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NEET NEET SOLVED PAPER 2017

  • question_answer
    A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is                                                               

    A) \[\frac{2\pi }{\sqrt{3}}\]                                               

    B)  \[\frac{\sqrt{5}}{\pi }\]

    C) \[\frac{\sqrt{5}}{2\pi }\]                                               

    D) \[\frac{4\pi }{\sqrt{5}}\]

    Correct Answer: D

    Solution :

                                    \[v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\]                 \[a=x{{\omega }^{2}}\]                 \[v=a\]                 \[\omega \sqrt{{{A}^{2}}-{{x}^{2}}}=x{{\omega }^{2}}\]                 \[\sqrt{{{(3)}^{2}}-{{(2)}^{2}}}=2\left( \frac{2\pi }{T} \right)\]                 \[\sqrt{5}=\frac{\pi }{T}\]                 \[T=\frac{4\pi }{\sqrt{5}}\]


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