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CMC Medical CMC-Medical Ludhiana Solved Paper-2010

  • question_answer
    Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants \[{{k}_{1}}\] and \[{{k}_{2}}\]respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of vibrations of M to that of N is

    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 velocities are equal Hence,         \[{{a}_{1}}{{\omega }_{1}}={{a}_{2}}{{\omega }_{2}}\] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}\] \[=\frac{2\pi \text{/}{{n}_{2}}}{2\pi \text{/}{{n}_{1}}}\] \[=\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{2\pi \sqrt{{{m}_{1}}\text{/}{{k}_{1}}}}{2\pi \sqrt{m\text{/}{{k}_{2}}}}\] \[=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}}\]


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