s
A) \[\frac{\pi }{3}\]
B) \[\frac{2\pi }{3}\]
C) \[\pi \]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
| Key Idea: Phase difference \[=\frac{2\pi }{\lambda }\times \] path difference |
| Path difference between two points, |
| \[\Delta x=15-10=5m\] |
| Time period, \[T=0.05\text{ }s\] |
| \[\Rightarrow \] frequency \[v=\frac{1}{T}=\frac{1}{0.05}=20Hz\] |
| Velocity, \[v=300\text{ }m/s\] |
| \[\therefore \] Wavelength, \[\lambda =\frac{v}{v}=\frac{300}{20}=15m\] |
| Hence, phase difference \[\Delta \phi =\frac{2\pi }{\lambda }\times \Delta x\] |
| \[=\frac{2\pi }{15}\times 5=\frac{2\pi }{3}\] |
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