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AMU Medical AMU Solved Paper-2009

  • question_answer
    Two bodies A and B of equal masses are suspended from two separate springs of force constants fe, and k respectively. If the two bodies oscillate such that their maximum velocities are equal, the ratio of the amplitudes of oscillation of A and B will be

    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 of SHM (or spring-mass motion) can be given by                 \[{{v}_{\max }}=A\,\omega \] As           \[\omega =\sqrt{\frac{k}{m}}\]                 \[{{v}_{\max }}=A\sqrt{\frac{k}{m}}\] Given, \[{{v}_{1\,(\max )}}={{v}_{2(\max )}}\] and \[{{m}_{1}}={{m}_{2}}\] \[\therefore \]  \[{{A}_{1}}\sqrt{\frac{{{k}_{1}}}{m}}={{A}_{2}}\sqrt{\frac{{{k}_{2}}}{m}}\] \[\Rightarrow \]               \[\frac{{{A}_{1}}}{{{A}_{2}}}=\sqrt{\frac{{{k}_{1}}}{{{k}_{2}}}}\]


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