Question

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

Answer 1

Let x be the distance where KE is equal to PE.

Kinetic energy of particles executing SHM is KE = \(\frac 12\)mω²(a² − x²)

Potential energy of  particles is given by PE = \(\frac 12\)mω²x²

Let x be the distance where

KE = PE

\(\frac 12\)​mω²(a² − x²) = \(\frac 12\)mω²x²

(a² − x²) = x²

x = \(\frac a{\sqrt 2}\)

Kinetic energy equal to its potential energy at x = \(\frac a{\sqrt 2}\).

Answer 2

Let the displacement of particle executing S.H.M = Y 

Amplitude of particle executing S.H.M = a Mass of particle = m 

Angular velocity = w