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Wave Motion, writing an equation.

  1. May 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A sinusoidal wave traveling in the -x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 29.0 cm, and a frequency of 15.0 Hz. The transverse position of an element of the medium at t = 0, x = 0 is y = -3.00 cm, and the element has a positive velocity here.

    Write an expression for the wave function y(x,t), where y and x are expressed in cm, and t is expressed in seconds.



    2. Relevant equations

    [tex]\omega = 2\pif[/tex]

    v=(lambda)(frequency)

    [tex] k = 2\pi/\lambda[/tex]

    [tex] y(x,t) = Asin[\left(2\pi/\lambda\right)(x - vt)][/tex]

    3. The attempt at a solution
    We are given the fact that:

    [tex]A = 20.0cm[/tex]

    [tex]\lambda = 29cm[/tex]

    [tex] v = (lambda)(frequency) = 435cm/s[/tex]

    I think the answer is the following, but I am not sure, and I only have one more submission left, so I want to be sure.

    [tex] y(x,t) = -20.0cos[.217(x-435t)][/tex]
     
    Last edited: May 1, 2008
  2. jcsd
  3. May 2, 2008 #2
    Anyone have any idea if I am correct?
     
  4. May 2, 2008 #3

    alphysicist

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    Hi Ithryndil,

    If you check your answer at x=0 and t=0, you get that y=-20 cm. However, this is not what the problem specifies; they say they want x==-3cm at x=0 and t=0.

    I think you need to include a phase constant in your expression that will give the correct initial condition.
     
  5. May 2, 2008 #4
    You mean y = 3 cm? And I think so too. I just realized that the general equation for y(x,t) is as follows:

    [tex]
    y(x,t) = Asin(kx - vt + \phi)]
    [/tex]

    I think the above equation is for a wave traveling to the right. I think because this wave is traveling to the left it needs to be:

    [tex]
    y(x,t) = Asin(kx + vt + \phi)]
    [/tex]
     
  6. May 2, 2008 #5

    alphysicist

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    Yes, I did mean y= -3 cm. Sorry!

    Your updated equation looks good. Once you plug in the numbers you found, you can check that it matches y=-3cm at x=0 and t=0 and that it also has a positive velocity (in the y direction at that point).
     
  7. May 2, 2008 #6
    Alright, thank you for the help.
     
  8. May 2, 2008 #7

    Redbelly98

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    That is almost correct. There should either be some more parentheses inside the sin argument, or replace vt with [tex]\omega t[/tex].
     
  9. May 3, 2008 #8
    You're right, it should be: [tex]y(x,t) = A\sin(kx + \omegat + \phi)[/tex]

    I wound up getting the following for the equation:

    [tex]y(x,t)20.0sin(0.217x+94.25t-0.1506)[/tex]

    Now I was stupid and input [tex]y(x,t)20.0cos(0.217x+94.25t-0.1506)[/tex]...note I put cosine and not sine. Yes, the cosine would be correct with the right phase angle, but it's not with that phase angle.
     
  10. May 3, 2008 #9

    Redbelly98

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    Your expression (using sin) looks good. :smile:
     
  11. May 3, 2008 #10
    Thank you. I just can't believe I inputted the expression with cos instead of sin.
     
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