JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Energy of Simple Harmonic Motion

  • question_answer
    For any S.H.M., amplitude is 6 cm. If instantaneous potential energy is half the total energy then distance of particle from its mean position is                    [RPET 2000]

    A)            3 cm                                         

    B)            4.2 cm

    C)            5.8 cm                                     

    D)            6 cm

    Correct Answer: B

    Solution :

                       If at any instant displacement is y then it is given that \[U=\frac{1}{2}\times E\] Þ \[\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}\times \left( \frac{1}{2}m{{\omega }^{2}}{{a}^{2}} \right)\] Þ \[y=\frac{a}{\sqrt{2}}=\frac{6}{\sqrt{2}}=4.2\,cm\]


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