Sound waves of frequency 600 ,Hz fall normall | vedclass

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Question

(B) When sound waves fall normally on a perfectly reflecting wall, a stationary wave is formed due to the superposition of incident and reflected wave

Vedclass · Accepted Answer

$1/8 \,m$

Vedclass · Answer

(B) When sound waves fall normally on a perfectly reflecting wall, a stationary wave is formed due to the superposition of incident and reflected waves.At a perfectly reflecting wall, a node is formed because the wall is a rigid boundary.The nodes are points of zero displacement, while antinodes are points of maximum displacement (maximum amplitude).The distance of the first antinode from the wall (node) is given by $d = \lambda/4$.First, calculate the wavelength $\lambda$ using the formula $\lambda = v/f$, where $v = 300 \,ms^{-1}$ and $f = 600 \,Hz$.$\lambda = 300/600 = 0.5 \,m$.Now, calculate the distance $d = \lambda/4 = 0.5/4 = 0.125 \,m = 1/8 \,m$.Thus, the shortest distance from the wall where particles have maximum amplitude is $1/8 \,m$.

Sound waves of frequency $600 \,Hz$ 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 $= 300 \,ms^{-1}$)

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