Question

A sound wave with frequency $ 256\,Hz $ 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 $ 336\,m/s$ )

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

Answer 1

Answer: (A)

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