RAJASTHAN PMT Rajasthan - PMT Solved Paper-1996

  • question_answer
    A bob of mass M is connected with two springs as shown in figure the time period of horizontal oscillation of M will be (no damping force is present):

    A) \[2\pi \sqrt{\frac{3k}{M}}\]                        

    B)        \[2\pi \sqrt{\frac{k}{M}}\]                          

    C)        \[2\pi \sqrt{\frac{M}{2k}}\]                        

    D)        \[2\pi \sqrt{\frac{M}{3k}}\]

    Correct Answer: D

    Solution :

    Time period \[T=2\pi \sqrt{\frac{m}{k+2k}}\] \[T=2\pi \sqrt{\frac{m}{3k}}\]


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