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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Displacement of S.H.M. and Phase

  • question_answer
    A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre. The shortest time it takes to reach a point \[\frac{a}{\sqrt{2}}m\] from its mean position in seconds is             [EAMCET (Med.) 2000]

    A)            T    

    B)            T/4

    C)            T/8

    D)            T/16

    Correct Answer: C

    Solution :

                       \[y=a\sin \frac{2\pi }{T}t\] Þ \[\frac{a}{\sqrt{2}}=a\sin \frac{2\pi }{T}\cdot t\]            Þ \[\sin \frac{2\pi }{T}t=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}\]Þ \[\frac{2\pi }{T}t=\frac{\pi }{4}\]Þ \[t=\frac{T}{8}\]


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