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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Mock Test - Simple Harmonic Motion

  • question_answer
    A particle is executing a motion in which its displacement as a function of time is given by \[x=3\,\sin \,(5\pi t+\pi /3)+cos(5\pi t+\pi /3)\]where \[x\] is in \[m\] and \[t\] is in s. Then the motion is

    A)  Simple harmonic with time period 0.2 s

    B)  Simple harmonic with time period 0.4 s

    C)  Simple harmonic with amplitude 3 m

    D)  Not a simple harmonic but a periodic motion

    Correct Answer: B

    Solution :

    [b] \[x=3\sin (5\pi t+\pi /3)+cos(5\pi t+\pi /3)\] Amplitude \[={{x}_{\max }}=\sqrt{{{3}^{2}}+{{1}^{2}}}=\sqrt{10,}\omega =5\pi \] \[T=\frac{2\pi }{\omega }=\frac{2\pi }{5\pi }=0.4s\]


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