JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Oscillations

  • question_answer
    What will be the force constant of the spring system shown in figure?     

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

    B) \[{{\left[ \frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]}^{-1}}\]

    C) \[\left[ \frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]\]

    D) \[{{\left[ \frac{2}{{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]}^{-1}}\]

    Correct Answer: B

    Solution :

    [b] Two springs of force constants \[{{k}_{1}}\] and \[{{k}_{1}}\] are in parallel. Hence \[k'={{k}_{1}}+{{k}_{1}}=2{{k}_{1}}\] The third spring \[{{k}_{2}}\] is in series with spring of force constant\[k'\]. \[\therefore \,\,\,\frac{1}{k}=\left[ \frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]or\,\,k={{\left[ \frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]}^{-1}}\]


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