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. Numerical answer type. Work it out, then open the answer.

Answer: 12. ω = 2π/T = (2× 3.14)/(3.14) = 2 rad/s vmax = Aω = 0.06× 2 = 0.12 m/s vmax = 12 cm/s

A particle is doing simple harmonic motion of amplitude 0.06 m and… | JEE Main 2024 (6 Apr, Shift 1)

Q26JEE Main 2024 · 6 APR · SHIFT 1Easy

Velocity & Acceleration in SHM

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 _________ $\text{cm/s}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

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

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