Simple Harmonic Motion MCQ Quiz - Objective Question with Answer for Simple Harmonic Motion - Download Free PDF

Last updated on Jul 8, 2025

Latest Simple Harmonic Motion MCQ Objective Questions

Simple Harmonic Motion Question 1:

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A simple pendulum has a time period T1 when it oscillates under normal gravity. Now, the point of suspension starts accelerating downward with a constant acceleration a = 2 m/s2. Let the new time period be T2. Find the ratio T12 / T22. (Take g = 10 m/s2)

  1. 6 / 4
  2. 5 / 4
  3. 1
  4. 4 / 5

Answer (Detailed Solution Below)

Option 4 : 4 / 5

Simple Harmonic Motion Question 1 Detailed Solution

Calculation:

In a non-inertial frame accelerating downward, the effective gravity is:

g' = g - a = 10 - 2 = 8 m/s2

Time period of a pendulum: T ∝ 1 / √g

So, T12 / T22 = g' / g = 8 / 10 = 4 / 5

Correct Option: (D)

Simple Harmonic Motion Question 2:

A particle executes simple harmonic motion between x = –A and x = +A. If time taken by particle to go from x = 0 to \(\frac{A}{2}\) is 2s; then time taken by particle in going from \(\rm x=\frac{A}{2}\) to A is :

  1. 3 s
  2. 2 s
  3. 1.5 s 
  4. 4 s

Answer (Detailed Solution Below)

Option 4 : 4 s

Simple Harmonic Motion Question 2 Detailed Solution

Calculation:

Let the time from 0 to A/2 be t1 and the time from A/2 to A be t2.

From the given motion, the angular displacement for t1 is ωt1 = π/6, and for t2, ωt2 = π/3.

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We can now find the ratio of the times:

t1 / t2 = 1 / 2

Using this ratio:

t2 = 2 × t1 = 2 × 2 = 4 sec

Final Answer: t2 = 4 sec

Simple Harmonic Motion Question 3:

The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R > r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T = \(\frac{2 \pi}{ω}\). The correct expression for ω is (g is the acceleration due to gravity): 

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  1. \(\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}\)
  2. \(\sqrt{\frac{2 g}{3(R-r)}+\frac{k}{m}}\)
  3. \(\sqrt{\frac{1}{6}\left(\frac{g}{R-r}+\frac{k}{m}\right)}\)
  4. \(\sqrt{\frac{1}{4}\left(\frac{g}{R-r}+\frac{k}{m}\right)}\)

Answer (Detailed Solution Below)

Option 1 : \(\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}\)

Simple Harmonic Motion Question 3 Detailed Solution

Calculation:

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From conservation of energy:

Total mechanical energy = Translational KE + Rotational KE + Spring PE + Gravitational PE

⇒ (1/2) m v2 + (1/2) (ICM + m(R − r)2) θ̇2 + (1/2) k ((R − r) θ)2 + mg (R − r)(1 − cosθ) = Constant

For rolling without slipping v = (R-r)θ̇ = rϕ̇   

Total energy:

E = (1/2) m(R−r)2θ̇2 + (1/2)((1/2)mr2 + m(R−r)2)θ̇2 + (1/2)k(R−r)2θ2 + mg(R−r)(1−cosθ) = constant

Differentiating w.r.t time:

⇒ [ (3/4) m(R−r)2 + (1/4) m r2 ] θ̈ + [ k(R−r)2/2 + mg(R−r) ] θ = 0

⇒ θ̈ + ω2θ = 0

⇒ ω2 = [k(R−r)2/2 + mg(R−r)] / [ (3/4) m(R−r)2 + (1/4) m r2 ]

Neglecting r2 as r << (R−r):

⇒ ω = √[(2/3)(k/m + g/(R−r))] ϕ 

Therefore, correct option is A.

Simple Harmonic Motion Question 4:

A particle is subjected two simple harmonic motions as
\(x_1 = \sqrt{7} \sin 5t \, \text{cm} \\ \text{and } x_2 = 2\sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \, \text{cm}\)
where x is displacement and t is time in seconds. The maximum acceleration of the particle is x × 10–2 ms–2. The value of x is : 
 

  1. 175
  2. \(25\sqrt{7} \)
  3. \(5\sqrt{7} \)
  4. 125

Answer (Detailed Solution Below)

Option 1 : 175

Simple Harmonic Motion Question 4 Detailed Solution

Calculation:

Given:

x1 = √7 sin 5t cm

x2 = 2√7 sin(5t + π/3) cm

Angular frequency, ω = 5 rad/s

Using phasor diagram:

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Angle between x1 and x2 = 60°

Resultant amplitude A:

A = √[ (√7)2 + (2√7)2 + 2 × √7 × 2√7 × cos 60° ]

⇒ A = √[ 7 + 4 × 7 + 2 × √7 × 2√7 × (1/2) ]

⇒ A = √49 = 7 cm

Maximum acceleration:

amax = A × ω2

⇒ amax = 7 × 25 cm/s2

⇒ amax = 175 × 10-2 m/s2

∴ The value of x is 175

Simple Harmonic Motion Question 5:

Two identical point masses P and Q, suspended from two separate massless springs of spring constants k₁ and k₂ respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQ / AP) of the amplitude AQ of mass Q to the amplitude AP of mass P is:

  1. \(\frac{k_2}{k_1}\)
  2. \(\frac{k_1}{k_2}\)
  3. \(\sqrt{\frac{k_2}{k_1}}\)
  4. \(\sqrt{\frac{k_1}{k_2}}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt{\frac{k_1}{k_2}}\)

Simple Harmonic Motion Question 5 Detailed Solution

Calculation:

Maximum speed in simple harmonic motion is given by:

vmax = A × ω

Since the masses are identical, angular frequency ω is given by:

ω = √(k / m)

Let AP and AQ be amplitudes and ω1 and ω2 be angular frequencies of P and Q respectively.

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Given: vmax(P) = vmax(Q)

⇒ AP × ω1 = AQ × ω2

⇒ AQ / AP = ω1 / ω2

⇒ AQ / AP = √(k1) / √(k2)

⇒ AQ / AP = √(k1 / k2)

Therefore, the correct answer is Option 4: √(k1 / k2)

Top Simple Harmonic Motion MCQ Objective Questions

If simple harmonic motion is represented by x = A cos(ωt + φ), then 'ω' is _____________.

  1. Displacement
  2. Amplitude
  3. Angular frequency
  4. Phase constant

Answer (Detailed Solution Below)

Option 3 : Angular frequency

Simple Harmonic Motion Question 6 Detailed Solution

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CONCEPT:

  • Simple harmonic motion occurs when the restoring force is directly proportional to the displacement from equilibrium.

F α -x

Where F = force and x = the displacement from equilibrium.

The equation of SHM is given by:

x = A sin(ωt + ϕ)

where x is the distance from the mean position at any time t, A is amplitude, t is time, ϕ is initial phase and ω is the angular frequency.

  • The amplitude of SHM (A): maximum displacement from the mean position.
  • frequency (f): no. of oscillations in one second.
  • Time period (T): time taken to complete one oscillation.

The relation between time period (T) and frequency (f) is given by:

\(f=\frac{1}{T}\)

Angular frequency (ω) of SHM is given by:

\(\omega=\frac{2π}{T}\)

where T is the time period.

EXPLANATION:

If the simple harmonic motion is represented by x = A cos(ωt + φ), then 'ω' is the angular frequency.

So the correct answer is option 3.

The body is said to move with Simple Harmonic Motion if its acceleration is ______.

  1. Always directed away from the centre, at the point of reference
  2. Proportional to square of the distance from the point of reference
  3. Proportional to the distance from the point of reference and directed towards it
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Proportional to the distance from the point of reference and directed towards it

Simple Harmonic Motion Question 7 Detailed Solution

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CONCEPT:

  • Simple Harmonic Motion (SHM)Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
    • Example: Motion of an undamped pendulum, undamped spring-mass system.

Force (F) = - k x

Acceleration (a) = - (k/m) x

Where a is acceleration, x is the displacement of the system from its equilibrium position, m is the mass of the system and k is a constant associated with the system.

EXPLANATION:

  • In simple harmonic motion, the acceleration is proportional to the distance from the point of reference and directed towards it. So option 3 is correct.

The acceleration of a particle performing simple harmonic motion is ____________ whose displacement is f(t) = A cos(ωt + φ).

  1. –ω2A cos(ωt + φ)
  2. –ωA cos(ωt + φ)
  3. –ω2A sin(ωt + φ)
  4. –ωA sin(ωt + φ)

Answer (Detailed Solution Below)

Option 1 : –ω2A cos(ωt + φ)

Simple Harmonic Motion Question 8 Detailed Solution

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CONCEPT:

Simple harmonic motion occurs when the restoring force is directly proportional to the displacement from equilibrium.

F α -x

Where F = force and x = the displacement from equilibrium.

The equation of displacement in SHM is given by:

x = A Cos(ωt+ϕ) .........(i)

where x is the distance from the mean position at any time t, A is amplitude, t is time, and ω is the angular frequency.

The equation of velocity in SHM is given by differentiating equation (i)

v = dx/dt = d (A Cos(ωt+ϕ)) / dt

v = - Aω Sin(ωt+ϕ)

where v is the velocity at any time t, A is amplitude, t is time, and ω is angular frequency.

In the same way, the equation of displacement can be obtained from the equation of velocity by integrating the equation of velocity.

CALCULATION:

v = - Aω Sin(ωt+ϕ)

The acceleration is given by:

a = dv/dt = d (- Aω Sin(ωt+ϕ))/dt = –ω2A cos(ωt + φ).

So option 1 is correct.

Equation of motion of a particle is given by a = -bx, where a is the acceleration, x is the displacement from the mean position and b any constant. The time period of the particle is

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

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{2\pi}{\sqrt{b}}\)

Simple Harmonic Motion Question 9 Detailed Solution

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CONCEPT:

  • Simple harmonic motion occurs when the restoring force is directly proportional to the displacement from equilibrium.

F α -x

Where F = force and x = the displacement from equilibrium.

  • For a simple harmonic motion equation of acceleration

a = -ω2x

where a is the acceleration ω is the angular frequency and x is the displacement.

  • Time period (T): time taken to complete one oscillation.

The relation between time period (T) and frequency (f) is given by:

\(f=\frac{1}{T}\)

  • Angular frequency (ω) of SHM is given by:

\(\omega=\frac{2π}{T}\)

where T is the time period.

CALCULATION:

Given that Equation of motion is a = -bx, where a is the acceleration, x is the displacement from the mean position, and b any constant.

Compare it with the equation of acceleration a = -ω2x

ω2 = b

ω = √b

\(T=\frac{2π}{\omega}\)

\(T=\dfrac{2\pi}{\sqrt{b}}\)

So the correct answer is option 3.

The velocity of a particle, executing S.H.M, is ________ at its mean position. 

  1. maximum
  2. minimum
  3. infinity
  4. zero

Answer (Detailed Solution Below)

Option 1 : maximum

Simple Harmonic Motion Question 10 Detailed Solution

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Concept

Simple Harmonic Motion or SHM is a specific type of oscillation in which the restoring force is directly proportional to the displacement of the particle from the mean position.

  • Velocity of SHM, \(v = ω \sqrt{A^2- x^2}\)
  • Where, x = displacement of the particle from the mean position,
  • A = maximum displacement of the particle from the mean position.
  • ω = Angular frequency

F1 Madhuri Defence 17.10.2022 D1

Calculation:

Velocity of SHM, \(v = ω \sqrt{A^2- x^2}\) --- (1)

At its mean position x = 0

Putting the value in equation 1,

⇒ \(v = ω \sqrt{A^2- 0^2}\)

⇒ v = ωA, which is maximum.

So, velocity is maximum at mean position.

At extreme position, x = ± A, v = 0

So, velocity is minimum or zero at extreme position.

Additional Information 

  • Acceleration, a = ω2x
  • Acceleration is maximum at the extreme position, x = ± A
  • Acceleration is minimum or zero at the mean position, a = 0

If simple harmonic motion is represented by x = A cos(ωt + φ), then 'φ' is _____________.

  1. Angular frequency
  2. Displacement
  3. Amplitude
  4. Phase constant

Answer (Detailed Solution Below)

Option 4 : Phase constant

Simple Harmonic Motion Question 11 Detailed Solution

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CONCEPT:

  • Simple Harmonic Motion (SHM)Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

Force (F) = - k x

Where k = restoring force, x = distance from the equilibrium position, F = force it experiences towards mean position

  • Example: Motion of an undamped pendulum, undamped spring-mass system.

​EXPLANATION:

  • Simple harmonic motion is represented by

⇒ x = A cos(ωt + φ)

Where A = amplitude, ω = Angular frequency, x = Displacement and φ = Phase constant

F1 A.K Madhu 26.06.20 D6

If 'V' is the magnitude of velocity, 'A' is the magnitude of acceleration and 'X' is the magnitude of displacement, then at the mean position of the particle performing simple harmonic motion, __________. 

  1. A is maximum and X and V are zero
  2. V is maximum and X and A are zero
  3. V and A are maximum and X is zero
  4. X and A are maximum and V is zero

Answer (Detailed Solution Below)

Option 2 : V is maximum and X and A are zero

Simple Harmonic Motion Question 12 Detailed Solution

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CONCEPT:

  • Wave: It is a disturbance that transfers energy from one place to another.
  • SHM (Simple harmonic motion): The type of oscillatory motion in which the restoring force on the system is directly proportional to the displacement of the system is called SHM.

The general expression for the simple harmonic equation is given by:

X = A Sin (ω t)

Where A is the amplitude of SHM, ω is the angular frequency and t is time

Velocity (V) of a particle at any position is given by:

\(V = \omega \sqrt {{A^2} - {x^2}}\)

Acceleration of the particle at any position is given by:

Acceleration (a) = ω2 x

EXPLANATION:

F1 J.K 16.4.20 Pallavi D4

  • From the above figure we can see that, at the mean position, the amplitude of S.H.M is zero 

i.e., x = A = 0 at mean position whereas at x = A at extreme position 

  • The acceleration of the particle will be minimum at mean position points (x = 0).

Maximum Acceleration (a) = ω2 x = 0

  • The velocity of the particle will be maximum at the mean position (x = 0).

\(Maximum\;velocity\;\left( V \right) = \omega \;\sqrt {{A^2} - {x^2}\;} = \omega \;\sqrt {{A^2} - {0^2}} = \omega \;A\)

  • Therefore option 2 is correct.

In the simple harmonic motion x = √2 sin(ωt – π/4), the phase constant can be shown as?

  1. √2
  2. 2π /ω
  3. ωt
  4. 7π /4

Answer (Detailed Solution Below)

Option 4 : 7π /4

Simple Harmonic Motion Question 13 Detailed Solution

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CONCEPT: 

  • Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
    • Example: Motion of an undamped pendulum, undamped spring-mass system.

The equation of shm is given by:

X = A Sin (ω t + θ)

Where A is amplitude, ω is the angular frequency, t is time and θ is the initial phase angle or phase constant.

EXPLANATION:

Given that:

x = √2 sin(ωt – π/4) = √2 sin( 2π + ωt – π/4) = √2 sin(ωt + 2π - π/4) = √2 sin(ωt + 7π/4)
Thus x = √2 sin(ωt + 7π/4)

Phase constant = 7π/4

So option 4 is correct.

If 'F' is the force acting on a particle of mass 'm' performing simple harmonic motion, then the time period of the simple harmonic motion is equal to ___________.

  1. 2π √(k /m)
  2. 2π √(m/k)
  3. 1/(2π) x √( m/k)
  4. 1/(2π) x √(k/m)

Answer (Detailed Solution Below)

Option 2 : 2π √(m/k)

Simple Harmonic Motion Question 14 Detailed Solution

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CONCEPT:

  • Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
    • Example: Motion of an undamped pendulum, undamped spring-mass system.
  • Time period (T): The time taken by the particle to completed one revolution is called time period.

The time period (T) of a spring block system doing the simple harmonic motion is given by:

Here force (F) is equal to the weight (mg) of the block.

03.10.2017.003

ma = -k x

a = -(k/m) x

\({\mathbf{T}} = 2\;{\mathbf{π }}\;\sqrt {\frac{{\mathbf{m}}}{{\mathbf{k}}}} \)

Where m is mass and k is spring constant.

EXPLANATION

The time period of the simple harmonic motion is given by

T = 2π √(m/k)

So option 2 is correct.

The two equations of Simple harmonic motions are Y1 = 12 Sin (50 t + π/2) and Y2 = 12 Sin (50 t + π/3). Find the phase difference between two motions.

  1. π/2
  2. π/3
  3. π/6
  4. π/8

Answer (Detailed Solution Below)

Option 3 : π/6

Simple Harmonic Motion Question 15 Detailed Solution

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CONCEPT:

  • Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
  • Example: Motion of an undamped pendulum, undamped spring-mass system.
  • The equation of shm is given by:

Y = A Sin (ω t + θ)

Where A is amplitude, ω is the angular frequency, t is time and θ is the initial phase angle

EXPLANATION:

Given that:

Y1 = 12 Sin (50 t + π/2)

Phase angle (θ1) = π/2

Y2 = 12 Sin (50 t + π/3)

Phase angle (θ2) = π/3

Phase difference (Δ θ) = θ1 – θ2 = π/2 – π/3 = π/6

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