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

  • question_answer
    The maximum speed of a particle executing S.H.M. is \[1m/s\] and its maximum acceleration is \[1.57m/se{{c}^{2}}\]. The time period of the particle will be                                                    [DPMT 2002]

    A)            \[\frac{1}{1.57}sec\]         

    B)            1.57 sec

    C)            2 sec                                        

    D)            4 sec

    Correct Answer: D

    Solution :

                       Given max velocity \[\omega a=1\]and maximum acceleration \[{{\omega }^{2}}a=1.57\]            \[\therefore \frac{{{\omega }^{2}}a}{\omega a}=1.57\Rightarrow \omega =1.57\] \[\Rightarrow \frac{2\pi }{T}=1.57\Rightarrow T=4\]


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