CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The equation of a simple harmonic wave is given by \[y=6\text{ }\sin \text{ }2\pi \,(2t-0.1x)\] where \[x\] and y are in mm and t is in seconds. The phase difference between two particles 2 mm apart at any instant is

    A) \[{{18}^{o}}\]                                    

    B)  \[{{36}^{o}}\]

    C)  \[{{54}^{o}}\]                                   

    D)  \[{{72}^{o}}\]

    Correct Answer: D

    Solution :

    From the given equation \[k=0.2\,\pi \] \[\Rightarrow \]               \[\frac{2\pi }{\lambda }=0.2\pi \] \[\Rightarrow \]               \[\lambda =10\,mm\]                 \[\Delta \phi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{10}\times 2=\frac{2\pi }{5}={{72}^{o}}\]


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