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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    A travelling wave represented by \[y=A\sin ((\omega t-kx)\]is superimposed on another wave represented by \[y=A\sin (\omega t+kx).\]The resultant is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  A wave travelling along + x direction

    B)  A wave travelling along - x direction

    C)  A standing wave having nodes at \[x=\frac{n\lambda }{2},n=0,1,2\]?..

    D)  A standing wave having nodes at\[x=\left( n+\frac{1}{2} \right)\frac{\lambda }{2};n=0,1,2\]??

    Correct Answer: D

    Solution :

                                    \[Y=A\sin (\omega t-kx)+Asin(\omega t+kx)\] \[Y=2A\sin \omega t\cos kx\]standing wave For nodes    \[\cos kx=0\]                 \[\frac{2\pi }{\lambda }.x=(2n+1)\frac{\pi }{2}\] \[\therefore \]  \[x=\frac{(2n+1)\lambda }{4},n=0,1,2,3,\]???..


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