Q 11-13-067JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
A particle of mass $250\ \text{g}$ executes a simple harmonic motion under a periodic force $F=(-25x)\ \text{N}$. The particle attains a maximum speed of $4\ \text{m s}^{-1}$ during its oscillation. The amplitude of the motion is ______ cm.
Numerical value type. Enter your answer.
Answer: 40
$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{25}{0.25}}=10\ \text{rad s}^{-1}$.
$v_{max}=A\omega\Rightarrow A=\dfrac{4}{10}=0.4\ \text{m}=40\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics