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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

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

    E)  \[\frac{\sqrt{3}}{\pi }\]

    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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