First slide

Simple hormonic motion

Question

The time period of a particle performing linear SHM is 12s. What is the time taken by it to make a displacement equal to half its amplitude from equilibrium position?

Moderate

A

1sec
B

2sec
C

3sec
D

4sec

Solution

$\large T\, = \,12\sec ;y\, = \,\frac{A}{2};t\, = \,?$
$\large y\, = \,A\sin \omega t$
$\large \frac{A}{2}\, = \,A\sin wt\, \Rightarrow \,\frac{1}{2}\, = \,\sin wt$
$\large \sin \left( {\frac{\pi }{6}} \right)\, = \,\sin \omega t\, \Rightarrow \,\frac{\pi }{6}\, = \,\omega t\, \Rightarrow \,\frac{\pi }{6}\, = \,\frac{{2\pi }}{T} \times t$
$\large \Rightarrow \,\frac{\pi }{6}\, = \,\frac{{2\pi }}{{12}} \times t\,\, \Rightarrow t\, = \,1\sec$

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