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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • question_answer
    The equation of a simple harmonic wave is given by y = 5 sin \[\frac{\pi }{2}\] (100t - x), where x and y are in metre and time is in second. The period of the wave in second will be:

    A)  0.04

    B)  0.01

    C)  1

    D)  5

    Correct Answer: A

    Solution :

    The given equation is \[y=5\sin \frac{\pi }{2}(100t-x)\]?(i) Comparing Eq. (i) with standard wave equation, given by\[y=A\sin (\omega t-kx)\]         ?(ii) we have \[\omega =\frac{100\pi }{2}=50\pi \] \[\Rightarrow \] \[\frac{2\pi }{T}=50\pi \] \[\Rightarrow \] \[T=\frac{2\pi }{50\pi }=0.04s\]


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