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Find wavelength, Phase Difference, and amplitude from min and max

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    See attached image


    2. Relevant equations

    Δphi = 2pi(Δx/λ) + Δphi0 = 2pi(m) -- at the maximum
    Δphi = (m + .5) (2pi) -- at the minimum


    3. The attempt at a solution

    The first step should be to solve for lambda, the wavelength.

    So I plug in the knowns:

    At 30 cm the amplitude is a maximum, so:

    Δphi = 2pi(Δx/λ) + Δphi0 = 2pi(m)

    Δphi = 2pi(30/λ) + Δphi0 = 2pi(m)

    I am not sure what to do next. I tried too many times, so it now displays the answer as λ = 80 cm, but I'm not sure why that is true. I need help on the rest.

    I'm just not sure what to do since I am missing more than one variable.
     

    Attached Files:

  2. jcsd
  3. Feb 13, 2013 #2

    tms

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    You can figure out the wavelength by inspection: while the speaker moves 40 cm the wave goes from a minimum to a maximum. That tells you that 40 cm is what fraction of the wavelength?
     
  4. Feb 13, 2013 #3

    rude man

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    Spkr 1 output = a*cos(kx - wt), k = 2π/λ and w = 2πf.
    Put spkr 1 at x = 0 and let t = 0 arbitrarily:
    Then spkr 1 = a.

    Similarly, Spkr 2 output is a*cos(kx - wt - ψ), allowing for a phase difference ψ between the two speakers when they're side-by-side (as you might get if spkr 2 had a much longer hookup cable than spkr 1).

    Now, what is the equation for the output of spkr 2 heard at x = 0 when placed a distance x0 in front of spkr 1? And what is the combined output at x = 0?

    From this you get 2 equations and 2 unknowns: k and ψ.
     
  5. Feb 14, 2013 #4
    Thanks!

    λ = 80 cm, since 40 cm = (1/2)λ.

    The equation for the output of speaker 2 at x = 0 would be:

    S2 = a*cos(-ψ)


    The equation for the the combined output would be:

    S1 + S2 = a*cos(-ψ2 + ψ1)

    k = 2π/80 = 0.0785

    So at x = -10 cm,

    S2 = a*cos(0.0785*(-10) + ψ2).


    I'm not sure what to do from here, though.
     
  6. Feb 14, 2013 #5

    tms

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    Aren't you leaving out something in the argument to the cosine?
    That's not how you add cosines.
     
  7. Feb 14, 2013 #6
    If t=0 and x=0 then both terms besides the psi should be 0, no?
     
  8. Feb 14, 2013 #7
    And the correct equation for the combined outputs is:

    S1 + S2 = 2a*cos(0.5*(ψ2+ψ1))*cos(0.5*(ψ2-ψ1))

    Yes?
     
  9. Feb 14, 2013 #8

    rude man

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    No, that would be the output of spkr 2 at its own position, not at x = 0. Call its position x' = 0. Along the x axis, x' = 0 is a distance x0 to the right of x = 0 if spkr 2 were located x0 to the right of spkr 1.
    So x' = x - x0.

    OK, remember, we are arbitrarily fixing t = 0 everywhere.

    So spkr 2 output = cos(kx' - ψ). But then x' = x - x2 if spkr 2 is in front of spkr 1 by x2 meters. And, if we put spkr 2 behind spkr 1 by x1 meters, spkr 2 output = cos(kx'' - ψ) where x'' = x + x1. Draw these three positions along an x axis with x = 0 at spkr 1 and x = -x1 and then x = + x2 for spkr 2 and this should be pretty clear.

    Now for each of the two position of spkr 2, add the signals from both spkrs and deduce what the cosine arguments must be in each case for the spkr signals to add up to 0 and 2a respectively. Remember we still can let x = 0 in these equations without loss of generality.
     
  10. Feb 14, 2013 #9

    rude man

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    Ugh! Nope!
    Why do you want to multiply? We are adding the signals from the two speakers.
    Also, there is no need for two phases ψ1 and ψ2. There is only one phase, namely the phase between the two speakers. As I said, think of this phase as being produced by a difference in cable lengths between the two speakers.
     
  11. Feb 14, 2013 #10

    tms

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    Just to clarify for the OP, the most general solution would have a phase for both waves, but one can adjust when time = 0 in order to eliminate one of the phases.
     
  12. Feb 14, 2013 #11

    tms

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    At that one time, at that one place, yes, but you want a more general solution, for part C at least.
     
  13. Feb 14, 2013 #12

    tms

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    Actually, yes, assuming x = t = 0. The formula for adding two cosines is:
    [tex]\cos\theta + \cos\phi = 2 \cos\left(\frac{\theta + \phi}{2}\right)
    \cos\left(\frac{\theta - \phi}{2}\right).[/tex]
     
  14. Feb 16, 2013 #13
    How would I find Psi from this?
     
  15. Feb 16, 2013 #14

    tms

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    From what?
     
  16. Feb 16, 2013 #15
    How would I find it from the equations I gave?
     
  17. Feb 16, 2013 #16

    tms

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    Think about the information you are given. You now know the wavelength, and you know about how far apart the speakers are when the total output is at a maximum (that is, when the speakers are in phase). From that you can figure out the phase shift.
     
  18. Feb 17, 2013 #17

    rude man

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    Or you can use the equation cos(kx - wt - psi) for spkr 2 in terms of the two speaker locations to find psi (and lamba).

    In other words, cos(kx' - wt - psi) for spkr 2 in front of spkr 1, and cos(kx'' - wt - psi) for spkr 2 behind spkr 1.

    With x' and x'' as I previously explained.
     
  19. Feb 18, 2013 #18
    Just to clarify, I have diagnosed myself Physics-retarded. I'm not bad at many other things, but Physics sometimes baffles me.

    So from this, I see that I might be able to use the equation I provided originally:



    Distance they are apart at a maximum = ΔPhi = 40

    = 2π*(30 cm/80 cm) + ψ

    40 = 2.35 + ψ

    ψ = 37.65

    Correct?
     
  20. Feb 18, 2013 #19

    rude man

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    Not what I got. tms?
     
  21. Feb 18, 2013 #20
    I think my problem is that delta phi, the distance they are apart needs to be in radians. How could I find that value?
     
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