# Intensity of Sound Waves

This topic is part of the HSC Physics course under the section *Sound Waves.*

### HSC Physics Syllabus

**investigate quantitatively the relationship between distance and intensity of sound**

### What is Intensity?

**Intensity** is a measure of the amount of energy a wave carries per unit time across a unit area. For sound waves, intensity can be thought of as the "power" of the sound as it travels through space.

When we discuss the intensity of sound in terms of power, we refer to the amount of energy the sound wave transfers per unit of time. Specifically, the intensity `I` of a sound wave can be defined as the power `P` of the sound wave divided by the area `A` over which it spreads:

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

Where:

- `I` is the intensity (typically measured in watts per square meter, $W/m2$).
- `P` is the power of the sound source (measured in watts, $W$).
- `A` is the area over which the sound energy spreads (measured in square metres, $m2$).

Intensity can also be expressed in terms of energy, it's the amount of energy that a sound wave carries to a given area in a specific time .

This can be expressed as:

$$I = \frac{E}{A \times t}$$

Where:

- `E` is the energy of the sound wave (typically measured in joules, $J$).
- `t` is the time over which this energy is measured (seconds, $s$).
- `A` is the area over which the energy is spread (measured in square meters, m
^{2}).

### What is a Decibel? (extension)

The intensity of sound is often expressed as decibels, which is a unit that measures sound intensity logarithmically. Decibels are used because the intensity of sound waves can vary over a great range.

$$L = 10 \times \log_{10}\left(\frac{I}{I_0}\right)$$

Where:

- `L` is the sound intensity level in decibels (dB).
- `I` is the intensity of the sound wave (usually in W/m
^{2}). - `I_0` is the reference intensity, typically taken as the quietest sound that the average human ear can hear ($`I_0`=1_{Wm2}$).

For example, a sound wave with an intensity of 10^{-5} W/m^{2} is expressed as

$$L = 10 \times \log_{10}\left(\frac{10^{-5}}{10^{-12}}\right)$$

$$L = 70 \, \text{dB}$$

Since decibels are calculated on a logarithmic scale, an intensity of 70 dB is 10 times greater than 60 dB; an intensity of 80 dB is 100 times greater than 60 dB.

### The Relationship Between Distance and Intensity of Sound

Sound travels through a medium, typically air, in the form of waves. As these waves move outward from a sound source, they spread over a larger area. This spreading causes the energy of the sound waves to be distributed over increasing areas, leading to a decrease in the intensity of the sound.

The relationship between the distance from the sound source and the intensity of the sound is described by the inverse square law.

### The Inverse Square Law for Sound

The inverse square law states that the intensity of a sound is inversely proportional to the square of the distance `r` from the source. In other words, as a sound wave moves further from its source, its intensity decreases.

Mathematically, this relationship can be expressed as:

$$I \propto \frac{1}{r^2}$$

For the same sound wave at two different distances from the source, the following is true:

$$I_1r_1^2 = I_2r_2^2$$

The inverse square law for intensity means that if you were to double the distance from the sound source (`r`), the intensity `I` would reduce to a quarter of its original value.

### Example

The intensity of sound detected at a distance of 6.0 m from a loudspeaker is 72 J s^{–1} m^{–2}.

What is the intensity of same sound wave measured by a person standing 10.0 away from the loudspeaker?

*Solution:*

$$I_1r_1^2 = I_2r_2^2$$

$$(72)(6)^2 = I_2(10)^2$$

$$I_2 = 26 \hspace{1 mm} J \hspace{1 mm} s^{-1} \hspace{1mm} m^{-2}$$