Question

A particle executing s.h.m linearly withbsn amplitude a the displacement of the particle from mean position when uts potential energy is 1/3 rd maximum energy is

Answer 1

We know,
potential energy in Simple harmonic motion is U = 1/2Kx²,

Let the amplitude of particle in simple harmonic motion is A.

Let the displacement of particle from the mean position is y
A/C to question,
potential energy of particle = 1/3 of maximum energy
e.g., 1/2 Ky² = 1/6 KA² [ maximum energy when displacement = amplitude]
y² = A²/3
so, y = ± A/√3

Hence, displacement of particle from the mean position is ±A/√3