During Ganesh Chaturthi a pandal plays music through a loudspeaker. Near the stage you must shout to be heard. Across the road the music is still loud. Two streets away it is only a murmur. The loudspeaker sends out the same power all the time. So why does the sound fade so quickly with distance?
A point source
Lesson 2 defined intensity as power per unit area. For a small source that emits sound uniformly, the power spreads over the surface of a sphere, of area \(A = 4\pi r^2\). This leads to the inverse-square law: intensity decreases with the square of the distance.
\[ I \propto \frac{1}{r^2} \]
Consider a point source emitting sound waves uniformly in all directions. A real source is never a point. But it acts as one when it is small compared with the distance to the listener. The surfaces on which the wave has the same phase are called wavefronts. Around a point source they are spheres. The figure below shows how the sound spreads outward.
As sound waves expand outward from the source, the same total power must pass through increasingly larger spherical surfaces. Since the area of a sphere grows as \(4\pi r^2\), the intensity, the power per unit area, must decrease as \(1/r^2\).
\[ P = I \times 4\pi r^2 \]
Why. no energy is lost on the way, so all the power \(P\) of the source crosses the sphere of radius \(r\). At every point of the sphere the intensity is the same, \(I\), because the source is uniform.
\[ I = \dfrac{P}{4\pi r^2} \]
Why. divide by the area of the sphere, \(4\pi r^2\).
\[ \dfrac{I_2}{I_1} = \dfrac{r_1^2}{r_2^2} \]
Read it. \(P\) and \(4\pi\) are the same for both spheres, so they cancel from the ratio of the intensities at \(r_1\) and \(r_2\).
Play the figure and watch the patch between the two dashed rays. It carries a fixed share of the power, because no energy leaves the cone. As the wavefront moves from \(r\) to \(2r\), both sides of the patch double. Seen face on, the one square becomes four. The same power on four times the area gives a quarter of the intensity, as the readout shows.
The intensity of sound from a point source is inversely proportional to the square of the distance from the source: \(I = P/(4\pi r^2)\). The law holds for a source that emits uniformly, in a medium that absorbs no energy.
You stand at a distance \(r\) from a small loudspeaker in an open field. You then walk back to a distance \(2r\). What happens to the intensity of the sound you receive?
Using the law
The formula links three quantities: the power of the source, the distance and the intensity. Any two give the third. The next example asks how close to a small source the sound becomes painful. An intensity of \(1.0\,\text{W/m}^2\) is the threshold of pain, as lesson 5 will show.
A point source emits sound uniformly with a power output of \(1.0\,\text{W}\). At what distance from the source is the intensity equal to the threshold of pain, \(I = 1.0\,\text{W/m}^2\)?
For a point source, the intensity at distance \(r\) is \(I = \text{Power}/\text{Area} = P/(4\pi r^2)\).
Set \(I = 1.0\,\text{W/m}^2\) and \(P = 1.0\,\text{W}\), and solve for \(r\): \(1.0 = 1.0/(4\pi r^2)\), so \(4\pi r^2 = 1.0\).
Then \(r^2 = 1.0/(4\pi)\) and \(r = \sqrt{1.0/(4\pi)}\).
Calculate: \(4\pi \approx 12.57\), so \(r^2 = 1.0/12.57 \approx 0.0796\,\text{m}^2\) and \(r \approx 0.282\,\text{m}\).
Final answer: \(0.28\,\text{m}\), rounded to two decimal places.
A source of only one watt is painful within \(28\,\text{cm}\), a little more than a hand's span. Farther away, the sound falls quickly. At ten times the distance, \(2.8\,\text{m}\), the intensity is a hundred times smaller.
A small loudspeaker at a pandal sends out \(2.0\,\text{W}\) of sound uniformly in all directions. What is the intensity \(10\,\text{m}\) away, in mW/m²?
The inverse-square law assumes that every joule of the source reaches each sphere. Outdoors, the ground reflects some sound and the air absorbs a little, most at high frequencies. Indoors, the walls reflect the sound back. So the law holds best in open air, not too far from the source. It is a model, and a very useful one.
Why does the same power cross a sphere of radius \(10\,\text{m}\) and a sphere of radius \(20\,\text{m}\) around a point source?
How far from the \(1.0\,\text{W}\) source of the example does its intensity fall to \(1.0 \times 10^{-6}\,\text{W/m}^2\), the intensity of normal conversation? Give the distance in metres, assuming the open air absorbs nothing.
A line source
Not every source is small. A busy highway, a long goods train or a row of loudspeakers stretches far in one direction. A long, straight source of this kind is a line source. Its wavefronts are not spheres. They are cylinders, with the source along their axis.
Take a length \(L\) of the source, which emits a power \(P\). Its sound spreads out sideways, over the curved surface of a cylinder of radius \(r\) and length \(L\). The area of that surface is \(2\pi rL\). So the intensity at a distance \(r\) is the power divided by this area.
The area of a cylinder grows only in proportion to its radius. Its length stays \(L\) as the radius grows. So the intensity of a line source falls only as \(1/r\), not \(1/r^2\). Doubling the distance from a highway halves the intensity. Doubling the distance from a single loudspeaker quarters it.
A long goods train passes on a straight track. At \(10\,\text{m}\) from the track its sound has an intensity of \(1.0 \times 10^{-3}\,\text{W/m}^2\). Treating the train as a line source, what is the intensity \(40\,\text{m}\) from the track, in mW/m²?
You double your distance first from a small loudspeaker, and then from a long, busy highway. How does the intensity change in each case?
A long railway platform has many small loudspeakers in a row, a few metres apart. A single loudspeaker at one end would be deafening nearby and too faint at the far end. The row acts as a line source instead. Across the platform its intensity falls only as \(1/r\), and along the platform it stays nearly the same. So every passenger hears the announcement at about the same level.
Now return to the pandal. Its loudspeaker is small, so it acts as a point source. Walk from \(5\,\text{m}\) to \(50\,\text{m}\) away, and the intensity falls by \(10^2 = 100\). At \(100\,\text{m}\), two streets away, it has fallen by \(20^2 = 400\). The houses in between also block and absorb part of the sound.
- Taking the intensity of a point source to fall as \(1/r\).The power spreads over a sphere of area \(4\pi r^2\), so \(I \propto 1/r^2\).
- Using \(4\pi r^2\) for a line source.A line source spreads over a cylinder of area \(2\pi rL\), so \(I \propto 1/r\).
- Confusing power and intensity.The power of the source is the same at every distance. The intensity, power per unit area, falls with distance.
- Forgetting to square the distance.In \(I = P/(4\pi r^2)\), square \(r\) before multiplying by \(4\pi\).