Assume that the displacement(s) of air is proportional to the pressure difference ($\Delta$p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air ($\rho$) and the frequency (f). If $\Delta$p ~ 10 Pa, v ~ 300 m/s, $\rho$ ~ 1 kg/m3 and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1) :
Solution
Given, S $\propto$ $\Delta$p
<br><br>and Proportionally constant = 1
<br><br>We know,
<br><br>$\Delta$p = S$\beta$k
<br><br>= $\rho$v<sup>2</sup> $\times$ ${\omega \over v} \times S$
<br><br>= ${\rho v\omega S}$
<br><br>$\therefore$ S = ${{\Delta p} \over {\rho v\omega }}$
<br><br>= ${{\Delta p} \over {\rho v2\pi f}}$
<br><br>= ${{\Delta p} \over {\rho vf}}$
<br><br>[As Proportionally constant = 1 so assume 2$\pi$ = 1]
<br><br>$= {{10} \over {1 \times 300 \times 1000}}$<br><br>$= {1 \over {30}}mm$<br><br>$\approx {3 \over {100}}mm$
About this question
Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types
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