Question

Sound source of frequnecy $170Hz$ is placed near a wall. $A$ man walking from the source towards the wall finds, that there is periodic rise and fall of sound intensity. If the speed of sound in air is $340m/s$, then the distance separating the two adjacent portions of minimum intensity is:

• A. 1 / 2 m
• B. 3 / 2 m
• C. 1 m
• D. 2 m

Answer 1

Answer: (C)

$\lambda =\frac{v}{v}=\frac{340}{170}=m$
$\lambda =2m$
$\therefore$ Distance between 2 portions of minimum intensity
$=\frac{\lambda }{2}=1m$