Question:
A particle executes S.H.M. with amplitude 'a' and time period $V$. 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:
$V=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$
$\mathrm{V}_{\max }=\mathrm{A} \omega$
$\frac{\mathrm{A} \omega}{2}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$
$\frac{\mathrm{A}^{2}}{4}=\mathrm{A}^{2}-\mathrm{x}^{2}$
$x^{2}=\frac{3 A^{2}}{4}$
$x=\frac{\sqrt{3}}{2} \mathrm{~A}$
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