Physics - SHM Question with Solution | TestHub

Amplitude $A=3$ cm.

Magnitude of velocity at distance $x$ from the mean position is $v=\omega \sqrt{A^2-x^2}$.

Magnitude of acceleration at distance $x$ is $a={\omega }^{2}x$.

Given that at $x=1$ cm, $v=\frac{1}{2}a$.

$\omega \sqrt{3^2-1^2}=\frac{1}{2}{\omega }^2(1)$

$\omega \sqrt{8}=\frac{1}{2}{\omega }^2$

$\omega =2\sqrt{8}=4\sqrt{2}$ rad/s.

Maximum velocity is $v_{\max }=A\omega$.

$v_{\max }=3\times 4\sqrt{2}=12\sqrt{2}$ cm/s.

Answer: $12\sqrt{2}$ cm/s