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  • A particle is performing angular SHM has amplitude as  .  What is its angular velocity when its angul

A particle is performing angular SHM has amplitude as \pi .  What is its angular velocity when its angular displacement is \frac{\sqrt{3}\pi }{2}  in term of T  where  T= Time period of SHM

Option: 1

\frac{T}{\pi^2}


Option: 2

\frac{T}{\pi}


Option: 3

\frac{\pi^2}{T}


Option: 4

\frac{\pi}{T}


Answers (1)

best_answer

 

 

 

 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}

So from the question

\dot{\theta }=\frac{2\pi}{T}\sqrt{\pi ^2-(\frac{\sqrt{3}\pi}{2})^2}=\frac{2\pi}{T}*\pi*\sqrt{1-\frac{3}{4}}=\frac{\pi^2}{T}

Posted by

Gaurav

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