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

Calculation:

Maximum speed in simple harmonic motion is given by:

v_max = A × ω

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

ω = √(k / m)

Let A_P and A_Q be amplitudes and ω₁ and ω₂ be angular frequencies of P and Q respectively.

Given: v_max(P) = v_max(Q)

⇒ A_P × ω₁ = A_Q × ω₂

⇒ A_Q / A_P = ω₁ / ω₂

⇒ A_Q / A_P = √(k₁) / √(k₂)

⇒ A_Q / A_P = √(k₁ / k₂)

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

Concept:

Simple Harmonic Oscillation: The oscillatory motion in which the restoring force on the particle at any position is directly proportional to the displacement of the particle from its mean position.

The magnitude of the restoring force is given as: \(F_{restoring} = kx\)

Given:

• m is the mass of the particle.
• k is the spring constant.
• l is the unstretched length of the spring.
• \(\omega\) is the angular speed of the spring.

Solution:

Since the body is rotating in a gravity-free space, the spring force provides the centripetal force.

\(F_{centripetal} = F_{spring} \)

\(m\omega^2r=kx\) ----(1)

Let the increase in the length of the spring be \(x\).

∴ the total the stretched length of the spring during the rotation will be 

\(r=l+x\), Substituting this in equation (1) we get,

\(m\omega^2(l+x)=kx\)

\(\frac{(l+x)}{x}=\frac{k}{m\omega^2}\)

\(\frac{l}{x}+1=\frac{k}{m\omega^2}\)

\(\frac{l}{x}=\frac{k}{m\omega^2}-1\)

\(\frac{l}{x}=\frac{k-m\omega^2}{m\omega^2}\)

\(\frac{x}{l}=\frac{m\omega^2}{k-m\omega^2}\)

\(x=\frac{m\omega^2l}{k-m\omega^2}\)

∴ The increase in the spring will be \(x=\frac{m\omega^2l}{k-m\omega^2}\)

Hence, the correct option is (2)

Answer : 4

Solution :

\(\mathrm{K} . \mathrm{E}=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\) and

\(\text { P.E }=\frac{1}{2} m \omega^2 x^2\)

∴ T.E = K.E + P.E

= \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2\)

= \(\frac{1}{2} m \omega^2 A^2\)

Given: K.E = P.E

∴ A² - x² = x²

∴ A² = 2x²

∴ \(x=\sqrt{\frac{A^2}{2}}=\frac{A}{\sqrt{2}}\)

∴ \(x=\frac{4}{\sqrt{2}}=2 \sqrt{2} \mathrm{~cm}\)

Concept:

The kinetic energy of the simple harmonic oscillator is written as:

\(K.E. = \frac{1}{2}m\omega ^2 (A^2 - x^2)\;\)

The potential energy of the simple harmonic oscillator is written as;

\(P.E. = \frac{1}{2}m\omega^2 x^2\;\)

Here we have m as the mass, \(\omega\) is the angular frequency and A is the amplitude and x is the displacement.

Calculation:

Given:

Kinetic Energy = Potential Energy

\(\frac{1}{2}m\omega ^2 (A^2 - x^2)= \frac{1}{2}m\omega^2 x^2\)

⇒ \(A^2-x^2 = x^2\)

⇒ \(A^2 = x^2 +x^2\)

⇒ \(A^2 = 2x^2\)

⇒ \(x^2 = \frac{A^2}{2}\)

⇒ \(x = ±\rm\frac{A}{\sqrt2}\)

Hence, option 3) is the correct answer.

Calculation:

In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE). At any point in the motion, the kinetic energy and potential energy can be expressed as:

The total energy (E) in SHM of amplitude A is

E = (1/2) k A²

where k is the spring constant.

The potential energy (PE) at a distance x from the mean position is given by:

PE = (1/2) k x²

The kinetic energy (KE) at a distance x from the mean position is:

KE = (1/2) k (A² - x²)

We need to find the distance from the mean position where the kinetic energy is equal to the potential energy. Therefore, we set the kinetic energy equal to the potential energy:

(1/2) k x² = (1/2) k (A² - x²)

By simplifying the equation, we get:

⇒ x² = A² - x²

⇒ 2x² = A²

⇒ x² = A² / 2

⇒ x = (A √2) / 2

We know that √2 ≈ 1.41, so:

x ≈ (1.41 / 2) A ≈ 0.71 A

Therefore, the distance from the mean position at which the kinetic energy is equal to its potential energy is approximately 0.71A. Hence, the correct answer is Option 2.

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.

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 (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.

• 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.

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

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

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).

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

• 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

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

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.

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