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Particle displacement by a speaker in a gas tube

  1. Jul 2, 2011 #1
    A speaker is generating waves in a gas that is closed in a tube. The tubes closed end is at x=0 and the speaker is at x=1. The motion of the membrane is given by

    [tex]s=s_0 \cdot sin(\omega t)[/tex]

    How does the particle displacement vary with x and t? The wavelength is [tex]\lambda[/tex]


    I do not know how to solve it. But i thougt of expressing it like this:

    [tex]s_1=s_0 sin(kx + \omega t)[/tex] (wave due to speaker oscillation, travelling from x=1 to x=0)

    [tex]s_2=s_0sin(kx - \omega t)[/tex] (reflected wave at x=0 travelling to x=1)

    The problem is that the reflected wave is reflected again and again....
     
    Last edited: Jul 2, 2011
  2. jcsd
  3. Jul 3, 2011 #2
    there is standing wave. do you know what it is?
     
  4. Jul 3, 2011 #3
    Yes i know what it is. If we solve it as a standing wave problem then we have:

    [tex]s_{tot}=2s_0cos(\omega t)sin(kx)[/tex]

    Boundary condition: [tex]s_{tot}=0 [/tex] at x=0


    According to the answer in my book it is a bit different, but x and t are separate from each other in the sinusodial functions as my suggestion above.

    Any hint for me?

    plz
     
  5. Jul 9, 2011 #4

    Philip Wood

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    Gold Member

    You've got the right idea, but you need to be a bit more flexible with your choice of phases of progressive waves. You have to choose them so that their sum conforms to the boundary conditions at x=0 and x=1.

    Yours fit at x=0, because sin(kx) =0, and s1 + s2 must indeed be zero next to the wall.

    But things aren't right at x=1. The speaker diaphragm moves according to sin(wt), whereas your s1 + s2 varies as cos(wt). May I suggest you try the progressive waves s1 = Acos(kx+wt) and s2 = -Acos(kx-wt). Then express the difference of two cosines as a product. It comes out as a product of sines, so you can meet both boundary conditions.
     
  6. Jul 19, 2011 #5
    I dont really know how to do...

    If we put a boundary condition at x=1 as:

    s=2scos(wt)sin(k)

    now what?

    But i recognized that this should also be a fact:

    2scos(wt)sin(k)=sin(wt) this should also help i think
     
  7. Jul 19, 2011 #6

    Philip Wood

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    Gold Member

    Did you add the two progressive waves that I recommended you try? Use the standard trig formula cos([itex]\vartheta[/itex]) - cos([itex]\varphi[/itex]) = 2sin[([itex]\vartheta[/itex] +[itex]\varphi[/itex])/2] sin[([itex]\vartheta[/itex] -[itex]\varphi[/itex])/2].

    You should then find that s1 +s2 is the product of an x-dependent sine and a t-dependent sine. s1 +s2 automatically disappears when x = 0, just as it should at a wall. It also agrees with the motion of the membrane at x = 1, as given in the question, provided that A =...
     
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