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  4. A particle having potential energy 1/3 of the maximum value of a distant of 4 cm from a mean position. The amplitude of the motion is:

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Q. A particle having potential energy 1/3 of the maximum value of a distant of 4 cm from a mean position. The amplitude of the motion is:

JIPMERJIPMER 2002

Solution:

The maximum energy during S.H.M. $ {{E}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}} $ ?(i) where A = amplitude of the S.H.M. now at x = 4 $ P.E.=\frac{{{E}_{\max }}}{3}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}} $ ?(ii) Then from equation (i) and (ii) we get $ \frac{1}{6}m{{\omega }^{2}}{{A}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}} $ $ {{x}^{2}}=\frac{{{A}^{2}}}{3}, $ so $ A=\sqrt{3}x=4\sqrt{3} $