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  4. A particle executes SHM of period 12 s. After two seconds, it passes through the centre of oscillation, the velocity is found to be 3.142 cm s-1. The amplitude of oscillations is

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Q. A particle executes $SHM$ of period $12\, s$. After two seconds, it passes through the centre of oscillation, the velocity is found to be $3.142\, cm\, s^{-1}$. The amplitude of oscillations is

Oscillations

Solution:

Let $y = A \,sin \omega t$
Then $v= \frac{dy}{dt} = \omega A \,cos\omega t$
$ = \frac{2\pi}{T} = A \,cos \frac{2\pi}{T}t \quad\left(\because\omega = \frac{2\pi}{T}\right) $
$\therefore 3.142 =\frac{ 2\times 3.142}{12}\times A\, cos \frac{2\pi}{12}\times 2$
$ \therefore A= 12 \,cm$