Physics - SHM Question with Solution | TestHub

The given equations of the waves are:

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

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

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:

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

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

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

The phase difference between the two waves is:

$\Delta \Phi ={\Phi }_2-{\Phi }_1=\left(\frac{2\pi x}{\lambda }-\omega t+\varphi +\frac{\pi }{2}\right)-\left(\frac{2\pi x}{\lambda }-\omega t\right)$

$\Delta \Phi =\varphi +\frac{\pi }{2}$

Answer: $\left(\varphi +\frac{\pi }{2}\right)$