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

  • question_answer
    A simple pendulum has time period T1. The point of suspension is now moved upward according to equation \[y=k{{t}^{2}}\] where\[k=1\,m/se{{c}^{2}}\]. If new time period is T2 then ratio \[\frac{T_{1}^{2}}{T_{2}^{2}}\] will be                                                                [IIT-JEE (Screening) 2005]

    A)            2/3 

    B)            5/6

    C)            6/5 

    D)            3/2

    Correct Answer: C

    Solution :

                       \[y=K{{t}^{2}}\]Þ\[\frac{{{d}^{2}}y}{d{{t}^{2}}}={{a}_{y}}=2K\]= 2 ´ 1=2m/s2 (\[\because \]K= 1m/s2) Now, \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] and \[{{T}_{2}}=2\pi \sqrt{\frac{l}{(g+{{a}_{y}})}}\] Dividing,  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{g+{{a}_{y}}}{g}}\] Þ \[\sqrt{\frac{6}{5}}\]Þ \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{6}{5}\]


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