Two waves are represented by the equations $y_1 = A \sin (\omega t + kx + 0.57)\text{ m}$ and $y_2 = A \cos…

Two waves are represented by the equations $y_1 = A \sin (\omega t + kx + 0.57)\text{ m}$ and $y_2 = A \cos (\omega t + kx)\text{ m}$, where $x$ is in metre and $t$ is in second. The phase difference between them is
  1. 1.25 rad
  2. 1.57 rad
  3. 0.57 rad
  4. 1.0 rad

Solution

Given, $y_1 = A \sin (\omega t + kx + 0.57)\text{ m}$ and $y_2 = A \cos (\omega t + kx)\text{ m}$ $\Rightarrow y_2 = A \sin \left(\frac{\pi}{2} + \omega t + kx\right)\text{ m}$ Phase difference, $\Delta\phi = \phi_2 - \phi_1 = \frac{\pi}{2} - 0.57$ $= 1.57 - 0.57 = 1\text{ rad}$

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