Q 11-13-056JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
A particle is doing simple harmonic motion with an amplitude of $0.06\ \text{m}$ and a time period of $3.14\ \text{s}$. The maximum velocity of the particle is ______ cm/s.
Numerical value type. Enter your answer.
Answer: 12
$\omega = \dfrac{2\pi}{3.14} = 2\ \text{rad/s}$, so $v_{max} = A\omega = 0.06\times2 = 0.12\ \text{m/s} = 12\ \text{cm/s}$.
Solution by Sreeraj P, M.Sc Physics