Wave Mechanics Question 705
Question: In stationary waves, distance between a node and its nearest antinode is 20 cm. The phase difference between two particles having a separation of 60 cm will be
[CMEET Bihar 1995]
Options:
A) Zero
B) p/2
C) p
D) 3p/2
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{\lambda }{4}=20\Rightarrow \lambda =,80,cm $ , also $ \Delta \varphi =\frac{\lambda }{2\pi }.,\Delta x $
Therefore $ \Delta \varphi = $
$ \frac{60}{80}\times 2\pi =\frac{3\pi }{2} $
BETA
