NEET Oscillation MCQs & PYQs

Practice NEET Oscillation Questions

Question 51:

moderate

A simple pendulum oscillates in a vertical plane. When it passes through the mean position, the tension in the string is \(3\) times the weight of the pendulum bob. What is the maximum angular displacement of the pendulum of the string with respect to the vertical ?

At the mean position, tension is \(T = mg + \frac{mv^2}{L}\). Given \(T = 3mg ⇒ \frac{mv^2}{L} = 2mg ⇒ v^2 = 2gL\). Using conservation of energy, \(mgL(1 - \cos\theta) = \frac{1}{2}mv^2 = mgL ⇒\cos\theta = 0 ⇒ \theta = 90^\circ\).

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Question 52:

easy

A particle executes SHM according to equation \(2\frac{d^2x}{dt^2} + 100x = 0\) (where \(x\) is in m and \(t\) is in second). Its time period of oscillation is

Divide the given equation by 2 to get \(\frac{d^2x}{dt^2} + 50x = 0\). Comparing this with the standard SHM differential equation \(\frac{d^2x}{dt^2} + \omega^2x = 0\) gives \(\omega = \sqrt{50} = 5\sqrt{2}\text{ rad/s}\). The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).

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Question 53:

easy

A body is executing simple harmonic motion with frequency \(n\), the frequency of its potential energy is

In SHM, potential energy oscillates with twice the frequency of displacement. Since the displacement frequency is \(n\), the potential energy frequency is \(2n\).

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Question 54:

easy

A spring is stretched by \(5\text{ cm}\) by a force \(10\text{ N}\). The time period of the oscillations when a mass of \(2\text{ kg}\) is suspended by it is

The spring constant is \(k = \frac{F}{x} = \frac{10}{0.05} = 200\text{ N/m}\). The time period of the spring-mass system is \(T = 2pi \sqrt{\frac{m}{k}} = 2pi \sqrt{\frac{2}{200}} = 0.2pi \approx 0.628\text{ s}\).

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Question 55:

easy

A body oscillates in SHM according to the equation \( x = 10 cos \left(2\pi t + \frac{\pi}{4} \right)\text{ cm}\). Its instantaneous displacement at \(t = 1\text{ s}\) is

Substituting \(t = 1\text{ s}\) in the given equation: \[x = 10 cos\left(2\pi(1) + \frac{\pi}{4}\right) = 10 cos\left(\frac{\pi}{4}
/right) = \frac{10}{\sqrt{2}} = 5\sqrt{2}\text{ cm}\].

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Question 56:

easy

Which of the following examples does not represent SHM?

Rotation of the Earth about its axis is a periodic motion but not oscillatory. Since it has no restoring force or back-and-forth motion about a mean position, it is not simple harmonic motion.

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Question 57:

easy

Assertion (A): In a SHM, kinetic and potential energies become equal when the displacement is \(\frac{1}{\sqrt{2}}\) times the amplitude.

Reason (R): In SHM, kinetic energy is zero when potential energy is maximum.

Equating \(KE = \frac{1}{2}k(A^2 - x^2)\) and \(PE = \frac{1}{2}kx^2\) gives \(x = \frac{A}{\sqrt{2}}\). Maximum PE occurs at extremes where KE is zero. Both statements are true, but (R) does not explain (A).

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Question 58:

easy

The distance covered by a particle undergoing SHM in one time period is (A = amplitude of oscillation)

In one complete time period, the particle travels from mean position to one extreme, back to mean, to the other extreme, and back to mean. Total distance is \(A + A + A + A = 4A\).

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Question 59:

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

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

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Question 60:

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

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

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