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

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)

Calculation:

Given:

Mass of the particle, m = m

Amplitude of SHM, a = a

Frequency of SHM, n = n

Using the formula for maximum velocity:

v_max = 2π n a

Substituting this into the formula for kinetic energy:

KE_max = (1/2) m v²

⇒ KE_max = (1/2) m (2π n a)²

⇒ KE_max = (1/2) m × 4π² n² a²

⇒ KE_max = 2m π² n² a²

Concept:

The frequency of a spring-mass system is inversely proportional to the square root of the mass attached to the spring.

Formula: f ∝ 1 / √m, where:

f = frequency of oscillation (in Hz)

m = mass attached to the spring (in kg)

When the mass changes, the relationship between the frequencies can be expressed as:
f₂ = f₁ × √(m₁ / m₂)

Calculation:

Given:

Initial frequency, f₁ = 4 Hz

Initial mass, m₁ = 1 kg

New mass, m₂ = 4 kg

Using the formula:

f₂ = f₁ × √(m₁ / m₂)

⇒ f₂ = 4 × √(1 / 4)

⇒ f₂ = 4 × (1 / 2)

⇒ f₂ = 2 Hz

∴ The frequency of oscillations is 2 Hz.

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.

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

Calculation:

From conservation of energy:

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

⇒ (1/2) m v² + (1/2) (I_CM + 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.

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:

$\omega =\frac{2\pi }{T}$

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.

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.

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.

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:

$\omega =\frac{2\pi }{T}$

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

$T=\frac{2\pi }{\omega }$

$T={\displaystyle \frac{2\pi }{\sqrt{b}}}$

So the correct answer is option 3.

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

Calculation:

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

At its mean position x = 0

Putting the value in equation 1,

⇒ $v=\omega \sqrt{A^2-0^2}$

⇒ 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

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

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:

• 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).

$Maximumvelocity\left(V\right)=\omega \sqrt{A^2-x^2}=\omega \sqrt{A^2-0^2}=\omega A$

• Therefore option 2 is correct.

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.

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.

ma = -k x

a = -(k/m) x

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

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.

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