Equation of SHM - NEET Physics Questions
Question 31: easy

A particle of mass 4 kg is moving along x-axis under the action of force \(F = -\pi^2 x\) (where F is in newton and x is in m). The time period of oscillation is

1. 1 s
2. 2 s
3. 3 s
4. 4 s
View Answer

Comparing \(F = -\pi^2 x\) with \(F = -k x\), we get \(k = \pi^2\). The time period is \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{4}{\pi^2}} = 2\pi \left(\frac{2}{\pi}\right) = 4 \text{ s}\).

Question 32: easy

The differential equation for a particle executing S.H.M. is given by \(\frac{d^2y}{dt^2} + 4y = 0\), where symbols have their usual meaning. The angular velocity of the particle is given by

1. 4 rad/s
2. 3 rad/s
3. 2 rad/s
4. 4π rad/s
View Answer

The standard equation of S.H.M. is \(\frac{d^2y}{dt^2} + \omega^2 y = 0\). Comparing this with the given equation, \(\omega^2 = 4 ⇒ \omega = 2\text{ rad/s}\).

Question 33: easy

If maximum speed and maximum acceleration of a particle executing SHM is found to be 5 cm/s and \( 50\pi \, \text{cm/s}^2 \) respectively, then its time period will be

1. 5 s
2. 2 s
3. \( \frac{1}{5} \, \text{s} \)
4. \( \frac{1}{10} \, \text{s} \)
View Answer

Using \( v_{\text{max}} = A \omega = 5 \) and \( a_{\text{max}} = A \omega^2 = 50\pi \), we get \( \omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{50\pi}{5} = 10\pi \, \text{rad/s} \). The time period is \( T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = \frac{1}{5} \, \text{s} \).

Question 34: easy

The differential equation for a particle executing S.H.M. is given by \( \frac{d^2 y}{dt^2} + 4y = 0 \), where symbols have their usual meaning. The angular velocity of the particle is given by

1. \( 4\text{ rad/s} \)
2. \( 3\text{ rad/s} \)
3. \( 2\text{ rad/s} \)
4. \( 4pi\text{ rad/s} \)
View Answer

The standard differential equation of S.H.M. is \( \frac{d^2 y}{dt^2} + \omega^2 y = 0 \). By comparison, \( \omega^2 = 4 ⇒ \omega = 2\text{ rad/s} \).

Question 35: easy

Assertion (A): A hole were drilled through the centre of earth and a ball is dropped into the hole at one end, it will not get out of other end of the hole.


Reason (R): Ball will execute simple harmonic motion inside the hole.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept of gravity inside earth and SHM. Inside Earth, \(F = -k r\). A ball dropped in a hole through the Earth's center oscillates in SHM. It reaches the other end with zero velocity and turns back, thus not 'getting out' (escaping) the hole.


Both A and R are true, and R explains A.

Question 36: easy

Assertion (A): A SHM may be assumed as composition of many SHM’s.


Reason (R): Superposition of many SHM’s (along same line) of same frequency will be a SHM.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) states that an SHM can be viewed as a composition of multiple SHMs. Reason (R) states that the superposition of multiple SHMs along the same line and with the same frequency results in another SHM.


Both statements are true, and (R) provides the explanation for how (A) can be possible.

Question 37: easy

Assertion (A): Displacement-time equation of a particle moving along \(x\)-axis is \(x = 4 + 6 sin\omega t\). Under this situation, motion of particle is not simple harmonic.


Reason (R): \(\frac{d^2x}{dt^2}\) for the given equation is not proportional to \(-x\).


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

For SHM, \(a = -\omega^2 (x-x_0)\). Given \(x = 4 + 6 sin\omega t\), the equilibrium position is \(x_0=4\). The acceleration is \(\frac{d^2x}{dt^2} = -\omega^2 (x-4)\). If SHM is strictly defined as \(a \propto -x\) (equilibrium at origin), then (A) is true. (R) is also true as \(\frac{d^2x}{dt^2}\) is proportional to \(-(x-4)\), not \(-x\). (R) explains (A).

Question 38: easy

Assertion (A): For a particle performing SHM, its speed decreases as it goes away from the mean position.


Reason (R): In SHM, the acceleration is always opposite to the velocity of the particle.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

As a particle in SHM moves from the mean to extreme position, its speed decreases as the restoring force opposes motion. So (A) is true. Acceleration is always directed towards the equilibrium. When moving towards equilibrium, velocity and acceleration are in the same direction, so (R) is false.

Question 39: easy

Assertion (A): Motion of a ball bouncing elastically in vertical direction on a smooth horizontal floor is a periodic motion but not an SHM.


Reason (R): Motion is SHM when restoring force is proportional to displacement from mean position.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Periodic motion repeats over time. SHM requires a restoring force proportional to displacement \(F = -kx\). A bouncing ball experiences gravitational force \(F=mg\) (constant) and impulsive forces upon impact, not a linear restoring force. Hence, it's periodic but not SHM, and the reason correctly defines SHM.

Question 40: easy

Assertion (A): A particle, simultaneously subjected to two simple harmonic motions of same frequency and same amplitude, will perform SHM only if the two SHM’s are in the same direction.


Reason (R): A particle, simultaneously subjected to two simple harmonic motions of same frequency and same amplitude, perpendicular to each other the particle can be in uniform circular motion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: Superposition of parallel SHMs results in SHM. Assertion (R) is true: Perpendicular SHMs of same frequency and amplitude with \(frac{pi}{2}\) phase difference result in UCM. However, (R) explains a scenario where SHM is not formed along a line, not why (A) is true.