JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    A particle is executing SHM. At a point x = A/3, kinetic energy of the particle is K, where A is    the    amplitude.    At    a    point x = 2A/3, kinetic energy of the particle will be :

    A) \[2K\]

    B) \[\sqrt{2}K\]

    C) \[\frac{5}{8}K\]

    D) \[\frac{5}{3}K\]

    Correct Answer: C

    Solution :

    \[V=\omega \sqrt{({{a}^{2}}-{{X}^{2}})}\]  \[V\,\propto \,\sqrt{({{A}^{2}}-{{x}^{2}})}\] At \[x=A/3\] \[\sqrt{({{A}^{2}}-{{x}^{2}})}=\sqrt{\left( {{A}^{2}}-\frac{{{A}^{2}}}{9} \right)}=A\sqrt{\frac{8}{9}}\] At \[x=2A/3\] \[\sqrt{({{A}^{2}}-{{x}^{2}})}=\sqrt{\left( {{A}^{2}}-\frac{4{{A}^{2}}}{9} \right)}=A\sqrt{\frac{5}{9}}\] \[K=1/2m\,{{v}^{2}}\] KE at\[x=2A/3=\frac{5}{9}\times \frac{9}{8}=\frac{5}{8}\,K.\]


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