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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    Equations of a stationary and a travelling waves are as follows; \[{{y}_{1}}=a\,\sin \,kx\cos \omega t\] and \[{{y}_{2}}=a\sin (\omega t-kx)\] The phase difference between two points \[{{x}_{1}}=\frac{\pi }{3k}\] and \[{{x}_{2}}=\frac{3\pi }{2k}\], are \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\] respectively for the two waves. The ratio \[\frac{{{\phi }_{1}}}{{{\phi }_{2}}}\] is:

    A) 1

    B) \[\frac{5}{6}\]

    C) \[\frac{3}{4}\]  

    D) \[\frac{6}{7}\]

    Correct Answer: D

    Solution :

    At \[{{x}_{1}}=\frac{\pi }{3k}\] and \[{{x}_{2}}=\frac{3\pi }{2k}\]
    \[{{\operatorname{sinkx}}_{1}}\] or \[{{\operatorname{sinkx}}_{2}}\] is not zero.
    Therefore, neither of \[{{x}_{1}}\] or \[{{x}_{2}}\]is a node
    \[\Delta x={{x}_{2}}-{{x}_{1}}=\left( \frac{3}{2}-\frac{1}{3} \right)\frac{\pi }{k}=\frac{7\pi }{6k}\]
    Since \[\frac{2\pi }{k}>\Delta x>\frac{\pi }{k};\,\lambda >\Delta x>\frac{\lambda }{2}\]
    \[\left( k=\frac{2\pi }{\lambda } \right)\]
    Therefore, \[{{\phi }_{1}}=\pi \,and\,{{\phi }_{2}}=k.\Delta x=\frac{7\pi }{6};\therefore \frac{{{\phi }_{1}}}{{{\phi }_{2}}}=\frac{6}{7}\]
    NOTE: In case of a stationary wave phase difference between any two points is either zero or \[\pi \].


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