A particle  execute SHM from the mean position. Its amplitude is A, its time period is ‘T’. At what displacement, its speed is half of its maximum speed.

$$\begin{gathered} \text{given oscillator speed is V}=\frac{1}{2}{\text{V}}_{\text{max}} \\ \text{=}\frac{\text{1}}{\text{2}}{\text{Aω.........(1) where maximum velocity V}}_{\text{max}}\text{=}A\omega \\ oscillator velocity is \text{V}=\text{ω}\sqrt{{\text{A}}^{\text{2}}{\text{-Y}}^{\text{2}}}\text{---(2)} \\ substitute \left(1\right) in \left(2\right) \\ \frac{\text{1}}{\text{2}}\text{Aω=ω}\sqrt{{\text{A}}^{\text{2}}{\text{-Y}}^{\text{2}}} \\ \Rightarrow \frac{\text{1}}{\text{2}}\text{A=}\sqrt{{\text{A}}^{\text{2}}{\text{-Y}}^{\text{2}}} \\ \Rightarrow \frac{\text{1}}{4}{\text{A}}^{\text{2}}{\text{=A}}^{\text{2}}{\text{-Y}}^{\text{2}} \\ \Rightarrow {\text{Y}}^{\text{2}}={\text{A}}^{\text{2}}-\frac{{\text{A}}^{\text{2}}}{4} \\ \Rightarrow {\text{Y}}^{\text{2}}=\frac{{\text{3A}}^{\text{2}}}{4} \\ \Rightarrow Y=\frac{\sqrt{3}\text{A}}{2} \end{gathered}$$