Question
The maximum velocity of a particle performing linear SHM is $0.16 m / s$. If its maximum acceleration is $0.64 m / s ^2$, calculate its period.
✓
Answer
$ \text { Data : } v_{\max }=0.16 m / s , a_{\max }=0.64 m / s ^2$
$v _{\max }=\omega A \text { and } a _{\max }=\omega^2 A$
$\therefore \frac{a_{\max }}{v_{\max }}=\frac{\omega^2 A}{\omega A}=\omega$
$\therefore \omega=\frac{0.64}{0.16}=4 rad / s$
$\therefore T=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2}=\frac{3.142}{2}=1.571 s $
$v _{\max }=\omega A \text { and } a _{\max }=\omega^2 A$
$\therefore \frac{a_{\max }}{v_{\max }}=\frac{\omega^2 A}{\omega A}=\omega$
$\therefore \omega=\frac{0.64}{0.16}=4 rad / s$
$\therefore T=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2}=\frac{3.142}{2}=1.571 s $
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