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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    A simple pendulum has time period\[{{T}_{1}}\]. The point of suspension is now moved upward according to the relation \[y=k{{t}^{2}}(k=1m{{s}^{2}})\]where\[y\]is the vertical displacement. The time period now become\[{{T}_{2}}.\] The ratio of\[\frac{T_{1}^{2}}{T_{2}^{2}}\]is\[g=10\,\,m/{{s}^{2}}\]

    A) \[6/5\] 

    B)                        \[5/6\]

    C) \[1\]                     

    D)        \[4/5\]

    Correct Answer: A

    Solution :

    \[y=k\,\,{{t}^{2}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2k\] \[a=2m/{{s}^{2}}\] \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g+{{a}_{y}}}}\] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g+{{a}_{y}}}{g}\] \[=\frac{10+2}{10}=\frac{6}{5}\]


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