Physics - Wave on String Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{y}}_1={\mathrm{a}}_1\sin \left(\omega \mathrm{t}-\frac{2\pi \mathrm{x}}{\lambda }\right)\\ & {\mathrm{y}}_2={\mathrm{a}}_2\cos \left(\omega \mathrm{t}-\frac{2\pi \mathrm{x}}{\lambda }+\varphi \right)\end{array}$ ${\mathrm{y}}_2$ can also be written as $\begin{array}{rl}{\mathrm{y}}_2& ={\mathrm{a}}_2\sin \left[\frac{\pi }{2}+\left(\omega \mathrm{t}-\frac{2\pi \mathrm{x}}{\lambda }+\varphi \right)\right]\\ & ={\mathrm{a}}_2\sin \left(\omega \mathrm{t}-\frac{2\pi \mathrm{x}}{\lambda }+\varphi +\frac{\pi }{2}\right)\end{array}$ The phase difference between the two waves is Path difference $=\frac{\lambda }{2\pi }\times$ Phase difference Path difference $=\frac{\lambda }{2\pi }\times \left(\varphi +\frac{\pi }{2}\right)$