NEET Sample Paper NEET Sample Test Paper-32

  • question_answer The ratio of K.E. of the particle executing S.H.M.at mean position to the K.E. at point whose distance is half of amplitude is :

    A)  \[\frac{1}{3}\]

    B)  \[\frac{2}{3}\]

    C)  \[\frac{4}{3}\]

    D)  \[\frac{3}{2}\]

    Correct Answer: C

    Solution :

    \[K.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\omega }^{2}}\,\,\,({{A}^{2}}-{{x}^{2}})\] at \[x=0,\]\[K.E.=\frac{1}{2}m{{v}^{2}}_{\max }=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] at \[x=\frac{A}{2},\,K.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\omega }^{2}}\left[ {{A}^{2}}-{{\left( \frac{A}{2} \right)}^{2}} \right]\] \[=\frac{3}{4}\left[ \frac{1}{2}m{{\omega }^{2}}{{A}^{2}} \right]\] \[\Rightarrow \]     \[\frac{K{{E}_{x=0}}}{K{{E}_{x=A/2}}}=\frac{4}{3}\]

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