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

  • question_answer
    The amplitude and the time period in a S.H.M. is 0.5 cm and 0.4 sec respectively. If the initial phase is \[\pi /2\] radian, then the equation of S.H.M. will be

    A)            \[y=0.5\sin 5\pi t\]            

    B)            \[y=0.5\sin 4\pi t\]

    C)            \[y=0.5\sin 2.5\pi t\]         

    D)            \[y=0.5\cos 5\pi t\]

    Correct Answer: D

    Solution :

                       \[y=a\sin (\omega \,t+\varphi )\] \[=a\sin \left( \frac{2\pi }{T}t+\varphi  \right)\]Þ\[y=0.5\sin \,\left( \frac{2\pi }{0.4}t+\frac{\pi }{2} \right)\] \[y=0.5\sin \,\left( 5\pi \,t+\frac{\pi }{2} \right)\]\[=0.5\cos 5\pi t\]


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