Question

at what distance from the mean position is the kinetic energy in simple harmonic oscillator equal to potential energy ?

Answer 1

A spring with spring constant K is oscillating on a frictionless horizontal surface with a mass m attached to its end. The distance x from its mean position may be specified as:

   x = A cos (wt+Ф),  A is the amplitude and Ф is the phase constant, t is time.

Force on mass m = - k x = ma                  But for SHM - w² x = a

So w = √(k/m)  or m = k/w²

v = dx/dt = - Aw sin (wt+Ф)

Potential Energy = 1/2 k x² = 1/2 k A² cos² (wt+Ф)
Kinetic Energy = 1/2 m v² = 1/2 m A² w² sin² (wt+Ф)  = 1/2 k A² sin² (wt+Ф)

So if PE = KE, then  sin² (wt+Ф) = cos² (wt+Ф)

But sin² (wt+Ф) + cos² (wt+Ф)  = 1

So, sin² (wt+Ф) = 1/2,  sin (wt+Ф) = 1/√2
         wt+Ф = π/4

NOW,  x = A /√2, So, when the body is at 1/√2 th of its amplitude then the PE = KE.

Answer 2

A spring with spring constant K is oscillating on a frictionless horizontal surface with a mass m attached to its end. The distance x from its mean position may be specified as:

  x = A cos (wt+Ф),  A is the amplitude and Ф is the phase constant, t is time.

Force on mass m = - k x = ma                  But for SHM - w² x = a

So w = √(k/m)  or m = k/w²

v = dx/dt = - Aw sin (wt+Ф)

Potential Energy = 1/2 k x² = 1/2 k A² cos² (wt+Ф)

Kinetic Energy = 1/2 m v² = 1/2 m A² w² sin² (wt+Ф)  = 1/2 k A² sin² (wt+Ф)

So if PE = KE, then  sin² (wt+Ф) = cos² (wt+Ф)

But sin² (wt+Ф) + cos² (wt+Ф)  = 1

So, sin² (wt+Ф) = 1/2,  sin (wt+Ф) = 1/√2

        wt+Ф = π/4

NOW,  x = A /√2, So, when the body is at 1/√2 th of its amplitude then the PE = KE.