Q. 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 $340\ m/s.$ then distance (in $metres$) separating the two adjacent position of minimum intensity is
A) $1/2$
B) 1
C) $3/2$
D) 2
Explanation
(b) $v = n\lambda \Rightarrow \lambda = \frac{v}{n} = \frac{340}{170} \Rightarrow \lambda = 2$
First case
[image:15_image65.emf]
Second case
[image:15_image66.emf]
$x = \frac{\lambda}{2} = 1\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.