# A particle excutes S.H.M with amplitude 'a' and time period T.

Question:

A particle excutes 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

Solution:

(3)

Fora particle excutes S.H.M

$V=\omega \sqrt{a^{2}-x^{2}}$

Given $V=\frac{V_{\max }}{2} \Rightarrow \frac{A \omega}{2}$

$\frac{A^{2} \omega^{2}}{4}=\omega^{2} a^{2}-\omega^{2} x^{2}$

$x=\frac{\sqrt{3}}{2} a$