A) \[\frac{\lambda }{2\pi }\varphi \]
B) \[\frac{\lambda }{2\pi }\left( \varphi +\frac{\pi }{2} \right)\]
C) \[\frac{2\pi }{\lambda }\left( \varphi -\frac{\pi }{2} \right)\]
D) \[\frac{2\pi }{\lambda }\varphi \]
Correct Answer: B
Solution :
\[{{y}_{1}}={{a}_{1}}\sin \,\left( \omega \,t-\frac{2\pi x}{\lambda } \right)\]and \[{{y}_{2}}={{a}_{2}}\cos \,\left( \omega \,t-\frac{2\pi x}{\lambda }+\varphi \right)\]\[={{a}_{2}}\sin \,\left( \omega \,t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2} \right)\] So phase difference =\[\varphi +\frac{\pi }{2}\] and D =\[\frac{\lambda }{2\pi }\left( \varphi +\frac{\pi }{2} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
