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VMMC VMMC Medical Solved Paper-2014

  • question_answer
    The displacement of a particle from its mean position (in metre) is given by \[y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )\] The motion of the particle is

    A)  periodic but not SHM

    B) non-periodic

    C)  simple harmonic motion with period 0.1 s

    D) simple harmonic motion with period 0.2 s

    Correct Answer: C

    Solution :

    Given, \[y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )\] \[y=0.1\sin 2(10\pi t+1.5\pi )\] [using 2 sin A cos A = sin2 A] \[y=0.1\sin (20\pi t+3\pi )\] This equation represents simple harmonic     motion of angular frequency \[20\pi .\] \[\therefore \]Time period \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }\] \[=\frac{1}{10}=0.1s\]


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