JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • question_answer
    What will be the force constant of the spring system shown in the figure                              [RPET 1996; Kerala (Med./ Engg.) 2005]

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

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

    C)            \[\frac{1}{2{{K}_{1}}}+\frac{1}{{{K}_{2}}}\]

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

    Correct Answer: B

    Solution :

                       In series combination                       \[\frac{1}{{{k}_{S}}}=\frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}}\] Þ \[{{k}_{S}}={{\left[ \frac{1}{2{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right]}^{-1}}\]


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