Physics - SHM Question with Solution | TestHub

Amplitude $A=4$ cm, displacement $x=3$ cm.

Magnitude of velocity is given by $v=\omega \sqrt{A^2-x^2}$

Magnitude of acceleration is given by $a={\omega }^{2}x$

Given that $|v|=|a|$ at $x=3$ cm:

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

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

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

$\sqrt{7}=3\omega$

$\omega =\frac{\sqrt{7}}{3}$

Time period $T=\frac{2\pi }{\omega }$

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

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