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WB JEE Medical WB JEE Medical Solved Paper-2008

  • question_answer
    The equation of a progressive wave can be  given by\[y=15\]\[\sin (660\pi t-0.02\pi x)\,cm.\] The frequency of the wave is

    A)  330 Hz                                   

    B)  342 Hz

    C)  365 Hz                                 

    D)  660 Hz

    Correct Answer: A

    Solution :

                     Given that y \[y=15\,\sin \,(660\pi t-0.02\,\pi x)\] Comparing with general equation of  progressive wave, we get \[y=(x,t)=A\sin \left( \frac{2\pi }{T}t-\frac{2\pi }{\lambda }x \right)\] \[\therefore \]  \[\frac{2\pi }{T}=660\pi \]or \[\frac{1}{T}=330\]pr \[v=330\,\text{Hz}\]


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