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Angular Simple Harmonic Motion MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 3 Questions around this concept.

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 Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is :

Concepts Covered - 1

Angular SHM

The general equation of linear SHM is given by x=Asin(\omega t+\alpha)

Similarly, The general equation of angular SHM is given by \theta =\theta _0sin(\omega t+ \phi )

where \theta \ and \ \theta_0 are angular displacement and angular amplitude of the bob respectively, as shown in the below figure

If l=length of the bob then we can write \theta =\frac{x}{l} \ \ and \ \theta_0 =\frac{A}{l}.

Similarly, The angular velocity if the bob which is in angular SHM is given by 

\\\dot{\theta }=\frac{d\theta }{dt}=\theta_0 \omega Cos(\omega t+\phi )\\ \\ or \ \ \dot{\theta }= \omega \sqrt{ {\theta_0 }^2-\theta ^2}

Similarly, The angular acceleration if the bob which is in angular SHM is given by 

\\ \alpha =\frac{d^2\theta }{dt}=-\theta_0 \omega^2Sin(\omega t+\phi )\\ \\ or \ \ \alpha =- \omega^2 \theta

And Thus restoring torque on the  body is given as 

\tau_{R}=-I \alpha=-I \omega^{2} \theta

Thus we can state that in angular SHM, the angular acceleration of the body and the restoring torque on the body are directly
proportional to the angular displacement of body from its mean position and are directed toward the mean position.

Similarly, a basic differential equation for angular SHM can be written as

\frac{d^{2} \theta}{d t^{2}}+\omega^{2} \theta=0

 

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Angular SHM

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