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Haryana PMT Haryana PMT Solved Paper-2002

  • question_answer
    The equation of a transverse wave is given by    \[y\] = 20 sin\[\pi \] (\[0.02x-2t\]) where \[x\] and \[y\] are        in cms and \[t\] is in seconds. The wavelength in   cm will be :

    A)  90                                         

    B)  100

    C)  200                                       

    D)  5

    Correct Answer: B

    Solution :

                    Equation of wave \[y=20\sin \pi (0.02x-2t\] comparing it with standard equation \[y=A\sin (kx-\omega t)\] we get \[k=0.02\pi \] Wavelength \[\lambda =\frac{2\pi }{k}=\frac{2\pi }{0.02\pi }=100cm\].


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