Haryana PMT Haryana PMT Solved Paper-2004

  • question_answer When the kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. The displacement of the body is \[x\] percent of the amplitude, where x is :

    A)  33                                         

    B)  87

    C)  67                                         

    D)  50

    Correct Answer: B

    Solution :

                                             The relation for kinetic energy of S.H.M. is given by                                          \[=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\]          ??.(1) Potential energy is given by                                          \[=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\]              .....(2) Now for the condition of question and from eqs. (1) and (2) \[\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=\frac{1}{3}\times \frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] or                                    \[\frac{4}{6}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] or                                    \[{{y}^{2}}=\frac{3}{4}{{a}^{2}}\] So,                                  \[y=\frac{a}{2}\sqrt{3}=0.866a\]                                          \[\approx 87%\] of amplitude


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