BCECE Medical BCECE Medical Solved Papers-2005

  • question_answer
    The speed of a wave is 360 m/s and frequency is 500 Hz. Phase difference between two consecutive particles is \[{{60}^{o}}\], 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, v = 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            \[\Delta \,x=\frac{0.72}{2\pi }\times \frac{\pi }{3}\] \[\left( \therefore \,\Delta \phi ={{60}^{o}}=\frac{\pi }{3} \right)\]                 = 0.12 m = 12 cm


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