For a particle executing SHM with an amplitude ‘r’, the kinetic energy will be equal to the potential energy when the displacement is equal to-

x = A sin w t           displacement in SHM

v = A w cos w t      velocity in SHM

PE = 1/2 k x^2 = 1/2 k A^2 sin^2 w t

KE = 1/2 m v^2 = 1/2 m w^2 A^2 cos^2 w t

If KE = PE then

k sin^2 w t = m w^2 cos^2 w t

sin^2 wt / cos^2 w t = tan^2 w t = m w^2 / k

but k / m = w^2

So tan^2 w t = 1  and tan w t = 1   or w t = pi / 4 or theta = 45 deg

Then  x = r sin w t  = r sin 45 = .707 r

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.