Physics - SHM Question with Solution | TestHub

Given amplitude $A=4$ cm and displacement $x=3$ cm.

The magnitude of velocity of a particle in S.H.M. is given by $v=\omega \sqrt{A^2-x^2}$.

The magnitude of acceleration is given by $a={\omega }^{2}x$.

According to the given condition, $v=a$ at $x=3$ cm.

$\omega \sqrt{A^2-x^2}={\omega }^{2}x$

$\sqrt{4^2-3^2}=\omega (3)$

$\sqrt{16-9}=3\omega$

$\sqrt{7}=3\omega$

$\omega =\frac{\sqrt{7}}{3}$ rad/s

The time period $T$ is given by $T=\frac{2\pi }{\omega }$.

$T=\frac{2\pi }{\frac{\sqrt{7}}{3}}=\frac{6\pi }{\sqrt{7}}$ s

Answer: $\frac{6\pi }{\sqrt{7}}$ s