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Oscillations question

2021 · Q129
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Oscillations question

2021 · Q129

NEETPhysicsOscillationsMCQ+4 / −1
A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is :
  1. A
    4n
  2. B
    n
  3. C
    2n
  4. D
    3n
View written solutionFree

Correct answer: C

Displacement equation of SHM of frequency 'n'

x = A sin (ω\omegaωt) = A sin (2π\piπnt)

Now,

Potential energy

U=12kx2=12KA2sin⁡2(2πnt)U = {1 \over 2}k{x^2} = {1 \over 2}K{A^2}{\sin ^2}(2\pi nt)U=21​kx2=21​KA2sin2(2πnt)

=12kA2[1−cos⁡(2π(2n)t)2] = {1 \over 2}k{A^2}\left[ {{{1 - \cos (2\pi (2n)t)} \over 2}} \right]=21​kA2[21−cos(2π(2n)t)​]

So frequency of potential energy = 2n

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