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NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    Two simple pendulums of length 0.5 m and 2.0 m respectively are given small linear displacement in one direction at die same time. They will again be in the same phase when the pendulum of shorter length has completed oscillations:                                                                                                                                           [AIPMT 1998]

    A)        5                    

    B)       1                    

    C)  2        

    D)                   3

    Correct Answer: C

    Solution :

    Key Idea: For the pendulum to be again in the same phase, there should be difference of 1 complete oscillation.
                If smaller pendulum completes n oscillations the larger pendulum will complete \[(n-1)\] oscillations, so
                Time period of n oscillations of first = Time period of \[(n-1)\] oscillations of second
                i.e.,       \[n{{T}_{1}}=(n-1){{T}_{2}}\]
    or         \[n\,2\,\pi \,\sqrt{\frac{{{L}_{1}}}{g}}=(n-1)\,2\,\pi \sqrt{\frac{{{L}_{2}}}{g}}\]
    or         \[n\sqrt{{{L}_{1}}}=(n-1)\,\sqrt{{{L}_{2}}}\]
    or         \[\frac{n}{n-1}=\sqrt{\frac{{{L}_{2}}}{{{L}_{1}}}}=\sqrt{\frac{2.0}{0.5}}\]
    or         \[\frac{n}{n-1}=2\]
    or         \[n=2n-2\]
    \[\therefore \]      \[n=2\]


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