The velocity of particle undergoing SHM at the mean position is 4m/s. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude

$$\begin{gathered} {\text{maximum velocity of simple harmonic oscillator is V}}_{\text{max}}\text{=w}\times \text{A} \\ where w=angular velocity, A=amplitude, given, \text{wA=4m/s} \\ we know that, Velocity of oscillator is \\ \text{V=w}\sqrt{{\text{A}}^{\text{2}}{\text{-x}}^{\text{2}}}\text{ here x= displacement of the oscillator} \\ \Rightarrow V\text{=w}\sqrt{{\text{A}}^{\text{2}}{\text{-A}}^{\text{2}}/4} \\ \Rightarrow V\text{=w}\sqrt{3{\text{A}}^2/4} \\ \Rightarrow V\text{=w}\sqrt{3}\frac{A}{2}\text{, } \\ substitute wA=4 \\ v=2\sqrt{3}m/s \end{gathered}$$