Q.

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

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a

Ux=K2a−x2

b

Ux=K1x+K2x2+K3x3

c

Ux=Ae−bx

d

Ux=a  constant

answer is A.

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Detailed Solution

PE of body in SHM at an instant PE=U=12mω2y2=12Ky2m is mass of oscillator, ω is angular velocity, y is displacement of oscillator, k is force constantIf the displacement , y=a−x then U=12Ka−x2=12Kx−a2
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