Velocity is distance per unit time. We can obtain the expression for velocity using the expression for acceleration.Let’s see how. Accelerationd2x/dt2= dv/dt = dv/dx× dx/dt. But dx/dt = velocity ‘v’

Therefore, acceleration = v(dv/dx) (II)

When we substitute equation II in equation I, we get,v(dv/dx) = – ω2x.

∴ vdv =– ω2xdx

After integrating both sides, we get,

∫vdv =∫-ω2xdx = -ω2∫xdx

Hence, v2/2 =-ω2x2/2 + C where C is the constant of integration. Now, to find the vaue of C, lets consider boundary value condition. When a particle performing SHM is at the extreme position, displacement of the particle is maximum and velocity is zero. (a is the amplitude of SHM)

Therefore, At x=± a, v= 0

And 0 =– ω2a2/2 + C

Hence, C =ω2a2/2

Let’s substitute this value of C in equationv2/2 =-ω2x2/2 + C

∴ v2/2 =-ω2x2/2 +ω2a2/2

∴ v2= ω2 (a2–x2)

Taking square root on both sides, we get,

v =± ω√(a2–x2) (III)

Equation III is the expression of the velocity of S.H.M. The double sign indicates that when a particle passes through a given point in the positive direction ofx, v is positive, and when it passes through the same point in opposite direction ofx, v is negative

so velocity is zero when the displacement is maximum