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

  • question_answer
    Two bodies M and N of equal masses are suspended from two separate massless springs of force constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude M to that of N is [IIT-JEE 1988; MP PET 1997, 2001; MP PMT 1997; BHU 1998; Pb. PMT 1998; MH CET 2000, 03; AIEEE 2003]

    A)            \[\frac{{{k}_{1}}}{{{k}_{2}}}\]

    B)            \[\sqrt{\frac{{{k}_{1}}}{{{k}_{2}}}}\]

    C)            \[\frac{{{k}_{2}}}{{{k}_{1}}}\]

    D)            \[\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}}\]

    Correct Answer: D

    Solution :

                       Maximum velocity \[=a\omega =a\sqrt{\frac{k}{m}}\] Given that \[{{a}_{1}}\sqrt{\frac{{{K}_{1}}}{m}}={{a}_{2}}\sqrt{\frac{{{K}_{2}}}{m}}\]Þ \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{{{K}_{2}}}{{{K}_{1}}}}\]


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