The displacement due to a wave moving in the positive $x$-direction is given by $y = \frac{1}{(1 + x^2)}$ at time $t = 0$ and by $y = \frac{1}{[1 + (x - 1)^2]}$ at $t = 2$ seconds,where $x$ and $y$ are in metres. The velocity of the wave in $m/s$ is
- A$0.5$
- B$1$
- C$2$
- D$4$
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The path difference between the two waves $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$ is
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View SolutionThe phase difference between the following two waves $y_1$ and $y_2$ is:
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