WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    Two identical springs are connected to mass m as shown (\[k=\]spring constant). If the period of the configuration in  is 2s, the period of the configuration in  is

    A)  \[\sqrt{2}\,s\]                  

    B)  1s

    C)  \[\frac{1}{\sqrt{2}}s\]                  

    D)  \[2\sqrt{2}\,s\]

    Correct Answer: B

    Solution :

                     When the springs are joined in series the spring constant would be equal to \[{{k}_{series}}=k+k=2k\] and when the springs are joined in parallel the spring constant would be equal to \[{{k}_{p}}=\frac{k\times k}{k+k}=\frac{k}{2}.\]Also the time fc+ k   2 period \[T\propto \sqrt{\frac{1}{k}},\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}},\frac{2}{T}=\sqrt{\frac{2k}{\frac{k}{2}}}=2\]\[\Rightarrow \]            \[T=1\,s.\]          


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