Question

The path difference between the two waves $y_1=a_{1}sin(\omega t-\frac{2\pi x}{\lambda })and{y}_2=a_{2}cos(\omega t-\frac{2\pi x}{\lambda }+\varphi )$ is

• A. $\frac{\lambda }{2\pi }\varphi$
• B. $\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$
• C. $\frac{2\pi }{\lambda }(\varphi -\frac{\pi }{2})$
• D. $\frac{2\pi }{\lambda }\left(\varphi \right)$

Answer 1

The correct option is B

$\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$

$y_1=a_{1}sin(\omega t-\frac{2\pi x}{\lambda })y_2=a_{2}sin(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2})$

Phase difference

$=(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2})-(\omega t-\frac{2\pi x}{\lambda })=(\varphi +\frac{\pi }{2})$

$Pathdifference=\frac{\lambda }{2\pi }\times Phasedifference=\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$