wave_mechanics basics_of_mechanical_waves Question 76
Question: A wave of frequency 500 Hz has velocity 360 m/sec. The distance between two nearest points 60° out of phase, is
[NCERT 1979; MP PET 1989; JIPMER 1997; RPMT 2002, 03; CPMT 1979, 90, 2003; BCECE 2005]
Options:
A) 0.6 cm
B) 12 cm
C) 60 cm
D) 120 cm
Show Answer
Answer:
Correct Answer: B
Solution:
The distance between two points i.e.
path difference between them $ \Delta =\frac{\lambda }{2\pi }\times \varphi =\frac{\lambda }{2\pi }\times \frac{\pi }{3}=\frac{\lambda }{6}=\frac{v}{6n} $
$ (\because v=n\lambda ) $
Therefore $ \Delta =\frac{360}{6\times 500}=0.12m=12cm $
BETA
