First slide
Simple harmonic motion
Question

The potential energy of a particle Ux executing SHM is given by

Moderate
Solution

PE of body in SHM at an instant
 PE=U=12mω2y2=12Ky2

m is mass of oscillator, ω is angular velocity, y is displacement of oscillator, k is force constant
If the displacement , y=ax then
 U=12Kax2=12Kxa2

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