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

Correct Answer: 12

$$v_{max} = A\omega = A \times \frac{2\pi}{T} = 0.06 \times \frac{2\pi}{3.14} = 0.06 \times 2 = 0.12$$ m/s = 12 cm/s.

The answer is 12.

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