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PDE (wave equation) used to find acoustic pressure in a a pipe

  1. Jun 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Assume that the wavelength of acoustic waves in an organ pipe is long relative to the width of the pipe so that the acoustic waves are one-dimensional (they travel only lengthwise in the pipe). Therefore, the equation governing the pressure in the wave is:

    2p/∂t2-c2*∂2p/∂x2 = 0

    p is the acoustic pressure. At a closed end of the pipe the air velocity (u) must be zero. Note that:

    -ρ*∂u/∂t = ∂p/∂x

    So that if u = 0, ∂p/∂x = 0. At an open end, p is approximately 0

    Find the acoustic modes in a pipe of length L with:

    (a) 2 closed ends
    (b) 2 open ends
    (c) one closed end and one open end

    3. The attempt at a solution

    From lecture we started by using separation of variables on the wave function

    2p/∂t2-c2*∂2p/∂x2 = 0

    let y(x,t) = c2*∂2p/∂x2

    ytt = f*g''

    yxx = f''*g

    f*g'' - c2*f''*g = 0

    g''/g = c2*f''/f = constant = -ω2

    ω = cn*∏/L k = n*∏/L

    g'' + ω2*g = 0 --> g = A1*cos(ωt) + A2*sin(ωt)

    f'' + ω2/c2*g = 0 --> f = B1*cos(kx + B2*sin(kx)

    f(0) = B1 f(L) = B2*sin(n*∏*x/L)

    y(0,t) = y(L,t) = 0

    y(x,t) = f(x)*g(x)

    = ∑ cn*cos(ωn*t+øn) * B2*sin(n*∏*x/L)

    The biggest problem I'm having is we aren't given an initial value for x. I'm sure you can find one with the second partial differential equation given but I don't know how to go about that.
    We have been covering Fourier series lately and this looks a lot like the form of Fourier series but I also don't know how to convert it. Finally, I don't know what a mode is. Does it have to do with the nodes and anti-nodes? Someone told me it's the amount of sets of different nodes/anti-nodes(if that makes sense) but wouldn't that be infinite?

    thanks
     
  2. jcsd
  3. Jun 20, 2014 #2

    AlephZero

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    What you did so far looks OK.

    Keeping it simple, an "acoustic mode" is just a non-zero solution of the PDE that satisfies the boundary conditions.

    The question says the pipe is of length L, so you want to find solutions for 0 <= x <= L.

    Or you could take -L/2 <= x <= L/2 if you prefer - it won't make the math any harder, or easier.

    You are told the boundary conditions at a closed and open end of the pipe. If you put those into your solution for f, you will only get non-zero values of B1 and B2 for certain values of ω, or k. There will be an infinite number of those values of ω = ωi, i = 1, 2, 3, ..., so the general solution of the PDE will be an infinite series.

    When you have done that, you will probably see the relation between all this and Fourier series, and how it relates to nodes and antinodes.

    If you know about eigenvalues and eigenvectors, you will probably also see that the ωi are rather like eigenvalues and the corresponding solutions for f are rather like eigenvectors.
     
  4. Jun 20, 2014 #3
    Sorry for some of the confusion, by initial value for x i meant there isn't a y(x,0) function given like similar wave functions that use a string.

    Also, how will having open versus closed ends affect the boundary conditions?
     
  5. Jun 20, 2014 #4

    Orodruin

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    The initial conditions will not affect the possible modes. The general solution is a superposition of the modes and can be adjusted to the initial conditions.

    Already in your first post you discussed the boundary conditions for a closed pipe, what changes if the pipe end is open?

    Note that the modes must fulfill both the differential equation and the boundary conditions (both homogenous). Thus, if you change the boundary conditions, the modes will change.
     
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