Assume that the displacement (s) of air is proportional to the pressure difference

Question:

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 $\Delta p \sim 10 \mathrm{~Pa}$, $v \sim 300 \mathrm{~m} / \mathrm{s}, p \sim q \mathrm{~kg} / \mathrm{m}^{3}$ and $f \sim 1000 \mathrm{~Hz}$, then $s$ will be of the order of (take the multiplicative constant to be 1 )

1. $\frac{3}{100} \mathrm{~mm}$

2. $10 \mathrm{~mm}$

3. $\frac{1}{10} \mathrm{~mm}$

4. $1 \mathrm{~mm}$

Correct Option: 1

Solution:

(1) As we know,

Pressure amplitude, $\Delta P_{0}=a K B=S_{0} K B=S_{0} \times \frac{\omega}{V} \times \rho V^{2}$

$\left[\because K=\frac{\omega}{V}, V=\sqrt{\frac{B}{\rho}}\right]$

$\Rightarrow S_{0}=\frac{\Delta P_{0}}{\rho V \omega} \approx \frac{10}{1 \times 300 \times 1000} \mathrm{~m}=\frac{1}{30} \mathrm{~mm} \approx \frac{3}{100} \mathrm{~mm}$