Question

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 ____________ .

Answer 1

Fora particle executes S.H.M
$V=\omega \sqrt{a^{2} - x^{2}}$
Given $V=\frac{V_{max}}{2}\Rightarrow \frac{a \omega }{2}$
$\frac{a^{2} \omega ^{2}}{4}=\omega ^{2}a^{2}-\omega ^{2}x^{2}$
$x=\frac{\sqrt{3}}{2}a$