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Finding the wavelength on a sinusoidal wave on a string

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, this is a problem regarding mathematical descriptions of waves. I've attached an imagine of the picture but I'll also type out the problem for clarity.
    "A sinusoidal wave is propagating along a stretched string that lies on the x-axis. The distplacement of the string as a function of time is graphed in Fig.E11 for particles x=0m and at x=0.09m. (a) What is the amplitude of the wave? (b) What is the period of the wave? (c) You are told that the two points x=0 and x=0.09 are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength and wave speed."
    2. Relevant equations
    [itex]k=2\pi/\lambda[/itex]
    [itex]y(x,t)=Acos(kx+\omega t)[/itex]

    3. The attempt at a solution
    (a) from reading off the graph you can see at the amplitude is 4mm or 0.004m

    (b) from reading off the graph you can see that the period = 0.04s

    (c) this is where I got confused. I thought about making simultaneous equations using x=0 and x=0.09. I figure that the answer involves using the fact that [itex]k=2\pi/\lambda[/itex] but I'm a little unsure.
    I started playing with the idea that when y=0 and t=0 [itex]0.004cos(kx)=0[/itex] and therefore [itex]cos(kx)=0[/itex] and [itex]kx=n\pi/2[/itex] however that's not particularly helpful since I don't know the value of x. If anyone could give me a hint or put me on the right track it'd be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2015 #2
    Wouldn't this mean the wavelength is 0.09m?
    How are frequency, wavelength and speed related?


    Correct
    Correct
     
  4. Mar 31, 2015 #3
    Well it just says that they're within one wavelength, so the wavelength could be bigger than 0.09m? I know that [itex]v=\lambda f[/itex] but this doesn't help since I only know the frequency.
     
  5. Mar 31, 2015 #4
    Oh, my bad, I thought the points are one wavelength away from each other. Anyway, v=λf=λ/T still holds.
     
    Last edited: Mar 31, 2015
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