Questions
The potential energy of a particle executing SHM is given by
a
Ux=K2a−x2
b
Ux=K1x+K2x2+K3x3
c
Ux=Ae−bx
d
Ux=a constant
detailed solution
Correct option is A
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−a2Talk to our academic expert!
Similar Questions
When the displacement is half the amplitude, the ratio of potential energy to the total energy is
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