WB JEE Medical WB JEE Medical Solved Paper-2008

  • question_answer
    Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are \[{{k}_{1}}\]and \[{{k}_{2}}\] respectively. The two springs are \[{{k}_{1}}\]and \[{{k}_{2}}\]respectively. The period of vertical oscillations of the system will  be

    A) \[\frac{1}{8\pi }\sqrt{{{k}_{1}}+{{k}_{2}}}\]                         

    B)  \[8\pi \sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}}}\]

    C)  \[\frac{\pi }{2}\sqrt{{{k}_{1}}-{{k}_{2}}}\]                            

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

    Correct Answer: B

    Solution :

                     The two springs are in series. Therefore, the  time period is \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{m\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}} \right)}\] As \[m=16\,kg;\] \[T=8\pi \sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}}}\]


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