Question

A sound wave with frequency $256Hz$ falls normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles will have maximum amplitude of vibrations is nearly ( velocity of sound in air is $336m/s$ )

• A. $32.8cm$
• B. $50cm$
• C. $65.8cm$
• D. $25cm$

Answer 1

Answer: (A)

Given, $f=256Hz$
$v_{\text{sound }}=336m{s}^{-1}$
$\lambda =\frac{v}{f}=\frac{336}{256}=1.3125m$
$=131.25cm$
The distance between a node $(N)$ and adjoining antinode $=\frac{\lambda }{4}=\frac{336}{256}=32.8cm$
the shortest distance from the wall at which the air particles will have maximum amplitude of vibration is $32.8cm$.