2. Solved Papers
  3. Haryana PMT

Haryana PMT Haryana PMT Solved Paper-2002

  • 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 amplitude 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: B

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner