JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \[10{{s}^{-1}}.\]   [JEE Online 08-04-2017] At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is\[\frac{\pi }{4}.\]

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

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

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

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

    Correct Answer: D

    Solution :

                    \[{{f}_{\max }}=\omega a\]                 \[{{V}_{\min }}=a\omega \]                 \[\frac{\omega a}{a\omega }=10\]                 W = 10                 \[x=a\sin (\omega +\pi /4)\]                 At f = 0                 \[5=a\sin (\pi /4)\]                 \[a=5\sqrt{2}\]                 Max acc. = w2a                 \[=100\times 5\sqrt{2}\]                 \[=500\sqrt{2}\]


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