WB JEE Medical WB JEE Medical Solved Paper-2007

  • question_answer
    A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

    A)  \[\frac{1}{2\pi \sqrt{3}}\]                           

    B)  \[2\pi \sqrt{3}\]

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

    D)  \[\frac{\sqrt{3}}{2\pi }\]

    Correct Answer: C

    Solution :

                     Velocity = acceleration \[\omega \sqrt{{{a}^{2}}-{{y}^{2}}}={{\omega }^{2}}y\] \[\sqrt{{{(2)}^{2}}-{{(1)}^{2}}}=\omega (1)\]                 \[\Rightarrow \]                               \[\omega =\sqrt{3}\]                                 \[T=\frac{2\pi }{\omega }\]                 \[\Rightarrow \]               \[T=\frac{2\pi }{\sqrt{3}}\]


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