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

  • question_answer
    A mass m is suspended from the two coupled springs connected in series. The force constant for springs are \[{{K}_{1}}\] and \[{{K}_{2}}\]. The time period of the suspended mass will be  [CBSE PMT 1990; Pb. PET 2002]

    A)            \[T=2\pi \sqrt{\left( \frac{m}{{{K}_{1}}+{{K}_{2}}} \right)}\]             

    B)            \[T=2\pi \sqrt{\left( \frac{m}{{{K}_{1}}+{{K}_{2}}} \right)}\]

    C)            \[T=2\pi \sqrt{\left( \frac{m({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}} \right)}\]      

    D)            \[T=2\pi \sqrt{\left( \frac{m{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)}\]

    Correct Answer: C

    Solution :

               In series \[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]so time period \[T=2\pi \sqrt{\frac{m({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}}\] 


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