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BCECE Engineering BCECE Engineering Solved Paper-2009

  • question_answer
    The displacement of a travelling wave is given by\[y=2\,\cos \,2\pi \,(10t-\,0.008\,x\,+\,.35)\] where, x and y are in centimetre and t in second. What is the phase difference between oscillatory motion at two points separated by a distance of 4 m?

    A)  \[2\pi \]                                              

    B) \[4\pi \]

    C)  \[6\pi \]             

    D)         \[8\pi \]

    Correct Answer: C

    Solution :

    Given, \[y=2\cos (20\pi t-2\pi \times 0.008x+0.7\pi )\] Comparing with general equation \[y=a\cos (\omega t\pm kx\pm {{\phi }_{0}})\] We get \[k=2\pi \times 0.008\] \[\frac{2\pi }{\lambda }=2\pi \times 0.008\]         \[\left[ As\,k=\frac{2\pi }{\lambda } \right]\] \[\Rightarrow \]               \[\lambda =\frac{1}{0.008}\]                 \[\lambda =125\,cm\]                 \[=1.25\,m\] Now, phase difference\[(\Delta \phi )\] \[=\frac{2\pi }{\lambda }\]path difference\[(\Delta x\,)\]                                 \[=\frac{2\pi }{1.25}\times 4=6\pi \]


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