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

  • question_answer
    A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is                                [MP PMT 1996]

    A)            \[37.5{{\pi }^{2}}ergs\]    

    B)            \[3.75{{\pi }^{2}}ergs\]

    C)            \[375{{\pi }^{2}}ergs\]     

    D)            \[0.375{{\pi }^{2}}ergs\]

    Correct Answer: C

    Solution :

                       Kinetic energy \[K=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\]    \[=\frac{1}{2}\times 10\times {{\left( \frac{2\pi }{2} \right)}^{2}}[{{10}^{2}}-{{5}^{2}}]\]\[=375\ {{\pi }^{2}}ergs\]


You need to login to perform this action.
You will be redirected in 3 sec spinner