Two waves are represented by the equation, $\mathrm{y}_1=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx}+0…

Two waves are represented by the equation, $\mathrm{y}_1=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx}+0.57) \mathrm{m}$ and $\mathrm{y}_2=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}) \mathrm{m}$, where $\mathrm{x}$ is in metre and $\mathrm{t}$ is in second. What is the phase difference between them?
  1. 0.57 radian
  2. 1.0 radian
  3. 1.57 radian
  4. 1.25 radian

Solution

$\begin{aligned} & \mathrm{y}_1=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx}+0.57) \\ & \mathrm{y}_2=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx})=\mathrm{A} \sin \left(\omega \mathrm{t}+\mathrm{kx}+\frac{\pi}{2}\right) \\ & \therefore \text { Phase difference }=\frac{\pi}{2}-0.57 \\ & =\frac{3.14}{2}-0.57 \\ & =1.57-0.57=1.0 \mathrm{rad}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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