Sound waves of frequency 660 , Hz fall normal | vedclass

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(A) Given frequency of sound waves $
u = 660 \, Hz$ and velocity $v = 330 \, m/s$.The wavelength of the sound waves is calculated as $\lambda = \frac

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(A) Given frequency of sound waves $
u = 660 \, Hz$ and velocity $v = 330 \, m/s$.The wavelength of the sound waves is calculated as $\lambda = \frac{v}{
u} = \frac{330}{660} = 0.5 \, m$.When sound waves reflect from a perfectly reflecting wall,stationary waves are formed due to the interference between the incident and reflected waves.$A$ displacement node is always formed at the reflecting surface (the wall).The air particles have maximum amplitude of vibration at the antinodes.The shortest distance from the wall (node) to the first point of maximum amplitude (antinode) is given by $d = \frac{\lambda}{4}$.Substituting the value of $\lambda$,we get $d = \frac{0.5}{4} = 0.125 \, m$.

Sound waves of frequency $660 \, Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is .... $m$ (velocity of sound in air is $330 \, m/s$)

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