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Homework Help: Longitudinal waves, thickness of an element

  1. Aug 9, 2004 #1
    I'm confused by something my book says. For reference, here is the relevant text:

    Figure 18-5b shows an oscillating element of air of cross-sectional area A and thickness [itex]\Delta x[/itex], with its center displaced from its equilibrium position by a distance s.

    From Eq. 18-2 we can write, for the pressure variation in the displaced element,

    [tex]\Delta p = - B \frac{\Delta V}{V}[/tex]

    The quantity V in Eq. 18-16 is the volume of the element, given by

    [tex]V = A ~ \Delta x[/tex]

    The quantity [itex]\Delta V[/itex] in Eq. 18-16 is the change in volume that occurs when the element is displaced. This volume change comes about because the displacements of the two faces of the element are not quite the same, differing by some amount [itex]\Delta s[/itex]. Thus, we can write the change in volume as

    [tex]\Delta V = A ~ \Delta s[/tex]


    They go on to make some substitutions and take partial derivatives to find some stuff. If that text would be helpful, just ask and I'll copy it into a post. My question, however, is this: Why aren't [itex]\Delta x[/itex] and [itex]\Delta s[/itex] the exact same thing? How is the "difference between the displacements of the two faces of the element" different from the "thickness" of the element?
  2. jcsd
  3. Aug 9, 2004 #2

    Doc Al

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    Staff: Mentor

    If I understand it correctly, [itex]\Delta x[/itex] is the thickness of the element, while [itex]\Delta s[/itex] is the change in thickness. If the two faces displace the same amount, [itex]\Delta s[/itex] would equal zero but [itex]\Delta x[/itex] would stay the same.
  4. Aug 9, 2004 #3
    That would imply that [itex]\Delta x[/itex] is a constant while [itex]\Delta s[/itex] varies, right? Would it make any sense to be taking [itex]\delta s / \delta x[/itex] if [itex]\delta x[/itex] was a constant?

    Just in case it's helpful, here's the rest of that section, starting from where I stopped quoting the book:

    Substituting Eqs. 18-17 and 18-18 into Eq. 18-16 and passing to the differential limit yield

    [tex]\Delta p = - B \frac{\Delta s}{\Delta x} = -B \frac {\delta s}{\delta x}[/tex]

    The symbols [itex]\delta[/itex] indicate that the derivative in Eq. 18-19 is a partial derivative, which tells us how s changes with x when the time t is fixed. From Eq. 18-13 we then have, treating t as a constant,

    [tex]\frac {\delta s}{\delta x} = \frac {\delta}{\delta x}[ s_m cos(kx - \omega t) ] = - ks_m sin(kx - \omega t)[/tex]

    Substituting this quantity for the partial derivative in Eq. 18-19 yields

    [tex]\Delta p = Bks_m sin(kx - \omega t)[/tex]

    Setting [itex]\Delta p_m = Bks_m[/itex], this yields Eq. 18-14, which we set out to prove.
    Using Eq. 18-3, we can now write

    [tex]\Delta p_m = (Bk)s_m = (v^2 \rho k)s_m[/tex]

    Equation 18-15, which we also promised to prove, follows at once if we substitute [itex]\omega/v[/itex] for k from Eq. 17-12.
  5. Aug 9, 2004 #4

    Doc Al

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    Staff: Mentor

    No it wouldn't. [itex]\Delta s[/itex] is the change in [itex]\Delta x[/itex]. If [itex]\Delta s[/itex] is non-zero, then [itex]\Delta x[/itex] is changing.
    Of course not. But [itex]\Delta x[/itex] is not a constant.
  6. Aug 9, 2004 #5
    Am I in error to think that implies that [itex]\Delta s[/itex] has units of length / time, and that [itex]\Delta x[/itex] has units of length?

    Edit: If they both have units of length, which seems to be the case looking at the equations, then what I don't understand is what [itex]\Delta s[/itex] signifies. It's the change in thickness - what does that mean? If I say, "This element has a thickness of 0.3 mm, and a change in thickness of 0.1 mm", then what would that mean? That it was going to increase by 0.1 mm in some period of time? That it had been increased from its "undisturbed thickness" by 0.1 mm? The latter doesn't seem to be meaningful if [itex]\Delta x[/itex] itself varies anyway.
    Last edited: Aug 9, 2004
  7. Aug 10, 2004 #6
    (I seem to be having a lot of technical problems with the LaTeX lately, so I'm going to substitute the letter D for capital delta in the following post.)

    The best interpretation I'm coming up with is this: you consider the thickness of some (very thin) arbitary slab of air if there were to be no longitudinal wave occuring. That thickness is a constant (I'll call it T.) Then let a longitudinal wave pass through all of the air, including our slab. We continue to consider the exact same atoms of air as we did before, but they now undergo motion, and the distance between the leftmost atom and the rightmost atom will sometimes but not usually be equal to T. Dx is the thickness at some given instant, and with time, it oscillates sinusoidally between being less or more than T. Ds represents the difference between Dx and T, such that Dx - Ds = T.

    Does that sound right?
    Last edited: Aug 10, 2004
  8. Aug 12, 2004 #7
    Have I made this thread too tangled to warrant a response, or are you all just as confused by the book's description as I am? :biggrin:

    I'd still like to know if my description in post #6 is the correct way to interpret what the book is saying.
  9. Aug 12, 2004 #8

    Doc Al

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    Staff: Mentor

    I think you are on the right track. Let me take another shot at it.

    Imagine dividing the undisturbed air into thin slices of thickness [itex]\Delta x[/itex]. We know (by assumption) that the air is getting displaced as the sound wave comes by. The amount of displacement from equilibrium position is given by [itex]s = s_m cos(kx -\omega t)[/itex]. Note that the equilibrium position is x. So now we want to get a formula in terms of pressure variation, given that we know the displacement formula. We know, from the bulk modulus, that the pressure (difference from equilibrium pressure) depends on [itex]\Delta V /V[/itex].

    Now look at a particular slice between point x and x + [itex]\Delta x[/itex]. We know how much each end is displaced (from the displacement formula). Each end of the slice displaces a different amount s and s + [itex]\Delta s[/itex]. So [itex]\Delta V/ V = A\Delta s/A\Delta x[/itex], which becomes [itex]\partial{s}/\partial{x}[/itex] in the limit as [itex]\Delta x[/itex] goes to zero.

    I'll bet this doesn't help at all! :yuck:
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