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Superposition: Sound pressure

  1. May 16, 2008 #1

    I am trying to determine the validity of a formula I came across... I have not seen it anywhere else and I can't be sure wether it's valid or not...

    Consider a number of sound pointsources positioned on a line. The soundwaves coming from each sound source are all in phase (at time t = 0) and they all have the same amplitude. (So they are perfectly coherent).

    Consider a point P at some distance away from the pointsources, where I want to use superposition to add up the sounds.
    Because the soundsources are not all in the same location, the soundwaves from one sound source will have travelled a short distance longer than the soundwave from another sound source; resulting in a phase difference.

    I have found a formula to calculate exactly the sound pressure on point P, but I don't know if it's valid...:

    [tex]p_{tot} = \left| \sum_{n=1}^N \frac{p_0}{r_n} e^{ikr_n} \right|[/tex]
    Here, [itex]p_0[/itex] is the (effective*) sound pressure of a single sound source at 1 meter from the source. The wavenumber [itex]k[/itex] is [itex]\frac{2 \pi}{\lambda}[/itex] with [itex]\lambda[/itex] the wavelength of the soundwave.
    N is the number of sound sources, and finally [itex]r_n[/itex] is the distance from point P to a single sound source n.

    Is this formula valid? If yes, how can it be derived?

    I know the general equation for a spherical wave is given by:
    [tex]\psi = \frac{A}{r} e^{i(kr - \omega t)}[/tex]

    First of all, I suppose we are only looking at a instanteneous time, for simplicity t = 0, which eliminates the omega*t term?

    Secondly, the sum is obvious, it is simply summing all waves, which each have a different phase [itex]kr_n[/itex] due to the difference in distance travelled.

    Then, the modulus (abs. value) is taken after summation. I guess that makes sence too, this will give us the maximum (effective) sound pressure (invariant of time) at the point P, right?

    But, the thing that troubles me most, is the units. They are not right, are they? Since the exponential is obviously dimensionless, we get:
    [tex]\text{[Pressure]} = \frac{ \text{[Pressure]} }{ \text{[Distance]}} \text{ } \rightarrow \text{ } \text{Pa} = \frac{ \text{Pa}}{\text{m}}[/tex]. Huh?

    Can someone help me out here? Thanks!

    *By effective sound pressure I mean [itex]p_{\text{eff}} = \frac{1}{2} \sqrt{2} p_{\text{max}}[/itex]
  2. jcsd
  3. May 16, 2008 #2


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    [tex]|p_{tot}| = \left| \sum_{n=1}^N \frac{p_0 r_0}{r_n} e^{-ik(r_n-r_0)} \right|[/tex]

    i think this is a little better dimensionally. be careful when you see some formulae with terms or factors that are set for 1 meter. nature doesn't give a rat's ass that some of us might measure lengths in meters.
  4. May 17, 2008 #3
    Thanks. What is r_0 in this case then? 1 meter? And why the - sign in the exponent?
  5. May 17, 2008 #4


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    sure, if you want it to be. r0 is whatever distance you define your reference sound pressure level, p0, to be.

    because i think it was wrong in your expression for a wave going outward from the source. we know that the pressure relationship in the spherical wave equation is inverse proportional (not inverse square, because there is also particle velocity which has an 1/r component in-phase with the pressure and another 1/r2 component 90o out of phase with the pressure wave, the two components in phase with each other multiply to give you intensity which must be inverse-square).

    [tex] p(t,r) = p(0,r_0) \frac{r_0}{r} e^{i(\omega t \ - \ c(r-r_0))} [/tex]

    that is the equation for an outward propagating spherical sinusoidal pressure wave.

    one could wire up a bunch of identical loudspeakers in a dodecahedran all pointing toward the center and get a pretty good spherical wave traveling inward creating, theoretically, an infinite SPL at the center (where you place a wine glass and watch it break). there are two solutions to the wave equations, one for a wave in either direction. if you have an inward traveling wave, you would change the sign on the c factor above.
  6. May 17, 2008 #5
    If the equation is what you say it is in your above post, I don't understand why my book tells me the general equation for a spherical wave is:
    [tex]\psi = \frac{A}{r} e^{i(kr-\omega t)}[/tex]

    I understand [itex]\psi[/itex] is ofcourse not the sound pressure, but merely some form of displacement, right?

    But that doesn't explain why the [itex]\omega t[/itex] and [itex]kr[/itex] are switched over... Does it?

    Surely, [tex]e^{i(\omega t - k(r-r_0))}[/tex] is not the same as [tex]e^{i(k(r-r_0) - \omega t)}[/tex], right..?

    So the formula I have been using (first post) is wrong, even though I have taken r_0 as 1 meter, since I did not substract the r_0 from the r in the kr term..? I was tempted to say "Oh well that shouldn't make much of a difference" but since the kr term is the only term deciding the phase difference (which is obviously the important issue when using superposition on soundwaves) there could potentially be a huge difference, right?

    Finally, how does a spherical wave travel inward? If you just have a pointsource, I can't see any way the wave would be going 'inwards'? Isn't inwards exactly the same as outwards in a point source?

    If you're interested, I need to know all this because I am trying to build a sound directing device; basically a speaker that no longer emits sound nearly spherical, but (in the most optimal case) in a straight line (more or less like a laser).
    If you put a number of sound sources (52 in my case) in a line (or pattern) with [itex]1/2 \lambda[/itex] between each speaker, you can get a nicely directed soundwave.
    (Actually it's not really a directed soundwave, it's merely an interference pattern that looks like a directed soundwave).
    We have built a device containing 52 speakers in a triangular pattern and managed to get quite an amazing effect. At about 5 meters from the speakers, if you stand to the side of the plane in which the speakers lie, you can 'only' hear a SPL of about 60 dB. If you stand right infront of the plane of speakers, you hear a SPL of about 110 dB, which is quite a weird feeling! There's also a very defined 'line' where the SPL drops significantly, only about 5-10 degrees out of the center... Pretty amazing...

    So I need this equation to theoretically calculate the SPL of our sound directing device at a fixed distance from the source, under any angle... I have used the formula from the first post and the result was quite nice, just like you would expect if you heard the device...
  7. May 17, 2008 #6


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    crap. i gotta check my equations better than i do.

    first, go to Wikipedia and see what they say.

    of course

    [tex] p(t,r) = p(0,r_0) \frac{r_0}{r} e^{i(\omega t \ - \ c(r-r_0))} [/tex]

    is incorrect. you do not multiply the wavespeed c by distance to get phase.

    the Wikipedia general solution is

    [tex] u(t,r) = \frac{1}{r} F(r-ct) + \frac{1}{r} G(r+ct) [/tex]

    so the wavespeed c multiplies time to get something that is dimensionally the same as distance r. F(.) is the wave moving outward. i think this works

    [tex] F(r-ct) = u(0,r_0) r_0 e^{-i2\pi((r-ct) - r_0)/\lambda} [/tex]

    then the outgoing wave is

    [tex] u(t,r) = \frac{1}{r} F(r-ct) = u(0,r_0) \frac{r_0}{r} e^{-i2\pi((r-ct) - r_0)/\lambda/} [/tex]

    which is

    [tex] u(t,r) = u(0,r_0) \frac{r_0}{r} e^{i 2\pi (c t \ - \ (r-r_0)) / \lambda} [/tex]

    [tex] u(t,r) = u(0,r_0) \frac{r_0}{r} e^{i 2\pi c (t \ - \ (r-r_0)/c)) / \lambda} [/tex]

    [tex] u(t,r) = u(0,r_0) \frac{r_0}{r} e^{i \omega (t \ - \ (r-r_0)/c)} [/tex]

    how does that look? somewhere i think i got a two pi factor off, but have trouble seeing it. does someone else see it? (edit, i think i found it.)
    Last edited: May 17, 2008
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