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

  • question_answer
     The displacement x (in metres) of a particle performing simple harmonic motion is related to time t (in seconds) as \[x=0.05\cos \left( 4\,\pi \,t+\frac{\pi }{4} \right)\]. The frequency of the motion will be                                                                   [MP PMT/PET 1998]

    A)            0.5 Hz                                      

    B)            1.0 Hz

    C)            1.5 Hz                                      

    D)            2.0 Hz

    Correct Answer: D

    Solution :

               From the given equitation \[\omega =2\pi n=4\pi \] Þ \[n=2Hz\]


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