If an object undergoes simple harmonic motion with an amplit

If an object undergoes simple harmonic motion with an amplitude of A and a maximum speed of v_max, what is its speed when its position is at one quarter of the amplitude? (1/2) V_max Squareroot [31/32] v_max Squareroot [3/4] V_max Squareroot [15/16] V_max (1/4) v_max

Solution

doing energy conservation

KE1 + PE1 = KE2 + PE2

=> 1/2*mVmax^2 + 0 =   1/2mv^2 + 1/2*K(A/4)^2 ( ( at time of max velocity X=0 => PE = 0)

=> 1/2*mVmax^2 = 1/2*K(A)^2 ( max KE = max PE)

=> v = sqrt(15/16)*Vmax

=> option D is correct