Question

The path difference between the two waves $y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$ and $y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\varphi )$

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

Answer 1

The correct option is B $\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$

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

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

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

phaze of $y_1=0$

phaze of $y_2=\frac{\pi }{2}+\varphi$

$\Delta \varphi =\frac{\pi }{2}+\varphi -0$

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

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

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

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

Option B