Simple harmonic motion

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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 ______

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Solution

$\begin{array}{l} V = ω \sqrt{A^{2} - x^{2}} \\ put v = \frac{1}{2} v_{max} = \frac{ω A}{2} \\ \Rightarrow \frac{A ω}{2} = ω \sqrt{A^{2} - x^{2}} \\ \Rightarrow \frac{A^{2}}{4} = A^{2} - x^{2} \\ \Rightarrow x^{2} = \frac{3 A^{2}}{4} \\ \Rightarrow x = \frac{\sqrt{3} A}{2} \end{array}$

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