The path difference between two waves, represented by $\mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega…

The path difference between two waves, represented by $\mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ and $\mathrm{y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right)$ is
  1. $\frac{\lambda}{2 \pi}(\phi)$
  2. $\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
  3. $\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$
  4. $\frac{2 \pi}{\lambda}(\phi)$

Solution

$\begin{aligned} & \mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right) \\ & \mathrm{y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right) \end{aligned}$ $\mathrm{y}_2$ can also be written as $\begin{aligned} \mathrm{y}_2 & =\mathrm{a}_2 \sin \left[\frac{\pi}{2}+\left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right)\right] \\ & =\mathrm{a}_2 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi+\frac{\pi}{2}\right) \end{aligned}$ The phase difference between the two waves is Path difference $=\frac{\lambda}{2 \pi} \times$ Phase difference Path difference $=\frac{\lambda}{2 \pi} \times\left(\phi+\frac{\pi}{2}\right)$

Asked in: MHT CET 2023 (11 May Shift 2)

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