JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Oscillations

  • question_answer
    The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \[10{{s}^{-1}}\]. At, t = 0 the displacement is 5 m. The initial phase is\[\frac{\pi }{4}\]. What is the maximum acceleration?

    A) \[500m/{{s}^{2}}\]

    B) \[500\,\sqrt{2}\,m/{{s}^{2}}\]

    C) \[750\,\,m/{{s}^{2}}\]

    D) \[750\,\sqrt{2}\,m/{{s}^{2}}\]

    Correct Answer: B

    Solution :

    [b] Given that, \[\frac{{{A}_{\max }}}{{{V}_{\max }}}=10i.c.,\,\omega =10{{s}^{-1}}\] Displacement is given by \[x=a\,\sin \,(\omega t+\pi /4)\] \[at\,t=0,\,x=5;5=a\,\sin \,{{45}^{o}}\Rightarrow a=5\sqrt{2}\] \[{{A}_{\max }}=a{{\omega }^{2}}=500\sqrt{2\,}m/{{s}^{2}}\]


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