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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

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


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