Physics - Wave on String Question with Solution | TestHub

The given equations of the waves are:

$y_1=5\sin \left(\frac{2\pi x}{\lambda }-\omega t\right)$

$y_2=5\cos \left(\frac{2\pi x}{\lambda }-\omega t+\frac{\pi }{6}\right)$

To find the phase difference, both equations must be expressed in terms of the same trigonometric function. Using the identity $\cos \theta =\sin \left(\theta +\frac{\pi }{2}\right)$, we can rewrite $y_2$ as:

$y_2=5\sin \left(\frac{2\pi x}{\lambda }-\omega t+\frac{\pi }{6}+\frac{\pi }{2}\right)$

$y_2=5\sin \left(\frac{2\pi x}{\lambda }-\omega t+\frac{2\pi }{3}\right)$

The phase of the first wave is ${\varphi }_1=\frac{2\pi x}{\lambda }-\omega t$

The phase of the second wave is ${\varphi }_2=\frac{2\pi x}{\lambda }-\omega t+\frac{2\pi }{3}$

The phase difference between the two waves is:

$\Delta \varphi ={\varphi }_2-{\varphi }_1=\frac{2\pi }{3}$

Answer: $\frac{2\pi }{3}$