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

  • question_answer
    A mass m =100 gms is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to 0.16 metre and time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement \[x=-0.16\] metre. The expression for the displacement of the mass at any time t is             [MP PMT 1995]

    A)            \[x=0.16\cos (\pi t)\]        

    B)            \[x=-\,0.16\cos (\pi t)\]

    C)            \[x=0.16\sin (\pi t+\pi )\]

    D)            \[x=-\,0.16\sin (\pi t+\pi )\]

    Correct Answer: B

    Solution :

                       Standard equation for given condition \[x=a\cos \frac{2\pi }{T}t\]Þ \[x=-0.16\cos (\pi \,t)\] [As a = ? 0.16 meter, T = 2 sec]


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