VMMC VMMC Medical Solved Paper-2004

  • question_answer
    Two springs of spring constants \[{{k}_{1}}\] and \[{{k}_{2}}\] have equal maximum velocities. When executing simple harmonic motion, the ratio of their amplitudes (masses are equal) will be

    A) \[{{\left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)}^{{1}/{2}\;}}\]              

    B)        \[{{\left( \frac{{{k}_{1}}}{{{k}_{2}}} \right)}^{{1}/{2}\;}}\]              

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

    D)        \[{{k}_{1}}{{k}_{2}}\]

    Correct Answer: A

    Solution :

    The angular frequency of spring is given by \[\omega =\frac{\sqrt{k}}{m}\propto \sqrt{k}\] (because \[m={{m}_{1}}={{m}_{2}}\]) For equal maximum velocities, we have \[{{a}_{1}}{{\omega }_{1}}={{a}_{2}}{{\omega }_{2}}\] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}=\frac{\sqrt{{{k}_{2}}}}{{{k}_{1}}}={{\left( \frac{{{k}_{2}}}{{{k}_{1}}} \right)}^{\frac{1}{2}}}\]


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