Intensity Of Sound Waves

Sound waves are an integral part of our daily lives, from the subtle rustling of leaves to the booming sound of a thunderstorm. The intensity of these sound waves plays a crucial role in how we perceive the world around us. Intensity, defined as the amount of energy a sound wave carries per unit area, determines the loudness of the sound we hear. In practical terms, it's the difference between a whisper and a shout. For instance, the gentle hum of a refrigerator operates at a lower intensity, while the blare of a siren in the streets is at a much higher intensity. In this article, we will understand sound wave intensity is not only essential in physics but also helps in various real-life applications like designing soundproof rooms, creating better audio equipment, and even protecting our hearing.

Intensity of Sound Waves

The intensity of sound waves refers to the amount of energy that a sound wave carries through a unit area perpendicular to the direction of the wave's propagation. It is a critical factor that determines how loud a sound is perceived by the human ear. Measured in watts per square meter (W/m²), sound intensity depends on both the amplitude of the sound wave and the distance from the source.

The Intensity of Periodic Sound Waves

The intensity I of a wave is defined as the power per unit area, as the rate at which the energy transported by the wave transfers through a unit area A perpendicular to the direction of travel of the wave.

$I=\frac{P}{A}$

In this case, the intensity is therefore $I=\frac{1}{2} \rho v(\omega A)^2$

Also, for any sound waves

$\begin{aligned} \Delta P_m & =A B k \\ A & =\frac{\Delta P_m}{B k}\end{aligned}$

Put this value in the equation of intensity

$
I=\frac{1}{2} \rho v \omega^2\left(\frac{\Delta P_m}{B k}\right)^2=\frac{1}{2} \rho v \omega^2 \frac{\Delta P_m^2}{B^2 k^2}
$

As $k=\omega / v$ and $B=v^2 \rho$
$
\therefore I=\frac{1}{2} \rho v \omega^2 \frac{\Delta P_m^2}{B^2 \frac{\omega^2}{v^2}}=\frac{v \Delta P_m^2}{2 B}=\frac{\Delta P_m^2}{2 \rho v}
$

Let us consider a source that emits sound equally in all directions, the result is a spherical wave. The figure given below shows these spherical waves as a series of circular arcs concentric with the source. Each circular arc represents a surface over which the phase of the wave is constant. We call such a surface of constant phase a wavefront. The distance between adjacent wavefronts that have the same phase is called the wavelength of the wave. The radial lines pointing outward from the source are called rays.

From the above figure, we can deduce that the $I=\frac{p_{\mathrm{avg}}}{A}=\frac{p_{\mathrm{avg}}}{4 \pi r^2}$

So, from this equation, we can say that it varies inversely with the square of the distance.

Now, the appearance of sound to the human ear is characterized by

a. Pitch

b. Loudness

c. quality

Pitch

The pitch of a sound is an attribute of the sound that tells us about its frequency. A sound that is at a high pitch, has a high frequency. A sound at a low pitch has a lower frequency.

Loudness

The loudness that a person senses is related to the intensity of sound though it is not directly proportional to it. Loudness can be defined and represented as

$\beta=10 \log _{10}\left(\frac{I}{I_o}\right)$

Where I = Intensity of the sound
$\mathrm{I}_{\mathrm{O}}=$ Reference intensity $\left(10^{-12} \mathrm{~W}-\mathrm{m}^{-2}\right)$

For $\mathrm{I}=\mathrm{I}_{\mathbf{0}}$, the sound level $\beta=0$

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Solved Examples Based on Intensity of Sound Waves

Example 1: A sound source emits sound waves in a uniform medium. If the energy density is E and the maximum speed of the particles of the medium is $V_{\max }$ is best represented by

1)

2)

3)

4)

Solution:

$\begin{aligned} & \text { Energy density }(\mathrm{E})=2 \pi^2 \rho \gamma^2 \rho \mathrm{n}^2 \mathrm{~A}^2 \\ & \quad V_{\max }=\omega A=2 \pi n A \\ & \Rightarrow \quad V_{\max }=2 \pi n A \\ & E \alpha\left(V_{\max }\right)^2\end{aligned}$

Hence, the answer is the option (2).

Example 2: A source (stationary) emits sound in a medium (non-absorbing). Two points A and B, are at distances 4 m & 9 m respectively from the source. The ratio of Amplitudes of waves at A and B is

1) $\frac{3}{2}$

2) $\frac{4}{9}$

3) $\frac{2}{3}$

4) $\frac{9}{4}$

Solution:

Intensity of periodic sound waves

$\begin{aligned} & I=\frac{\Delta P_m^2}{2 \rho v} \\ & \text { wherein } \\ & \Delta P_m=\text { Maximum pressure change } \\ & \rho=\text { density } \\ & v=\text { velocity of sound } \\ & I=\frac{p}{4 \pi r^2} \text { since } \mathrm{P}=\text { constant } \\ & I \alpha \frac{1}{r^2} \text { and } I \alpha A^2 \\ & \frac{A_1}{A_2}=\sqrt{\frac{I_1}{I_2}}=\frac{9}{4}\end{aligned}$

Hence, the answer is the option (4).

Example 3: A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

1) 100

2) 10000

3) 10

4) 1000

Solution:

Sound level in decibels

$
\begin{aligned}
& \beta=10 \log _{10}\left(I / I_0\right), \\
& \text { for } I=I_0 \\
& \beta=0
\end{aligned}
$
wherein
Where $I$ is the intensity of the sound and $I_0$ is a constant reference intensity $=10^{-12} \mathrm{~W} / \mathrm{m}$
$
B_1=10 \log \left[\frac{I}{I_o}\right] ; B_2=10 \log \left[\frac{I^{\prime}}{I_o}\right]
$

Given $B_2-B_1=20$
$
20=10 \log \left[\frac{I^{\prime}}{I}\right] \Rightarrow I^{\prime}=100 I
$

Hence, the answer is the option (1).

Example 4: A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of:

1) 100

2) 1000

3) 10000

4) 10

Solution:

$\begin{aligned} & L_1=10 \log \left(\frac{I_1}{I_0}\right), L_2=10 \log \left(\frac{I_2}{I_0}\right) \\ & \therefore L_1-L_2=10 \log \left(\frac{I_1}{I_0}\right)-10 \log \left(\frac{I_2}{I_0}\right) \\ & \text { or } \Delta L=10 \log \left(\frac{I_1}{I_2}\right) \\ & \text { or } 20 d B=10 \log \left(\frac{I_1}{I_2}\right) \\ & \text { or } 10^2=\frac{I_1}{I_2} \text { or } I_2=\frac{I_1}{100}\end{aligned}$

Where L₁ and L₂ are loudness and I₁ and I₂ are Intensities.

Hence, the answer is the option (1).

Example 5: If the frequency of the sound produced by a siren increases from 400 Hz to 1200 Hz. while the wave amplitude remains constant, the ratio of the intensity of the 1200 Hz to that of the $400 \mathrm{H}_2$ wave will be:

1) $1: 1$
2) $1: 3$
3) $3: 1$
4) $9: 1$

Solution:

$\begin{aligned} & \text { Use, } I=\frac{1}{2} \rho \omega^2 A^2 \gamma \text { or } I \alpha \omega^2 \text { or } I \alpha f^2 \\ & \frac{I_2}{I_1}=\left(\frac{f_2}{f_1}\right)^2=\left(\frac{1200}{400}\right)^2 \\ & \frac{I_2}{I_1}=\left(\frac{12}{4}\right)^2=(3)^2 \\ & \frac{I_2}{I_1}=\frac{9}{1} \\ & I_2=I_1=9: 1\end{aligned}$

Hence, the answer is the option (4).

Summary

The intensity of sound waves is a measure of the energy a sound wave carries per unit area and is crucial in determining how loud a sound is perceived. It is influenced by factors such as amplitude, frequency, and distance from the source. Understanding sound intensity has practical applications in acoustics, hearing protection, and audio engineering. The discussed examples illustrate how sound intensity relates to real-world scenarios, emphasizing the importance of this concept in various fields.