For a particle executing simple harmonic motion, find the distance from the mean position at which its potential and kinetic energies are equal.

Explanation

Let `x_(0)` = required distance from the mean position.
Then, `KE=E_(x)=1/2momega^(2)(a^(2)-x_(0)^(2))`
and `PE=E_(p)=1/2momega^(2)x_(0)^(2)`
From the given condition, `E_(k)=E_(p)`
`therefore" "1/2momega^(2)(a^(2)-x_(0)^(2))=1/2momega^(2)x_(0)^(2)or,a^(2)-x_(0)^(2)=x_(0)^(2)`
or, `2x_(0)^(2)=a^(2)or,x_(0)^(2)=a^(2)/2or,x_(0)=pma/sqrt2`