Q.The maximum speed of a particle performing linear S.H.M. is 0.08 m/s. If its maximum acceleration is 0.32 m/s², calculate its

Speed in SHM at a given displacement, using the second's-pendulum period.

A second's pendulum has $T=2$ s, so $\omega = \dfrac{2\pi}{T} = \pi$ rad/s. Speed at displacement $x$ from mean position: …

Differentiate the displacement equation to get velocity.

Given $x = 5\sin\left(\dfrac{\pi t}{3}\right)$ m, the velocity is

$v = \frac{dx}{dt} = 5\cdot\frac{\pi}{3}\cos\left(\frac{\pi t}{3}\right) = \frac{5\pi}{3}\cos\left(\frac{\pi t}{3}\right)\ \text{m/s}$

At $t = 1$ s: …