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

given velocity at mean position,which is maximum velocity as ${\text{V}}_{\text{max}}\text{=4m/s}$

$$\begin{gathered} \text{Velocity of simple harmonic oscillator is V=ω}\sqrt{{\text{A}}^{\text{2}}{\text{-x}}^{\text{2}}} \\ \Rightarrow at x=0 , mean position, {\text{V}}_{\text{max}}\text{=ωA} \\ given \text{ωA=4m/s} \\ \text{V=ω}\sqrt{{\text{A}}^{\text{2}}{\text{-x}}^{\text{2}}} \\ at x=\frac{A}{2}, velocity\text{=ω}\sqrt{{\text{A}}^{\text{2}}{\text{-A}}^{\text{2}}/4}\text{=ω}\sqrt{3{\text{A}}^2/4} \\ \text{V=ω}\frac{\text{A}\sqrt{\text{3}}}{\text{2}}\text{, substitute ωA=4,} \\ V=\frac{4\sqrt{\text{3}}}{2}=2\sqrt{3}\text{m/s} \end{gathered}$$