BHU PMT BHU PMT Solved Paper-2001

  • question_answer
    The distance between two point?s difference in phase by \[{{60}^{\circ }}\] on a wave having a wave velocity \[360\,\,m\text{/}s\] and frequency \[500\,\,Hz\] is:

    A)  0.36 m                 

    B)  0.12 m

    C)  0.18 m                                 

    D)  0.72 m

    Correct Answer: B

    Solution :

    We know that Velocity \[\left( v \right)=\]frequency \[\left( n \right)\times \]wavelength\[\left( \lambda  \right)\] Given, \[v=360\,m/s,\,n=500\,Hz\] \[\therefore \]  \[\lambda =\frac{v}{n}=\frac{360}{500}=0.72\,m\] Also Phase difference \[\left( \Delta \phi  \right)\] \[=\frac{2\pi }{\lambda }\times \]path difference\[\left( \Delta \,x \right)\] Given,   \[\Delta \,\phi =\frac{\pi }{3}rad\] \[\therefore \]  \[\Delta \,x=\frac{\lambda \,\Delta \,\phi }{2\pi }\] \[\Delta \,x=\frac{\lambda \,}{2\pi }\times \frac{\pi }{3}=\frac{\lambda }{6}=\frac{0.72}{6}=0.12\,m\]


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