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

  • question_answer
    In arrangement given in figure, if the block of mass m is displaced, the frequency is given by                                                                          [BHU 1994; Pb. PET 2001]       

    A)                                                      \[n=\frac{1}{2\pi }\sqrt{\left( \frac{{{k}_{1}}-{{k}_{2}}}{m} \right)}\]            

    B)            \[n=\frac{1}{2\pi }\sqrt{\left( \frac{{{k}_{1}}+{{k}_{2}}}{m} \right)}\]

    C)            \[n=\frac{1}{2\pi }\sqrt{\left( \frac{m}{{{k}_{1}}+{{k}_{2}}} \right)}\]           

    D)            \[n=\frac{1}{2\pi }\sqrt{\left( \frac{m}{{{k}_{1}}-{{k}_{2}}} \right)}\]

    Correct Answer: B

    Solution :

                       With respect to the block the springs are connected in parallel combination. \[\therefore \] Combined stiffness k = k1+ k2  and \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\]


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