• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q81P

Expert-verified Found in: Page 511 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # (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 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 $59.7$ .
2. The ratio of the intensities of the waves if the pressure amplitudes are equal is $2.81×{10}^{-4}$ .
See the step by step solution

## Step 1: The given data

1. The density of fresh water, ${\mathrm{\rho }}_{\mathrm{w}}=1×{10}^{3}\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$
2. The density of air, ${\mathrm{\rho }}_{\mathrm{a}}=1.21\text{\hspace{0.17em}}\text{kg}/{\text{m}}^{\text{3}}$
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, ${\mathrm{v}}_{\mathrm{w}}=1482\text{\hspace{0.17em}m}/\text{s}$

## Step 2: Understanding the concept of intensity

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,

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

## Step 3: a) Calculation of ratio of the pressure amplitudes

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

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

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 $59.7$

## Step 4: b) Calculation of ratio of intensities

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

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

Therefore, the ratio of the intensities of the waves if the pressure amplitudes are equal is $2.81×{10}^{-4}$ . ### Want to see more solutions like these? 