JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    A body of mass \[m\] is attached between two springs of force constants \[{{k}_{1}}\] and \[{{k}_{2}}\] as Shown in figure. The other ends of the springs are fixed to firm supports. The frequency of the oscillation is:

    A) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}-{{k}_{2}}}{m}}\]

    B)  \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{m}}\]

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

    D)   none of the above

    Correct Answer: C

    Solution :

    The two springs in the arrangement form parallel combination. So, spring constant in parallel combination is given by                 \[k={{k}_{1}}+{{k}_{2}}\] where \[{{k}_{1}}\] and \[{{k}_{2}}\] are force constants of springs. Frequency of oscillation of the system is given by                 \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] or            \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\]


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