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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Critical Thinking

  • question_answer
    A simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in horizontal position from the wall and is left free to move. The bob hits on the wall the coefficient of restitution is \[\frac{2}{\sqrt{5}}\]. After how many collisions the amplitude of vibration will become less than 60° [UPSEAT 1999]

    A)            6    

    B)            3

    C)            5    

    D)            4

    Correct Answer: B

    Solution :

                       From the relation of restitution \[\frac{{{h}_{n}}}{{{h}_{0}}}={{e}^{2n}}\] and \[{{h}_{n}}={{h}_{0}}(1-\cos 60{}^\circ )\] Þ \[\frac{{{h}_{n}}}{{{h}_{0}}}=1-\cos 60{}^\circ ={{\left( \frac{2}{\sqrt{5}} \right)}^{2n}}\] Þ \[1-\frac{1}{2}={{\left( \frac{4}{5} \right)}^{n}}\]Þ \[\frac{1}{2}={{\left( \frac{4}{5} \right)}^{n}}\] Taking log of both sides we get \[\log 1-\log 2=n(\log 4-\log 5)\] \[0-0.3010=n(0.6020-0.6990)\] \[-\,0.3010=-n\times 0.097\] Þ \[n=\frac{0.3010}{0.097}=3.1\approx 3\]


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