Physics - Wave on String Question with Solution | TestHub

The given equations of the waves are:

$y_1=a_1\sin \left(\omega t-\frac{2\pi x}{\lambda }\right)$

$y_2=a_2\cos \left(\omega t-\frac{2\pi x}{\lambda }+\frac{\pi }{6}\right)$

To find the phase difference, we express $y_2$ in terms of the sine function:

$y_2=a_2\sin \left(\omega t-\frac{2\pi x}{\lambda }+\frac{\pi }{6}+\frac{\pi }{2}\right)$

$y_2=a_2\sin \left(\omega t-\frac{2\pi x}{\lambda }+\frac{2\pi }{3}\right)$

The phase difference $\Delta \varphi$ between the two waves is:

$\Delta \varphi =\frac{2\pi }{3}$

The relation between path difference $\Delta x$ and phase difference $\Delta \varphi$ is given by:

$\Delta x=\frac{\lambda }{2\pi }\Delta \varphi$

Substituting the value of $\Delta \varphi$:

$\Delta x=\frac{\lambda }{2\pi }\times \frac{2\pi }{3}=\frac{\lambda }{3}$

Answer: $\frac{\lambda }{3}$