Answer
The period is $$ T=2.0 \mathrm{ms} \text { (or } 0.0020 \mathrm{s})$$
and the amplitude is $\Delta p_{m}=8.0 \mathrm{mPa}$ (which is equivalent to $ 0.0080 \mathrm{N} / \mathrm{m}^{2}) $
From Eq. $17-15$ we get
$$
s_{m}=\frac{\Delta p_{m}}{v \rho \omega}=\frac{\Delta p_{m}}{v \rho(2 \pi / T)}=6.1 \times 10^{-9} \mathrm{m}
$$
where $$\rho=1.21 \mathrm{kg} / \mathrm{m}^{3} \quad \quad \text{ and } \quad \quad v=343 \mathrm{m} / \mathrm{s}$$
Work Step by Step
The period is $$ T=2.0 \mathrm{ms} \text { (or } 0.0020 \mathrm{s})$$
and the amplitude is $\Delta p_{m}=8.0 \mathrm{mPa}$ (which is equivalent to $ 0.0080 \mathrm{N} / \mathrm{m}^{2}) $
From Eq. $17-15$ we get
$$
s_{m}=\frac{\Delta p_{m}}{v \rho \omega}=\frac{\Delta p_{m}}{v \rho(2 \pi / T)}=6.1 \times 10^{-9} \mathrm{m}
$$
where $$\rho=1.21 \mathrm{kg} / \mathrm{m}^{3} \quad \quad \text{ and } \quad \quad v=343 \mathrm{m} / \mathrm{s}$$
