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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Time Period and Frequency

  • question_answer
    The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as                     \[x=0.01\cos \left( \pi \,t+\frac{\pi }{4} \right)\]            The frequency of the motion will be       [UPSEAT 2004]

    A)            0.5 Hz                                      

    B)            1.0 Hz

    C)            \[\frac{\pi }{2}Hz\]            

    D)            \[\pi \,Hz\]

    Correct Answer: A

    Solution :

                       Comparing given equation with standard equation, \[x=a\cos \,(\omega t+\varphi )\] we get, \[a=0.01\] and\[\omega =\pi \]  \[\Rightarrow 2\pi n=\pi \Rightarrow n=0.5\,Hz\]


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