Question

A sound wave of frquency 330 Hz is incident normally at reflected wall then minimum distance from wall at which partical vibrate very much :- (Vsound = 330 m/s)

• A. 0.25m
• B. 0.125 m
• C. 1 m
• D. 0.5 m

Answer 1

The correct option is A 0.25m

Treating second waves as standing pressure waves * we get

$\lambda =\frac{v}{f}=\frac{330m/s}{330Hz}1m$

Minimum distance from wall at which particle vibrate very much$=$ Minimum distance from the wall where the pressure in zero.

Figure sideways represents variation pressure of the wave with distance. So, required distance$=\frac{\lambda }{4}=\frac{1}{4}m=0.25$m

Option A is correct.