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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2010

  • question_answer
    Two simple harmonic motions A and B are given respectively by the following equations \[{{y}_{1}}=a\,\sin \,\left( \omega t+\frac{\pi }{6} \right),\]         \[{{y}_{2}}=a\,\sin \,\left( \omega t+\frac{3\pi }{6} \right)\] The phase difference between the waves is

    A) \[\frac{\pi }{2}\]                                              

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

    C) \[\frac{\pi }{3}\]                                              

    D)  zero

    Correct Answer: C

    Solution :

    Phase difference \[\Delta \phi ={{\phi }_{1}}-{{\phi }_{2}}\] \[=\frac{3\pi }{6}-\frac{\pi }{6}\] \[=\frac{2\pi }{6}-\frac{\pi }{3}\]


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