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 $V=\frac{V_{\max }}{2} \Rightarrow \frac{A \omega}{2}$
$\frac{A^{2} \omega^{2}}{4}=\omega^{2} a^{2}-\omega^{2} x^{2}$
$x=\frac{\sqrt{3}}{2} a$
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