Q 11-13-057JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
A particle executes simple harmonic motion with an amplitude of $4\ \text{cm}$. At the mean position, velocity of the particle is $10\ \text{cm s}^{-1}$. The distance of the particle from the mean position when its speed becomes $5\ \text{cm s}^{-1}$ is $\sqrt\alpha\ \text{cm}$, where $\alpha =$ ______.
Numerical value type. Enter your answer.
Answer: 12
$v_{max} = A\omega \Rightarrow \omega = \dfrac{10}{4} = 2.5\ \text{rad s}^{-1}$
Using $v = \omega\sqrt{A^2 - x^2}$:
$$5 = 2.5\sqrt{16 - x^2} \;\Rightarrow\; 16 - x^2 = 4 \;\Rightarrow\; x = \sqrt{12}\ \text{cm}$$
So $\alpha = 12$.
Solution by Sreeraj P, M.Sc Physics