Oscillations MCQ Quiz - Objective Question with Answer for Oscillations - 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)

Explanation:

At any point of time, time period is given by

T = 2π √(m / k)

Here m is decreasing, so time period T will be decreasing

Since ω = 2π / T

Hence as mass leaks, ω will increase

Now, at any instant

mg = kx₀

So, equilibrium length x₀ = mg / k, where m is decreasing

So, equilibrium length will decrease.

So, amplitude also go on decreasing.

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.

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 correct answer is Infinite.

CONCEPT:

• Simple pendulum: A point mass attached to a light inextensible string and suspended from fixed support is called a simple pendulum.
  • The vertical line passing through the fixed support is the mean position of a simple pendulum.

The vertical distance between the point of suspension and the center of mass of the suspended body is called the length of the simple pendulum denoted by L.

The time period of the pendulum:

\(T = 2\pi\sqrt{L/g}\)

where, L = length of string, g = acceleration due to gravity (m/s²)

EXPLANATION:

• As we know that \(T \alpha \frac{1}{g}\)
• On an artificial satellite, orbiting the earth, net gravity is zero.

As g = 0, so T = ∞ 

The correct option is ∞ .

Additional Information

The value of g is zero in orbiting satellite:

• Natural satellites have their own acceleration due to gravity. Artificial satellites don't possess any natural g.
  • Artificial satellites experience a gravitational pull towards earth due to the earth's gravitational force.
  • But as the velocity of artificial satellites is extremely large, so, they have a centrifugal force acting tangentially to the orbit.
  • As the gravitational pull of the earth on the artificial satellite is equal to the centrifugal force acting on the artificial satellite, the net force cancels.

CONCEPT:

Simple Pendulum:

• An ideal simple pendulum consists of a heavy point mass body (bob) suspended by a weightless, inextensible, and perfectly flexible string from rigid support about which it is free to oscillate.
• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.

\(\Rightarrow T = 2\;{\rm{\Pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} \)

The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = length of the pendulum, and g = gravitational acceleration 

• The force which acts to bring a body to its mean position/equilibrium position is called a restoring force. 
• Example: The spring force is a restoring force because it always applies a force towards the equilibrium point.

EXPLANATION:

• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.

\(\Rightarrow T = 2\;{\rm{\Pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}}\)

• From the above equation, it is clear that the period of oscillation is directly proportional to the length of the arm and inversely proportional to acceleration due to gravity.
• Thus, an increase in the length of a pendulum arm results in a subsequent increase in the period of oscillation given a constant gravitational acceleration. Thus option 2 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 (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 Pendulum

• An ideal simple pendulum consists of a heavy point mass body (bob) suspended by a weightless, inextensible, and perfectly flexible string from rigid support about which it is free to oscillate.
• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity,

\(\Rightarrow T=2\pi\sqrt{\frac{l}{g}}\,\)

• The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = Effective length of the pendulum, and g = gravitational acceleration

Explanation:

• The swinging girl can be considered as a simple pendulum.

The time period of oscillation of the simple pendulum is,

\(\Rightarrow T=2\pi\sqrt{\frac{l}{g}}\,\)

where l is the effective length of the simple pendulum.

• The length of the simple pendulum is equal to the distance from point of suspension to the centre of gravity (CG) of the oscillating body.
• When the girl stands up, the distance of CG from the point of suspension decreases.
• Therefore, the time period of oscillation also decreases. 

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:

• Simple pendulum: When a point mass is suspended with the help of a string or rod of negligible mass and does the to and fro motion about its mean position is called as a simple pendulum.
• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.

\(T = 2{\rm{\Pi }}\sqrt{ {\frac{{\rm{l}}}{{\rm{g}}}}}\)

The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = length of the pendulum, and g = gravitational acceleration

CALCULATION:

Given that:

Acceleration due to gravity on the moon (g') = acceleration due to gravity on earth (g)/6

\(T = 2{\rm{\Pi }}\sqrt{{\frac{{\rm{l}}}{{\rm{g}}}}}\) : The time period on earth.

The time period on the moon (T') is given by:

\(T' = 2{\rm{\Pi }}\sqrt{\frac{{\rm{l}}}{{\rm{g'}}}} =2{\rm{\Pi }}\sqrt{\frac{{\rm{l}}}{{\rm{g/6}}}}\)

\(T' = 2{\rm{\Pi }}\sqrt{\frac{{\rm{6l}}}{{\rm{g}}}}\)

T' = √6 T

So option 3 is correct