WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    A particle of mass m is attached to three identical massless springs of spring constant k  as shown in the figure. The time period of vertical oscillation of the particle is

    A)  \[2\pi \sqrt{\frac{m}{k}}\]                          

    B)  \[2\pi \sqrt{\frac{m}{2k}}\]

    C)  \[2\pi \sqrt{\frac{m}{3k}}\]                       

    D)  \[\pi \sqrt{\frac{m}{k}}\]

    Correct Answer: B

    Solution :

                     When the particle of mass m at O is pushed by y in the direction of A, the spring A will be compressed by y while spring B and C will be stretched by \[y=y\cos {{45}^{o}}.\]So, that the total restoring force on the mass m is along OA \[{{F}_{net}}={{F}_{A}}+{{F}_{B}}\cos {{45}^{o}}+{{F}_{C}}\cos {{45}^{o}}\] \[=ky+2ky\cos {{45}^{o}}\] \[=ky+2k(y\,cos{{45}^{o}})cos{{45}^{o}}\]\[=2ky\]                 Also, \[{{F}_{net}}=ky\Rightarrow \,ky=2ky\Rightarrow k2k\] \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{2k}}\]


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