfrac{\pi}{2}
\]

\[
v=-\omega A
\]

The negative sign means the particle moves toward the negative direction.

A spring-mass system has time period T. If the mass is made four times larger while the spring constant remains unchanged, the new time period becomes:

1. \(\dfrac{T}{2}\)
2. T
3. 4T
4. 2T

Explanation

\[
T=2\pi\sqrt{\dfrac{m}{k}}
\]

If

\[
m’=4m
\]

then

\[
T’=2\pi\sqrt{\dfrac{4m}{k}}
\]

\[
T’=2T
\]

A particle in SHM has amplitude A. At what displacement is its kinetic energy equal to its potential energy?

1. \(\dfrac{A}{4}\)
2. \(\dfrac{A}{\sqrt{2}}\)
3. \(\dfrac{A}{2}\)
4. \(\dfrac{\sqrt{3}A}{2}\)

Explanation

Total energy:

\[
E=\dfrac{1}{2}kA^2
\]

Potential energy:

\[
U=\dfrac{1}{2}kx^2
\]

If kinetic energy equals potential energy,

\[
U=\dfrac{E}{2}
\]

\[
\dfrac{1}{2}kx^2
=
\dfrac{1}{4}kA^2
\]

\[
x=\dfrac{A}{\sqrt{2}}
\]

A simple pendulum has length L and time period T. If its length is increased to 9L, what is the new time period?

1. 3T
2. 6T
3. 9T
4. \(\dfrac{T}{3}\)

Explanation

For a simple pendulum,

\[
T=2\pi\sqrt{\dfrac{L}{g}}
\]

If

\[
L’=9L
\]

then

\[
T’=2\pi\sqrt{\dfrac{9L}{g}}
\]

\[
T’=3T
\]

A block attached to a vertical spring is in equilibrium. It is pulled downward slightly and released. Which statement is correct?

1. Gravity changes the time period directly
2. The block cannot execute SHM because gravity acts on it
3. The block executes SHM about its new equilibrium position
4. The equilibrium position is at the natural length of the spring

Explanation

Gravity stretches the spring and creates a new equilibrium position.

If displacement is measured from that equilibrium point, the restoring force becomes

\[
F=-kx
\]

Therefore, the block performs SHM with

\[
T=2\pi\sqrt{\dfrac{m}{k}}
\]

Gravity changes the equilibrium position, not the time period.

A particle executes SHM with angular frequency \(\omega\). What is the magnitude of its acceleration when its displacement is \(\dfrac{A}{3}\)?

1. \(\omega A\)
2. \(\dfrac{\omega A}{3}\)
3. \(\dfrac{\omega^2A}{3}\)
4. \(\omega^2A\)

Explanation

In SHM,

\[
a=-\omega^2x
\]

At

\[
x=\dfrac{A}{3}
\]

\[
|a|=\omega^2\dfrac{A}{3}
\]

\[
|a|=\dfrac{\omega^2A}{3}
\]

A particle completes 25 oscillations in 5 seconds. What is its angular frequency?

1. \(5\pi\) rad/s
2. \(7\pi\) rad/s
3. \(10\pi\) rad/s
4. \(25\pi\) rad/s

Explanation

Frequency:

\[
f=\dfrac{25}{5}=5\text{ Hz}
\]

Angular frequency:

\[
\omega=2\pi f
\]

\[
\omega=2\pi(5)
\]

\[
\omega=10\pi\text{ rad/s}
\]

A particle in SHM has total energy E. What is its kinetic energy when its displacement is \(\dfrac{\sqrt{3}A}{2}\)?

1. \(\dfrac{E}{6}\)
2. \(\dfrac{E}{4}\)
3. \(\dfrac{E}{2}\)
4. \(\dfrac{3E}{4}\)

Explanation

Potential energy is

\[
U=\dfrac{1}{2}kx^2
\]

Total energy is

\[
E=\dfrac{1}{2}kA^2
\]

At

\[
x=\dfrac{\sqrt{3}A}{2}
\]

\[
U
=
\dfrac{1}{2}k
\left(
\dfrac{\sqrt{3}A}{2}
\right)^2
\]

\[
U=\dfrac{3E}{4}
\]

Therefore,

\[
K=E-U
\]

\[
K=\dfrac{E}{4}
\]

A block of mass \(m\) is placed between two identical springs, each with spring constant \(k\), on a frictionless horizontal surface. Both springs are unstretched when the block is at equilibrium. What is the time period for small oscillations?

1. \(2\pi\sqrt{\dfrac{m}{k}}\)
2. \(\pi\sqrt{\dfrac{m}{k}}\)
3. \(2\pi\sqrt{\dfrac{m}{2k}}\)
4. \(2\pi\sqrt{\dfrac{2m}{k}}\)

Explanation

If the block moves by a small distance \(x\), one spring stretches by \(x\) and the other compresses by \(x\).

Both springs provide restoring force toward equilibrium.

\[
F=-kx-kx=-2kx
\]

Therefore, the effective spring constant is

\[
k_{\text{eff}}=2k
\]

Hence,

\[
T=2\pi\sqrt{\frac{m}{k_{\text{eff}}}}
\]

\[
T=2\pi\sqrt{\frac{m}{2k}}
\]

A particle of mass \(m\) moves in the potential energy field \(U(x)=U_0+\alpha x^2\), where \(\alpha>0\). What is the angular frequency of small oscillations about \(x=0\)?

1. \(\sqrt{\dfrac{\alpha}{m}}\)
2. \(\sqrt{\dfrac{2\alpha}{m}}\)
3. \(\sqrt{\dfrac{m}{2\alpha}}\)
4. \(\dfrac{2\alpha}{m}\)

Explanation

Force is

\[
F=-\frac{dU}{dx}
\]

\[
F=-2\alpha x
\]

Comparing with the SHM form

\[
F=-m\omega^2x
\]

gives

\[
m\omega^2=2\alpha
\]

Therefore,

\[
\omega=\sqrt{\frac{2\alpha}{m}}
\]

A simple pendulum of length \(L\) is placed inside a lift accelerating upward with acceleration \(a\). What is its time period for small oscillations?

1. \(2\pi\sqrt{\dfrac{L}{g-a}}\)
2. \(2\pi\sqrt{\dfrac{L}{g}}\)
3. \(2\pi\sqrt{\dfrac{L}{a}}\)
4. \(2\pi\sqrt{\dfrac{L}{g+a}}\)

Explanation

Inside an upward accelerating lift, the effective gravitational acceleration becomes

\[
g_{\text{eff}}=g+a
\]

For a simple pendulum,

\[
T=2\pi\sqrt{\frac{L}{g_{\text{eff}}}}
\]

Thus,

\[
T=2\pi\sqrt{\frac{L}{g+a}}
\]

The period becomes smaller because the restoring effect is stronger.

A particle executes SHM with amplitude \(A\) and angular frequency \(\omega\). At what displacement is the magnitude of acceleration equal to the magnitude of velocity divided by a time constant \(\tau=\dfrac{1}{\omega}\)?

1. \(\dfrac{A}{\sqrt{2}}\)
2. \(\dfrac{A}{2}\)
3. \(\dfrac{A}{2\sqrt{2}}\)
4. \(\dfrac{\sqrt{3}A}{2}\)

Explanation

The condition is

\[
|a|=\frac{|v|}{\tau}
\]

Since

\[
\tau=\frac{1}{\omega}
\]

\[
|a|=\omega|v|
\]

For SHM,

\[
|a|=\omega^2|x|
\]

and

\[
|v|=\omega\sqrt{A^2-x^2}
\]

Therefore,

\[
\omega^2x
=
\omega^2\sqrt{A^2-x^2}
\]

\[
x=\sqrt{A^2-x^2}
\]

\[
2x^2=A^2
\]

\[
x=\frac{A}{\sqrt{2}}
\]

A mass \(m\) is attached to a spring of spring constant \(k\). The mass is initially at equilibrium. A constant horizontal force \(F_0\) is suddenly applied and kept constant. What is the maximum displacement of the mass from its original equilibrium position?

1. \(\dfrac{F_0}{k}\)
2. \(\dfrac{2F_0}{k}\)
3. \(\dfrac{F_0}{2k}\)
4. \(\dfrac{4F_0}{k}\)

Explanation

The new equilibrium position is

\[
x_{\text{eq}}=\frac{F_0}{k}
\]

Immediately after the force is applied, the mass is still at the old equilibrium position.

So its initial displacement from the new equilibrium is

\[
-\frac{F_0}{k}
\]

It oscillates around the new equilibrium with amplitude

\[
\frac{F_0}{k}
\]

Therefore, its farthest position from the original equilibrium is

\[
\frac{F_0}{k}+\frac{F_0}{k}
\]

\[
x_{\max}=\frac{2F_0}{k}
\]

A uniform rod of length \(L\) is free to slide without friction inside a narrow vertical tube. The tube is drilled through a uniform spherical planet of radius \(R\) and mass \(M\), passing through its center. The rod remains entirely inside the planet and is constrained to move along the tube. What is the time period of its oscillation?

1. \(2\pi\sqrt{\dfrac{R^3}{GM}}\)
2. \(2\pi\sqrt{\dfrac{2R^3}{GM}}\)
3. \(2\pi\sqrt{\dfrac{R^3}{GM}}\)
4. \(2\pi\sqrt{\dfrac{GM}{R^3}}\)

Explanation

Inside a uniform sphere, gravitational acceleration at position \(x\) is

\[
g(x)=\frac{GM}{R^3}x
\]

Every small element of the rod experiences a force proportional to its position.

The net force on the rod is equal to the total mass multiplied by the gravitational field at its center of mass.

Thus,

\[
F=-m_{\text{rod}}\frac{GM}{R^3}x
\]

Comparing with

\[
F=-m_{\text{rod}}\omega^2x
\]

gives

\[
\omega=\sqrt{\frac{GM}{R^3}}
\]

Therefore,

\[
T=2\pi\sqrt{\frac{R^3}{GM}}
\]

The rod length does not affect the period as long as it remains inside the uniform planet.

A particle moves in one dimension under the force \(F=-kx-\lambda x^3\), where \(k>0\) and \(\lambda>0\). For very small amplitude oscillations, which statement is correct?

1. The motion cannot be oscillatory
2. The angular frequency is \(\sqrt{\dfrac{\lambda}{m}}\)
3. The angular frequency approaches \(\sqrt{\dfrac{k}{m}}\)
4. The time period is exactly independent of amplitude for all amplitudes

Explanation

For very small displacement,

\[
\lambda x^3
\]

is much smaller than

\[
kx
\]

Therefore,

\[
F\approx-kx
\]

The motion approaches SHM with

\[
\omega=\sqrt{\frac{k}{m}}
\]

For larger amplitudes, the cubic term changes the motion and the period can depend on amplitude.

A block of mass \(m\) is attached to a spring of constant \(k\) on a frictionless surface. At equilibrium, the spring is suddenly cut into two equal halves, and the block remains attached to one half. What is the new time period?

1. \(T\)
2. \(2T\)
3. \(\dfrac{T}{\sqrt{2}}\)
4. \(\sqrt{2}T\)

Explanation

For a uniform spring, spring constant is inversely proportional to length.

Cutting the spring into two equal halves doubles the spring constant:

\[
k’=2k
\]

Original period:

\[
T=2\pi\sqrt{\frac{m}{k}}
\]

New period:

\[
T’=2\pi\sqrt{\frac{m}{2k}}
\]

\[
T’=\frac{T}{\sqrt{2}}
\]

A particle executes SHM. At an instant, its displacement is \(\dfrac{3A}{5}\) and its speed is \(\dfrac{4}{5}v_{\max}\). Which statement is correct?

1. The particle is not in SHM because both values cannot occur together
2. The given displacement and speed are consistent with SHM
3. The amplitude must be \(\dfrac{5A}{3}\)
4. The speed should be \(\dfrac{3}{5}v_{\max}\)

Explanation

For SHM,

\[
\frac{v^2}{v_{\max}^2}
+
\frac{x^2}{A^2}
=
1
\]

Given,

\[
\frac{x^2}{A^2}
=
\left(\frac{3}{5}\right)^2
=
\frac{9}{25}
\]

and

\[
\frac{v^2}{v_{\max}^2}
=
\left(\frac{4}{5}\right)^2
=
\frac{16}{25}
\]

Thus,

\[
\frac{9}{25}+\frac{16}{25}=1
\]

Therefore, the values are fully consistent with SHM.

A particle of mass \(m\) is attached to a spring of constant \(k\). While it passes through equilibrium with speed \(v\), it sticks to another particle of equal mass initially at rest. What is the amplitude of the subsequent SHM?

1. \(\dfrac{v}{\omega}\)
2. \(\dfrac{v}{2\omega}\)
3. \(\dfrac{v}{\sqrt{2}\omega}\)
4. \(\dfrac{\sqrt{2}v}{\omega}\)

Explanation

Before collision, the original angular frequency is

\[
\omega=\sqrt{\frac{k}{m}}
\]

At equilibrium, the spring force is zero, so momentum is conserved during the collision.

\[
mv=(2m)V
\]

\[
V=\frac{v}{2}
\]

After collision, the new mass is

\[
2m
\]

Therefore, the new angular frequency is

\[
\omega’=\sqrt{\frac{k}{2m}}=\frac{\omega}{\sqrt{2}}
\]

The collision occurs at equilibrium, so the new amplitude is

\[
A’=\frac{V}{\omega’}
\]

\[
A’=
\frac{v/2}{\omega/\sqrt{2}}
\]

\[
A’=\frac{v}{\sqrt{2}\omega}
\]

A particle of mass \(m\) moves in the potential energy field \(U(x)=U_0+\dfrac{1}{2}kx^2+\beta x^4\), where \(k>0\) and \(\beta>0\). For very small oscillations about \(x=0\), what is the time period?

1. \(2\pi\sqrt{\dfrac{m}{k+4\beta}}\)
2. \(2\pi\sqrt{\dfrac{m}{4\beta}}\)
3. \(2\pi\sqrt{\dfrac{m}{k}}\)
4. \(2\pi\sqrt{\dfrac{k}{m}}\)

Explanation

For small oscillations, only the quadratic term near equilibrium determines the SHM frequency.

\[
F=-\frac{dU}{dx}
\]

\[
F=-kx-4\beta x^3
\]

When \(x\) is very small, the \(x^3\) term is negligible.

Therefore,

\[
F\approx-kx
\]

Hence,

\[
\omega=\sqrt{\frac{k}{m}}
\]

and

\[
T=2\pi\sqrt{\frac{m}{k}}
\]

A particle executes SHM with amplitude \(A\). What fraction of one complete period does it spend with \(|x|>\dfrac{A}{2}\)?

1. \(\dfrac{1}{6}\)
2. \(\dfrac{1}{3}\)
3. \(\dfrac{1}{2}\)
4. \(\dfrac{2}{3}\)

Explanation

Take

\[
x=A\cos\theta
\]

The condition is

\[
|\cos\theta|>\frac{1}{2}
\]

This occurs over a total phase interval

\[
\frac{2\pi}{3}
\]

during one full cycle of phase \(2\pi\).

Therefore, the required fraction is

\[
\frac{2\pi/3}{2\pi}
=
\frac{1}{3}
\]

The particle moves slowly near the extreme positions, so it spends a noticeable fraction of time there.

A block of mass \(m\) is attached to a spring of constant \(k\) on a smooth horizontal surface. At the instant it passes through equilibrium with speed \(v\), the spring constant is suddenly changed to \(4k\). What is the new amplitude of oscillation?

1. \(\dfrac{v}{\omega}\)
2. \(\dfrac{v}{4\omega}\)
3. \(\dfrac{v}{2\omega}\)
4. \(\dfrac{2v}{\omega}\)

Explanation

Before the change,

\[
\omega=\sqrt{\frac{k}{m}}
\]

At equilibrium, the displacement is zero and the speed remains \(v\) immediately after the sudden change.

The new angular frequency is

\[
\omega’=\sqrt{\frac{4k}{m}}=2\omega
\]

At equilibrium, the speed is maximum, so

\[
v=\omega’A’
\]

Thus,

\[
A’=\frac{v}{2\omega}
\]

A particle moves along the x-axis under the force \(F=-ax\) for \(x>0\) and \(F=-4ax\) for \(x<0\), where \(a>0\). If it is released from rest at \(x=A\), what is the time period of its oscillation?

1. \(2\pi\sqrt{\dfrac{m}{a}}\)
2. \(\pi\sqrt{\dfrac{m}{a}}\)
3. \(\dfrac{3\pi}{4}\sqrt{\dfrac{m}{a}}\)
4. \(\dfrac{3\pi}{2}\sqrt{\dfrac{m}{a}}\)

Explanation

For \(x>0\),

\[
\omega_1=\sqrt{\frac{a}{m}}
\]

The time from \(+A\) to \(0\) is one-quarter period:

\[
t_1=\frac{\pi}{2\omega_1}
\]

For \(x<0\), \[ \omega_2=\sqrt{\frac{4a}{m}}=2\omega_1 \] The time from \(0\) to the negative extreme is \[ t_2=\frac{\pi}{2\omega_2} = \frac{\pi}{4\omega_1} \] The full cycle includes two positive-side quarters and two negative-side quarters: \[ T=2t_1+2t_2 \] \[ T= \frac{\pi}{\omega_1} + \frac{\pi}{2\omega_1} \] \[ T= \frac{3\pi}{2}\sqrt{\frac{m}{a}} \]

A simple pendulum has period \(T\) on Earth. It is placed inside a spacecraft orbiting Earth, where the spacecraft and pendulum are in free fall. What is the period of small oscillations of the pendulum relative to the spacecraft?

1. \(T\)
2. \(\dfrac{T}{2}\)
3. 2T
4. The pendulum does not execute ordinary gravity-based SHM

Explanation

Inside an orbiting spacecraft, both the support and pendulum are in continuous free fall.

The effective gravitational field inside the spacecraft is approximately zero.

For a simple pendulum,

\[
T=2\pi\sqrt{\frac{L}{g_{\text{eff}}}}
\]

Since

\[
g_{\text{eff}}\approx0
\]

there is no gravity-based restoring force.

Therefore, the pendulum does not perform ordinary simple pendulum SHM.

A mass \(m\) is attached to two springs of constants \(k\) and \(3k\) in series on a frictionless horizontal surface. What is the angular frequency of small oscillations?

1. \(\sqrt{\dfrac{4k}{m}}\)
2. \(\sqrt{\dfrac{k}{m}}\)
3. \(\sqrt{\dfrac{3k}{4m}}\)
4. \(\sqrt{\dfrac{4k}{3m}}\)

Explanation

For springs in series,

\[
\frac{1}{k_{\text{eff}}}
=
\frac{1}{k}
+
\frac{1}{3k}
\]

\[
\frac{1}{k_{\text{eff}}}
=
\frac{4}{3k}
\]

\[
k_{\text{eff}}=\frac{3k}{4}
\]

Therefore,

\[
\omega
=
\sqrt{\frac{k_{\text{eff}}}{m}}
\]

\[
\omega
=
\sqrt{\frac{3k}{4m}}
\]

A particle executes SHM with amplitude \(A\). At what displacement is the magnitude of its velocity equal to the magnitude of its acceleration multiplied by \(\dfrac{T}{2\pi}\)?

1. \(\dfrac{A}{2}\)
2. \(\dfrac{A}{2\sqrt{2}}\)
3. \(\dfrac{A}{\sqrt{2}}\)
4. \(\dfrac{\sqrt{3}A}{2}\)

Explanation

Since

\[
\omega=\frac{2\pi}{T}
\]

the given condition becomes

\[
|v|=\frac{|a|}{\omega}
\]

For SHM,

\[
|v|=\omega\sqrt{A^2-x^2}
\]

and

\[
|a|=\omega^2|x|
\]

Thus,

\[
\omega\sqrt{A^2-x^2}
=
\omega|x|
\]

\[
A^2-x^2=x^2
\]

\[
x=\frac{A}{\sqrt{2}}
\]

A block of mass \(m\) attached to a spring oscillates with amplitude \(A\). At the instant the block is at \(x=\dfrac{A}{2}\), it receives a sudden impulse in the direction opposite to its velocity, reducing its speed to zero. What is the new amplitude?

1. \(\dfrac{A}{2}\)
2. \(\dfrac{A}{3}\)
3. \(\dfrac{\sqrt{3}A}{2}\)
4. A

Explanation

Immediately after the impulse, the block is at

\[
x=\frac{A}{2}
\]

and its velocity becomes zero.

A point with zero velocity is an extreme position of the new oscillation.

Therefore, the new amplitude is simply

\[
A’=\frac{A}{2}
\]

The impulse removes part of the mechanical energy.

A particle moves in the potential \(U(x)=\alpha(x^2-a^2)^2\), where \(\alpha>0\). What is the angular frequency of small oscillations about either stable equilibrium position?

1. \(\sqrt{\dfrac{2\alpha a^2}{m}}\)
2. \(\sqrt{\dfrac{4\alpha a^2}{m}}\)
3. \(\sqrt{\dfrac{6\alpha a^2}{m}}\)
4. \(\sqrt{\dfrac{8\alpha a^2}{m}}\)

Explanation

Stable equilibrium positions occur at

\[
x=\pm a
\]

For small oscillations,

\[
\omega^2=\frac{1}{m}\left.\frac{d^2U}{dx^2}\right|_{x=\pm a}
\]

First,

\[
U=\alpha(x^4-2a^2x^2+a^4)
\]

\[
\frac{d^2U}{dx^2}
=
12\alpha x^2-4\alpha a^2
\]

At

\[
x=\pm a
\]

\[
\frac{d^2U}{dx^2}
=
8\alpha a^2
\]

Therefore,

\[
\omega
=
\sqrt{\frac{8\alpha a^2}{m}}
\]

A particle executes SHM with period \(T\). Starting from the mean position and moving toward positive x, how long does it take to reach \(x=\frac{\sqrt{3}A}{2}\) for the first time?

1. \(\dfrac{T}{12}\)
2. \(\dfrac{T}{8}\)
3. \(\dfrac{T}{6}\)
4. \(\dfrac{T}{4}\)

Explanation

Starting from the mean position,

\[
x=A\sin(\omega t)
\]

Given,

\[
A\sin(\omega t)=\frac{\sqrt{3}A}{2}
\]

\[
\sin(\omega t)=\frac{\sqrt{3}}{2}
\]

For the first time,

\[
\omega t=\frac{\pi}{3}
\]

Since

\[
\omega=\frac{2\pi}{T}
\]

\[
t=
\frac{\pi/3}{2\pi/T}
\]

\[
t=\frac{T}{6}
\]