Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

(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 $20 ° C$ . (See Table 14-1.)
(b) If the pressure amplitudes are equal instead, what is the ratio of the intensities of the waves?

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 $59.7$ .
The ratio of the intensities of the waves if the pressure amplitudes are equal is $2.81 \times 10^{- 4}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The density of fresh water, $ρ_{w} = 1 \times 10^{3} kg / m^{3}$
The density of air, $ρ_{a} = 1.21 kg / m^{3}$
The speed of the sound wave in air, $v_{a} = 343 m / s$
The speed of the sound wave in water, $v_{w} = 1482 m / s$

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,

$I = \frac{1}{2} {ρvω}^{2} s_{m}^{2}$ …(i)

The displacement amplitude of the wave,

role="math" localid="1661508014075" $S_{w} = \sqrt{\frac{2 I}{{ρvω}^{2}}}$ …(ii)

The pressure amplitude of the wave,

$Δp = ρvω \times s_{w}$ …(iii)

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

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

$ΔP = vρω \sqrt{\frac{2 I}{{ρvω}^{2}}}$

…(a)

$ΔP = \sqrt{ρv 2 I}$

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

$\frac{{ΔP}_{w}}{{ΔP}_{a}} = \frac{\sqrt{ρ_{w} v_{w} 2 I_{w}}}{\sqrt{ρ_{a} v_{a} 2 I_{a}}}$

As $I_{a} = I_{w}$

$\begin{matrix} \frac{{ΔP}_{w}}{{ΔP}_{a}} = \sqrt{\frac{ρ_{w} v_{w}}{ρ_{a} v_{a}}} \\ = \sqrt{\frac{(1 \times 10^{3} kg / m^{3}) (1482 m / s)}{(1.21 kg / m^{3}) (343 m / s)}} \\ = 59.7 \end{matrix}$

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:

$I = \frac{{ΔP}^{2}}{ρv}$ …(b)

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

$\begin{matrix} \frac{I_{w}}{I_{a}} = \frac{\frac{{ΔP}_{w}^{2}}{ρ_{w} v_{w}}}{\frac{{ΔP}_{a}^{2}}{ρ_{a} v_{a}}} \\ = \frac{{ΔP}_{w}^{2}}{{ΔP}_{a}^{2}} \frac{ρ_{a} v_{a}}{ρ_{w} v_{w}} \end{matrix}$

As ${ΔP}_{w} = {ΔP}_{a}$

$\begin{matrix} \frac{I_{w}}{I_{a}} = \frac{(1.21 kg / m^{3}) (343 m / s)}{(1 \times 10^{3} kg / m^{3}) (1482 m / s)} \\ = 2.81 \times 10^{- 4} \end{matrix}$

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of intensity

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

Step 4: b) Calculation of ratio of intensities

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Fundamentals Of Physics

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