# Find the distance from the centre at which the velocity in s.h.m will be half of the maximum.

$\begin{array}{l}\text{We know that,}\\ \text{speed in S.H.M,}\\ v=\omega\sqrt{a^2-x^2}\\ v_{\max}=a\omega\\ \text{Given in question,}\\ v=\frac{v_{\max}}{2}=\omega\sqrt{a^2-x^2}\\ \Rightarrow\frac{a\omega}{2}=\omega\sqrt{a^2-x^2}\\ \Rightarrow x^2=a^2-\frac{a^2}{4}\\ \Rightarrow x=\frac{\sqrt{3}}{2}a\\ \text{At,}\ x=\frac{\sqrt{3}}{2}a,\ \text{speed is half of its maximum. }\end{array}$

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