Q 11-13-117JEE MainJEE Main 2021 (26 Feb, Shift 2)Easy
A particle executes S.H.M. with amplitude $A$ and time period $T$. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x}A}{2}$. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 3
$v = \omega\sqrt{A^2 - y^2}$ and $v_{max} = \omega A$.
$$\omega\sqrt{A^2 - y^2} = \frac{\omega A}{2} \Rightarrow y^2 = \frac{3A^2}{4} \Rightarrow y = \frac{\sqrt{3}A}{2}$$
So $x = 3$.
Solution by Sreeraj P, M.Sc Physics