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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • question_answer
    Two springs with spring constants \[{{K}_{1}}=1500\,N/m\] and \[{{K}_{2}}=3000\,N/m\] are stretched by the same force. The ratio of potential energy stored in spring will be           [RPET 2001]

    A)            2 : 1                                          

    B)            1 : 2

    C)            4 : 1                                          

    D)            1 : 4

    Correct Answer: A

    Solution :

               \[U=\frac{{{F}^{2}}}{2K}\Rightarrow U\propto \frac{1}{K}\Rightarrow \frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{K}_{2}}}{{{K}_{1}}}=2\]


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