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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    Two springs, of force constants \[{{k}_{1}}\] and \[{{k}_{2}}\] are connected to a mass m as shown. The frequency of oscillation of the mass is f If both \[{{k}_{1}}\] and \[{{k}_{2}}\] are made four times their original values, the frequency of oscillation becomes

    A) \[2f\]

    B)        \[f/2\]

    C) \[f/4\]

    D)                    \[4f\]

    Correct Answer: A

    Solution :

    The two springs are in parallel. \[\therefore \]  Effective spring constant, \[K={{K}_{1}}+{{K}_{2}}\] Now, frequency of oscillation is given by \[f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}\] or, \[f=\frac{1}{2\pi }\sqrt{\frac{{{K}_{1}}+{{K}_{2}}}{m}}\]    ?(i) When both \[{{k}_{1}}\]and \[{{k}_{2}}\] are made four times their original values, the new frequency is given by \[f'=\frac{1}{2\pi }\sqrt{\frac{4{{K}_{1}}+4{{K}_{2}}}{m}}\] \[=\frac{1}{2\pi }\sqrt{\frac{4({{K}_{1}}+4{{K}_{2}})}{m}}=2\left( \frac{1}{2\pi }\sqrt{\frac{{{K}_{1}}+{{K}_{2}}}{m}} \right)\]  \[=2f;\]                        from eqn. (i)


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