- #1
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A quote from a book:
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.
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?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.
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.