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JEE Main & Advanced Physics Wave Mechanics Question Bank Interference and Superposition of Waves

  • question_answer
    Two sound waves (expressed in CGS units) given by \[{{y}_{1}}=0.3\sin \frac{2\pi }{\lambda }(vt-x)\] and \[{{y}_{2}}=0.4\sin \frac{2\pi }{\lambda }(vt-x+\theta )\]  interfere. The resultant amplitude at a place where phase difference is \[\pi /2\] will be                                [MP PET 1991]

    A)            0.7 cm                                     

    B)            0.1 cm

    C)            0.5 cm                                     

    D)            \[\frac{1}{10}\sqrt{7}\,cm\]

    Correct Answer: C

    Solution :

               Resultant amplitude = \[\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \varphi }\]            = \[\sqrt{{{0.3}^{2}}+{{0.4}^{2}}+2\times 0.3\times 0.4\times \cos \frac{\pi }{2}}\]= \[0.5\,\,cm\]


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