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

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by swetha89

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by swetha89

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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.**

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,

wt+Ф = π/4