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

maximum velocity of simple harmonic oscillator is Vmax=w×A where w=angular velocity, A=amplitude, given, wA=4m/s we know that, Velocity of oscillator is V=wA2-x2 here x= displacement of the oscillator ⇒V=wA2-A2/4 ⇒V=w3A2/4 ⇒V=w3A2,  substitute wA=4 v=23m/s