wave_mechanics critical_thinking Question 15
Question: A source producing sound of frequency 170 Hz is approaching a stationary observer with a velocity 17 ms?1. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 ms?1)
[EAMCET (Engg.) 2000]
Options:
A) 0.1m
B) 0.2m
C) 0.4m
D) 0.5m
Show Answer
Answer:
Correct Answer: A
Solution:
$ \lambda =\frac{v}{n}=\frac{340}{170}=2m,n’=\frac{340}{340-17}\times 170\Rightarrow n’=178.9Hz $ Now $ {\lambda }’=\frac{v}{{{n}’}}=\frac{340}{178.9}=1.9 $
Therefore $ \lambda -{\lambda }’=2-1.9=0.1 $
BETA
