Question

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _______ cm/s.

Answer 1

We know:

$v_{\text{max}} = \omega A \quad \text{at mean position}$

$\omega = \frac{2\pi}{T} \quad \text{and} \quad v_{\text{max}} = \frac{2\pi}{T} \times A$

$v_{\text{max}} = \frac{2\pi}{3.14} \times 0.06 = 0.12 \, \text{m/s}$

$v_{\text{max}} = 12 \, \text{cm/s}$