(a) If two sound waves, one in air and one in (fresh) water, are equal in intensity and angular frequency, what is the ratio of the pressure amplitude of the wave in water to that of the wave in air? Assume the water and the air are at $\mathbf{20}\mathbf{°}\mathbf{C}$. (See Table 14-1.)

(b) If the pressure amplitudes are equal instead, what is the ratio of the intensities of the waves?

1. The density of fresh water,${\mathrm{\rho }}_{\mathrm{w}}=1\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$
2. The density of air,${\mathrm{\rho }}_{\mathrm{a}}=1.21\text{kg}/{\text{m}}^{\text{3}}$
3. The speed of the sound wave in air,${\mathrm{v}}_{\mathrm{a}}=343\text{\hspace{0.17em}m}/\text{s}$
4. The speed of the sound wave in water,${\mathrm{v}}_{\mathrm{w}}=1482\text{\hspace{0.17em}m}/\text{s}$

We can use the equation of intensity of the wave in the relation between displacement amplitude and pressure amplitude to find the ratio of the pressure amplitude of the wave in water to that of the wave in the air if two sound waves are equal in intensity and angular frequency.

Formula:

The intensity of the wave,

$\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho v\omega }}^{\mathbf{2}}{\mathbf{s}}_{\mathbf{m}}^{\mathbf{2}}$ …(i)

The displacement amplitude of the wave,

role="math" localid="1661508014075" ${\mathbf{S}}_{\mathbf{w}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{I}}{{\mathbf{\rho v\omega }}^{\mathbf{2}}}}$ …(ii)

The pressure amplitude of the wave,

$\mathbf{\Delta p}\mathbf{=}\mathbf{\rho v\omega }\mathbf{\times }{\mathbf{s}}_{\mathbf{w}}$ …(iii)

Using equation (ii) in equation (iii), we get the pressure amplitude as:

$\mathrm{\Delta P}=\mathrm{v\rho \omega }\sqrt{\frac{2\mathrm{I}}{{\mathrm{\rho v\omega }}^2}}$

…(a)

$\mathrm{\Delta P}=\sqrt{\mathrm{\rho v}2\mathrm{I}}$

So, the ratio of the pressure amplitudes of water to air using equation (a) is given as:

$\frac{{\mathrm{\Delta P}}_{\mathrm{w}}}{{\mathrm{\Delta P}}_{\mathrm{a}}}=\frac{\sqrt{{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_{\mathrm{w}}2{\mathrm{I}}_{\mathrm{w}}}}{\sqrt{{\mathrm{\rho }}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}2{\mathrm{I}}_{\mathrm{a}}}}$

As${\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{w}}$

$\begin{array}{c}\frac{{\mathrm{\Delta P}}_{\mathrm{w}}}{{\mathrm{\Delta P}}_{\mathrm{a}}}=\sqrt{\frac{{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_{\mathrm{w}}}{{\mathrm{\rho }}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}}}\\ =\sqrt{\frac{(1\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}})(1482\text{\hspace{0.17em}m}/\text{s})}{(1.21\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}})(343\text{\hspace{0.17em}m}/\text{s})}}\\ =59.7\end{array}$

Therefore, the ratio of the pressure amplitude of the wave in water to that of the wave in air if two sound waves are equal in intensity and angular frequency is $59.7$

From equation (a), we get the intensity of the wave as:

$\mathrm{I}=\frac{{\mathrm{\Delta P}}^2}{\mathrm{\rho v}}$ …(b)

So, the ratio of the intensities of water to air using equation (b) is given as:

$\begin{array}{c}\frac{{\mathrm{I}}_{\mathrm{w}}}{{\mathrm{I}}_{\mathrm{a}}}=\frac{\frac{{\mathrm{\Delta P}}_{\mathrm{w}}^2}{{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_{\mathrm{w}}}}{\frac{{\mathrm{\Delta P}}_{\mathrm{a}}^2}{{\mathrm{\rho }}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}}}\\ =\frac{{\mathrm{\Delta P}}_{\mathrm{w}}^2}{{\mathrm{\Delta P}}_{\mathrm{a}}^2}\frac{{\mathrm{\rho }}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}}{{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_{\mathrm{w}}}\end{array}$

As${\mathrm{\Delta P}}_{\mathrm{w}}={\mathrm{\Delta P}}_{\mathrm{a}}$

$\begin{array}{c}\frac{{\mathrm{I}}_{\mathrm{w}}}{{\mathrm{I}}_{\mathrm{a}}}=\frac{(1.21\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}})(343\text{\hspace{0.17em}m}/\text{s})}{(1\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}})(1482\text{\hspace{0.17em}m}/\text{s})}\\ =2.81\times {10}^{-4}\end{array}$

Therefore, the ratio of the intensities of the waves if the pressure amplitudes are equal is $2.81\times {10}^{-4}$.