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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    The displacement y of wave travelling in the x-direction is given by \[y={{10}^{-4}}\sin \left( 600t-2x+\frac{\pi }{3} \right)metre,\] where, x is expressed in metres and t in seconds. The speed of the wave motion, in Ms-1 is

    A)  300                                       

    B)  600

    C)  1200                     

    D)         200

    Correct Answer: A

    Solution :

    The given equation of wave,                 \[y={{10}^{-4}}\sin \left( 600t-2x+\frac{\pi }{3} \right)\] ... (i) Standard equation of wave                 \[y=a\sin (\omega t-kx+\phi )\]                ... (ii) Now comparing Eqs. (i) and (ii), we get,                 \[\omega =600\]and\[k=2\] \[\therefore \]Velocity of wave\[=\frac{\omega }{k}=\frac{600}{2}=300\,\,m/s\]


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