MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    A mass m is suspended from the two springs of spring constants\[{{k}_{1}}\]and\[{{k}_{2}}\]as shown. The time period of vertical oscillations of the mass will be

    A)  \[2\pi \sqrt{\left( \frac{{{k}_{1}}+{{k}_{2}}}{m} \right)}\]             

    B)  \[2\pi \sqrt{\left( \frac{m}{{{k}_{1}}+{{k}_{2}}} \right)}\]

    C)  \[2\pi \sqrt{\frac{m({{k}_{1}}{{k}_{2}})}{{{k}_{1}}+{{k}_{2}}}}\] 

    D)  \[2\pi \sqrt{\frac{m({{k}_{1}}+{{k}_{2}})}{({{k}_{1}}{{k}_{2}})}}\]

    Correct Answer: D

    Solution :

                    \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}\Rightarrow k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]\[t=2\pi \frac{\sqrt{m({{k}_{1}}+{{k}_{2}})}}{{{k}_{1}}+{{k}_{2}}}\]


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