Sound waves of frequency $600 \mathrm{~Hz}$ fall normally on a perfectly reflecting wall. The shortest…
Sound waves of frequency $600 \mathrm{~Hz}$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have maximum amplitude of vibration is (speed of sound $=300 \mathrm{~ms}^{-1}$ )
$\frac{1}{4} \mathrm{~m}$
$\frac{1}{8} \mathrm{~m}$
$\frac{3}{8} \mathrm{~m}$
$\frac{7}{8} \mathrm{~m}$
Solution
The maximum displacement of the wave will occur at antinode of the wave. The first antinode will be a point where, $\mathrm{d}=\frac{\lambda}{4}$
The wavelength is given by,
$\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{300}{600}$
So, $d=\frac{0.5}{4}=\frac{1}{8} m$