wave_mechanics organ_pipe_vibration_of_air_column Question 36
Question: Stationary waves are set up in air column. Velocity of sound in air is 330 m/s and frequency is 165 Hz. Then distance between the nodes is
[EAMCET (Engg.) 1995; CPMT 1999]
Options:
A) 2 m
B) 1 m
C) 0.5 m
D) 4 m
Show Answer
Answer:
Correct Answer: B
Solution:
$ v=330 $
$ m/s $ ; $ n=165 $
$ Hz $ .
Distance between two successive nodes = $ \frac{\lambda }{2} $
$ =\frac{v}{2n}=\frac{330}{2\times 165}=1m $
BETA
