1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  5. Jul 9, 2011 #4

    Philip Wood

    User Avatar
    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:


    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

    User Avatar
    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 =...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook