BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    The speed of a wave is 360 m/s and frequency is 500 Hz. Phase difference between two consecutive particles is \[60{}^\circ \], then path difference between them will be:

    A)  0.72 cm                               

    B)         120 cm

    C)         12 cm                                  

    D)         7.2 cm

    Correct Answer: C

    Solution :

    The relation between velocity, frequency and wavelength is given by \[v=n\lambda \]                 or            \[\lambda =\frac{v}{n}\]                 Given, \[=360\,m/s,\,n=500\,Hz\] \[\therefore \]  \[\lambda =\frac{360}{500}=0.72\,m\] Path difference \[=\frac{\lambda }{2\pi }\times \]phase difference       i.e.,        \[\Delta x=\frac{\lambda }{2\pi }\times \Delta \,\phi \] or            \[\Delta x=\frac{0.72}{2\pi }\times \frac{\pi }{3}\]                 \[\left( \therefore \,\Delta \phi \,\text{=}\,\text{6}{{\text{0}}^{o}}=\frac{\pi }{3} \right)\]                 \[=0.12\,m\] \[=12\,cm\]


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