Two waves are represented as y_1 = a_1 sin le | vedclass

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(C) First, express both waves in the same trigonometric function (sine). $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ $y_2 = a_2

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$\frac{\lambda}{3}$

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(C) First, express both waves in the same trigonometric function (sine). $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ $y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \frac{\pi}{6} \right) = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \frac{\pi}{6} + \frac{\pi}{2} \right)$ $y_2 = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \frac{2\pi}{3} \right)$ The phase difference $\phi$ between the two waves is the difference in their arguments: $\phi = \left( \omega t - \frac{2\pi x}{\lambda} + \frac{2\pi}{3} \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \frac{2\pi}{3}$ The relationship between path difference $\Delta x$ and phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$. Substituting the values: $\frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x$ $\Delta x = \frac{\lambda}{3}$

Two waves are represented as $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \frac{\pi}{6} \right)$. The path difference between the two waves is:

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