Question 342: Waves

Question

The path difference between two waves $\mathrm{Y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ and $\mathrm{Y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right)$ is

Accepted Answer

The path difference between two waves given by $Y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right)$ and $Y_2 = a_2 \cos\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right)$ can be calculated as follows:
First, express the second wave in terms of sine:
$ y_2 = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) $
This shows that the phase of $Y_2$ is $\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}$.
The phase difference between the two waves is thus given by:
$ \delta = \phi + \frac{\pi}{2} $
Using the relation for path difference $\Delta x = \frac{\lambda}{2\pi} \cdot \delta$, we substitute the phase difference:
$ \Delta x = \frac{\lambda}{2\pi} \left(\phi + \frac{\pi}{2}\right) $

Answer

$\frac{\lambda \phi}{2 \pi}$

Answer

$\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$

Answer

$\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$

Answer

$\frac{2 \pi}{\lambda} \phi$

Waves

The path difference between two waves \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

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