Question

Find the path difference between the following waves.
$y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$

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

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


Given:
$y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$
$y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\varphi )$
$\Rightarrow y_2=a_2\sin (\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2})$

The phase difference between the waves is given by,
$\Delta \varphi =(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2})-(\omega t-\frac{2\pi x}{\lambda })$
$\Delta \varphi =(\varphi +\frac{\pi }{2})$

Therefore, the path difference between the waves will be,
$\Delta x=\frac{\lambda }{2\pi }\times \Delta \varphi =\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$

Hence, $\left(B\right)$ is the correct answer.