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JCECE Engineering JCECE Engineering Solved Paper-2009

  • 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                   \[\omega =\frac{100\pi }{2}=50\pi \] \[\Rightarrow \]               \[\frac{2\pi }{T}=50\pi \]                    \[T=\frac{2\pi }{50\pi }=0.04\,\,s\]


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