Equation of SHM - NEET Physics Questions
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Equation of SHM

Practice Questions:
Question 11: moderate

A periodic time of a body executing simple harmonic motion is 3s. After how much interval from time t = 0, its displacement from mean position will be half of its amplitude ?

1. 1/8 s
2. 1/6 s
3. 1/4 s
4. 1/3 s
View Answer

\[ x= A sin\left( \omega t \right) \]

\[ \frac{A}{2}= A sin\left( \omega t \right) \]

\[ \frac{1}{2}= sin\left( \omega t \right) \]

\[ \frac{\Pi}{6}= \omega t =\frac{2\Pi}{T}t = \frac{2\Pi}{3}t \]

\[ t= \frac{1}{4}s \]

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Question 12: moderate

Displacement-time graph of a particle executing SHM is as shown below :-

The corresponding force-time graph of the particle can be :

1.
2.
3.
4.
View Answer

Equation of displacement x = A sin (ωt) so, acceleration a= -Aω sin(ωt) force will have same nature so,

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Question 13: 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 ?

1. \(30^\circ\)
2. \(45^\circ\)
3. \(60^\circ\)
4. \(90^\circ\)
View Answer

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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