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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant \[{{k}_{1}}\]and \[{{k}_{2}}\] respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B 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 :

    Since maximum velocity is same, so maximum KE ie, TE would be same. Also maximum PE would also be same. Let \[{{A}_{1}}\] and  \[{{A}_{2}}\] be the amplitudes of two bodies A and B respectively, \[\frac{{{k}_{1}}A_{1}^{2}}{2}=\frac{{{k}_{2}}A_{2}^{2}}{2}\]                 \[\frac{A_{1}^{{}}}{{{A}_{2}}}=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}}\]


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