JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    Two waves are represented as\[{{y}_{1}}=2a\,\sin \left( \omega t+\frac{\pi }{6} \right)\]and\[{{y}_{2}}=-2a\,\cos \left( \omega t+\frac{\pi }{6} \right)\]. The phase difference between the two waves is

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

    B)  \[\frac{4\pi }{3}\]

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

    D)  \[\frac{5\pi }{6}\]

    Correct Answer: D

    Solution :

     Idea we can compare the phase of two waves only when they both are represented by 'sin' or 'cosine' functions. \[{{y}_{1}}=2a\sin \left( \omega t+\frac{\pi }{6} \right)=-2a\cos \left( \frac{\pi }{2}+\omega t+\frac{\pi }{6} \right)\] \[\Rightarrow \]\[{{\phi }_{1}}=\frac{\pi }{2}+\omega t+\frac{\pi }{6}\] and\[{{y}_{2}}=-2a\cos \left( \omega t-\frac{\pi }{6} \right)\] \[\Rightarrow \]\[{{\phi }_{2}}=\omega t-\frac{\pi }{6}\] \[\therefore \]Phase difference \[={{\phi }_{1}}-{{\phi }_{2}}=\frac{\pi }{2}+\omega t+\frac{\pi }{6}-\left( \omega t-\frac{\pi }{6} \right)\] \[=\frac{\pi }{2}+\frac{\pi }{3}=\frac{5\pi }{6}\] TEST Edge For example the phase of a wave can be represented by\[\left( \omega t+\frac{\pi }{3} \right)\]. Here\[\frac{\pi }{3}\]is 'phase constant. The value of phase at t = 0 is known as phase constant.

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