Question

At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P.E.?

Answer 1

When the displacement of a particle executing S.H.M. is y, then its

K.E. = \(\frac{1}{2}\)mω²(A² − y²)

And P.E. = \(\frac{1}{2}m\omega^2y^2\)

If K.E. = P.E.

then, \(\frac{1}{2}m\omega^2(A^2\,-\,y^2)=\frac{1}{2}m\omega^2y^2\)

Or 2y² = A²

Or y = \(\frac{A}{\sqrt2}\)