wave_mechanics progressive_waves Question 5
Question: At a moment in a progressive wave, the phase of a particle executing S.H.M. is $ \frac{\pi }{3} $ . Then the phase of the particle 15 cm ahead and at the time $ \frac{T}{2} $ will be, if the wavelength is 60 cm
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{2\pi }{3} $
C) Zero
D) $ \frac{5\pi }{6} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let the phase of second particle be $ \varphi $ .
Hence phase difference between two particles is $ \Delta \varphi =\frac{2\pi }{\lambda }\Delta x $
Therefore $ ( \varphi -\frac{\pi }{3} )=\frac{2\pi }{60}\times 15 $
$ \Rightarrow \varphi -\frac{\pi }{3}=\frac{\pi }{2}\Rightarrow \varphi =\frac{5\pi }{6} $
BETA
