wave_mechanics Beats Question 51
Question: A sound source of frequency 170 Hz is placed near a wall. A man walking from a source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is 340 m/s the distance (in metres) separating the two adjacent positions of minimum intensity is
[MNR 1992; UPSEAT 2000; CPMT 2002]
Options:
1/2
1
3/2
2
Show Answer
Answer:
Correct Answer: B
Solution:
$ v=n\lambda $
Therefore $ \lambda =\frac{v}{n}=\frac{340}{170} $
Therefore $ \lambda =2 $ Distance separating the position of minimum intensity = $ \frac{\lambda }{2}=\frac{2}{2}=1 $ m
BETA
