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My text says that in a plane sound wave (or in the far field), particle velocity and pressure is in phase. As we move closer to the sound source (to near field and more spherical waves), the phase angle between these two quantities will gradually shift towards 90˚.

Why does this phase shift occur? I haven't been able to find an intuitive explanation for why there is a phase shift in spherical waves, but not in plane waves, and vice versa.

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How did you actually derive sound waves in your text? If you want intuitive thinking solution, it is necessary to know what you already know. –  Pygmalion Apr 20 '12 at 16:59
    
en.wikipedia.org/wiki/Acoustic_wave give a thorough explanation. the issue is traveling waves vs standing waves. –  user27777 Aug 16 '13 at 22:04
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3 Answers 3

I generally agree with John Rennie's nice derivation (apart for the few details I'm afraid to ask not to start a flame ;). However, if I had to tell it engineering students, I'd started with more specific formulas for mechanical waves they probably already know:

$$\Delta p = - B \frac{\partial s}{\partial x}$$

$$v = \frac{\partial s}{\partial t}$$

where $B$ is bulk modulus of material, $s$ is displacement and $\frac{\partial}{\partial x}$ is derivation in the direction of the wave.

It is pretty obvious that displacement function for plane wave must be multiplied by $1/r$ for spherical wave in order to conserve energy, which falls as $1/r^2$ and is proportional to square of displacement, so

$$s(r,t) = \frac{\sin(kr - \omega t)}{r}.$$

(Exact derivation is not very difficult and can be seen at http://scienceworld.wolfram.com/physics/SphericalWave.html)

If you make calculations above you end up with one term for velocity and two terms for pressure variation:

$$v(r,t) = -\omega \frac{\cos(kr - \omega t)}{r}$$

$$\Delta p(r,t) = -B k \frac{\cos(kr - \omega t)}{r} + B \frac{\sin(kr - \omega t)}{r^2}.$$

At larger distances, first term is predominant, and velocity and pressure variation are in phase. However, the second term becomes predominant at smaller distances $kr \le 1$, so the velocity and pressure variation come out of phase.

It is interesting to note that at critical distance $kr = 1$ pressure variation is always zero! It is really curious thing, because looking at compressions and rarefactions it would seem that all start at that distance, as if surface of the sphere $r = 1/k$ would actually be source of pressure variations. Also, within this sphere, displacement and pressure variation are in phase, which means that we have standing wave within the sphere in terms of energy.

I think this all comes to one thing: Sphere of dimension $r = 1/k = \lambda/2\pi$ is the smallest dimension of aperture you need that you actually create waves of wavelength $\lambda$. You need some moving part that creates compressions and rarefactions and within dimensionality of those movements you cannot have classical traveling wave. It seems that solution for $r \le 1/k$ is not physical considering traveling wave as such.

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Actually I must admit I think your answer makes the physics clearer. I think the reason we get an out of phase contribution to the pressure is due to the curvature of the wavefront, though I need to sit down and think about this to be able to phrase it in a way that makes sense. –  John Rennie Apr 20 '12 at 17:38
    
Thanks for your opinion. Also, I am curious what your explanation would be. If possible, please explain what exactly means "curvature of the wavefront" too, as I don't understand it. –  Pygmalion Apr 21 '12 at 12:53
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Remembering back to my undergrad days, and courtesy of a quick Google: if you define the acoustic potential $\phi$ by:

$$u = - \nabla \phi$$

then the pressure is:

$$p = \rho \frac{\partial\phi}{\partial t}$$

For a plane wave in one dimension the wave equation is:

$$\frac{\partial^2\phi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2\phi}{\partial t^2}$$

and the solution is:

$$\phi = f(ct \pm x)$$

for some function $f$. So $u$ and $p$ are in phase. However for a spherical wave the wave equation looks rather different:

$$\frac{\partial^2 (r\phi)}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 (r\phi)}{\partial t^2}$$

and you get:

$$\phi = \frac {f(ct \mp x)}{r}$$

where if I recall correctly the minus sign means a wave travelling outwards and the plus means a wave travelling inwards. So:

$$u = \frac {f(ct \mp x)}{r^2} \pm \frac {f'(ct \mp x)}{r}$$ $$p = \rho c\frac {f'(ct \mp x)}{r}$$

It's because $u$ and $p$ have a different $r$ dependance that you get the change in phase for distances below a wavelength or two.

Phew! Now you're going to ask me to put this into simple physical terms, but you'll have to let me go away and think about that. I guess it's basically because with a spherical wave the acoustic energy falls with distance in a sort of 1/$r^2$ way, but for a plane wave the acoustic energy is constant with distance.

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Whoa! I clearly have a little gap in my maths. Please do come back if you can think of a way to put this into simple physical terms. Meanwhile I will try to decipher the above maths and maybe read up on some multivariable calculus... –  jodles Apr 20 '12 at 11:03
    
@JohnRennie It can be very frustrating, when the mathematical level of the answer does not match the expected level of the OP. This is happening to me all the time (at least on SE), but in the opposite direction than in this case... –  Pygmalion Apr 20 '12 at 12:54
    
I don't think that this (very nice) derivation explains the difference in phase - per se. Based on this derivation it suggets that the phase between velocity and pressure depends on the functional form of $f(ct\mp x)$. For instance, if $f$ was a gaussian there would be no phase-difference; while if it was $sin()$ or $cos()$ you would pick up a 90 deg phase difference. Did I miss something @JohnRennie ? –  zhermes Apr 20 '12 at 15:29
    
@Pygmalion I realise that in questions like these it is not always possible to go simple enough to align the math with the level of the OP. I am perfectly happy with that; that just signals to me what I need to learn. What I would like to see though, is an intuitive physical explanation (if there is one). –  jodles Apr 20 '12 at 16:08
    
@zhermes: $\phi$ is the acoustic potential and $u$ is d$\phi$/dx (in 1D) while p is d$\phi$/dt. To take your example of $\phi$ = sin(ct + x) we get u = cos(ct +x) while p is c.cos(ct + x) so they're in phase. Because the argument of $f$ is (ct + x) differentiation wrt x and t will give the same function. –  John Rennie Apr 20 '12 at 16:46
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It's a little easier to understand if you look at the electrical case with cylindrical geometry: a long wire carrying AC current. If you are far from the wire, the E and B fields are in phase, because they represent radiated power flowing away from the wire. But close to the wire, you are in the range where inductive current flows, and the energy mostly just washes back and forth in and out of the wire.

The total energy radiated by the wire is only a small fraction of the very large energies that are exchanged inductively in the region near the wire. That's why there has to be a mismatch in the phases...because in the near field, the energy is not radiated but just temporarily stored.

It's the same basic idea with the acoustic case of the spherical source. Near the source, the amplitudes have to fall off as 1/r-squared, from simple geometric arguments. If this represented in-phase acoustic power, your total radiated power would be falling of as 1/r^4, which wouldn't add up.

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