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

  • question_answer
    A particle in SHM is described by the displacement equation \[x(t)=A\cos (\omega t+\theta ).\]If the initial (t = 0) position of the particle is 1 cm and its initial velocity is \[\pi \]cm/s, what is its amplitude? The angular frequency of the particle is \[\pi {{s}^{-1}}\] [DPMT 2004]

    A)            1 cm                                         

    B)            \[\sqrt{2}\]cm

    C)            2 cm                                         

    D)            2.5 cm                                             

    Correct Answer: B

    Solution :

                       Given, \[v=\pi \,cm/\sec ,\] \[x=1\,cm\] and \[\omega =\pi {{s}^{-1}}\] using \[v=\omega \sqrt{{{a}^{2}}-{{x}^{2}}}\]Þ \[\pi =\pi \sqrt{{{a}^{2}}-1}\]            \[\Rightarrow 1={{A}^{2}}-1\]\[\Rightarrow A=\sqrt{2}\,cm.\]


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