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

  • question_answer
    The equation of motion of a particle is \[\frac{{{d}^{2}}y}{d{{t}^{2}}}+Ky=0\], where K is positive constant. The time period of the motion is given by                                                                [AIEEE 2005]

    A)            \[\frac{2\pi }{K}\]               

    B)            \[2\pi K\]                               

    C)            \[\frac{2\pi }{\sqrt{K}}\] 

    D)            \[2\pi \sqrt{K}\]

    Correct Answer: C

    Solution :

               On comparing with standard equation \[\frac{{{d}^{2}}y}{d{{t}^{2}}}+{{\omega }^{2}}y=0\] we get \[{{\omega }^{2}}=K\Rightarrow \omega =\frac{2\pi }{T}=\sqrt{K}\Rightarrow T=\frac{2\pi }{\sqrt{K}}\].


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