2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    If the displacement equation of a particle from its mean position is given as \[y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )\] then, the motion of particle is

    A)  non-periodic

    B)  periodic but not SHM

    C)  SHM with period 0.2 s

    D)  SHM with period 0.1 s

    Correct Answer: D

    Solution :

    Given, \[y=0.2\sin (10\pi t+1.5\pi )cos(10\pi t+1.5\pi )\] \[=0.1\sin 2(10\pi t+1.5\pi )\]                 \[[sin\,2A=2sin\,A\,cos\,A]\] \[=0.1\,\sin (20\pi t+3.0\pi )\] So, the motion of particle is SHM Time period, \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }=\frac{1}{10}=0.1\,s\]


You need to login to perform this action.
You will be redirected in 3 sec spinner