Is the Amplitude of Sound Waves Constant with Distance?

In summary, the displacement amplitude is constant, but the frequency and intensity decrease with distance.
  • #1
Nathanael
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A quote from a book:
Fundamentals of Physics said:
A sound wave causes a longitudinal displacement s of a mass element in a medium as given by [itex]s=s_mcos(kx-\omega t)[/itex] where [itex]s_m[/itex] is the displacement amplitude (maximum displacement) from equilibrium, [itex]k=\frac{2\pi}{\lambda}[/itex], and [itex]\omega=2\pi f[/itex] [itex][/itex], [itex]\lambda[/itex] and [itex]f[/itex] being the wavelength and frequency, respectively, of the sound wave.
At first, I was under the impression that [itex]s_m[/itex] was constant. But shouldn't [itex]s_m[/itex] be inversely proportional to the distance?

The book never says this anywhere directly, but it seems to be implied by two other equations (for the intensity) given in the chapter:
$$I=\frac{P_{source}}{4\pi R^2}=\frac{1}{2}\rho v\omega ^2s_m^2$$
Wouldn't this mean that either the frequency or the amplitude decreases with distance? But the frequency can't change without a corresponding change in wavelength, so doesn't this mean the displacement-amplitude (and therefore also the pressure-amplitude) decreases with the distance?

I just want to make sure, because all the writing in the chapter seemed to imply to me that the amplitude was constant, yet these two equations for the intensity seem to say otherwise.
 
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  • #2
The wave provided is a plane wave solution, which does not decrease with distance traveled unless damped. It only describes a wave generated at a point source far away from the source (much further than the wavelength and the distance over which you are considering the wave).

Edit: Of course, if you solve the wave equation for a spherical wave, you will end up with a decreasing amplitude.
 
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  • #3
Thank you very much.
 
  • #4
One more question; When you say that equation describes a planar wave, is this just because the amplitude is constant?

In other words, would the equation that describes a spherical wave be the same except with a variable amplitude?
(Something like [itex]\frac{c}{R}cos(kR-\omega t)[/itex]?)
 
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  • #5
Just as an exercise, you can do the following: Assume ##s(r,t) = f(r) \exp(i(kr - \omega t))## and insert into the wave equation (for r>0). The radial part of the Laplace operator is ##\partial_r^2 + (2/r) \partial_r##. Insert this into the wave equation and check that it solves it (again, for r > 0, you would need a source in r = 0 to keep it going).
 
  • #6
Sorry, that exercise is beyond me; I know nothing of Laplace operators. Thanks anyway.
 

What is amplitude in sound waves?

Amplitude refers to the maximum displacement or distance that a sound wave travels from its resting position. It is a measure of the strength or intensity of the sound wave.

How does amplitude affect the loudness of a sound?

The higher the amplitude, the louder the sound will be. This is because a larger amplitude means a stronger vibration, which results in a louder sound.

What is the unit of measurement for amplitude in sound waves?

Amplitude is measured in decibels (dB). The higher the number of decibels, the greater the amplitude and the louder the sound.

How does amplitude relate to the pitch of a sound?

Amplitude does not directly affect the pitch of a sound. Pitch is determined by the frequency of the sound wave, which is the number of vibrations per second. However, a sound with a higher amplitude may be perceived as having a higher pitch due to its stronger and more intense vibrations.

How can amplitude be changed or manipulated in sound waves?

Amplitude can be changed by adjusting the strength or energy of the source of the sound wave, or by manipulating the medium through which the sound wave travels. In digital sound, amplitude can also be altered through digital signal processing techniques.

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