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

Correct Answer: 12

We know:

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

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

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

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