Transverse Wave Calculations

  1. A transverse wave is described by the expression y= 0.05cos(4.19x-1260t). You may assume all measurements are in the correct SI units.
    A) What is the amplitude of this wave?
    B) What is the wavelength of this wave?
    C) How fast is the wave travelling?
    D) What is the maximum tranverse velocity of the wave?


    Relevant equations: y= a sin (b(x-c)) +d
    Lamda= velocity/frequency
    Velocity= delta x/ time

    A) the amplitude is 0.05 as "a" in the formula is always amplitude.

    I am having trouble figuring out the last three questions but. How you do figure out the wavelength? Do you have to draw the wave on a quadrant graph and get the information off that?
     
    Last edited: Mar 25, 2011
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Welcome to PF!

    hi KatieKT! Welcome to PF! :smile:

    (have a lambda: λ and a delta: ∆ :wink:)

    We measure the wavelength at a fixed time

    for fixed t, how much do we change x for y to be repeated?​

    (And we measure the speed at a fixed height

    for fixed y, what is the ratio of x to t for y to stay constant?)​
     
  4. Yeah so in y=a cos(bx + c) + d, a is amplitude and b would be the one that effects the period/wavelength. So is it 4.19m's long? Usually I thought that along the bottom of the graph it's like 45, 90, etc for thesse types of graphs but I don't think that's right.
     
  5. ideasrule

    ideasrule 2,323
    Homework Helper

    Think about what the wavelength means. 4.19x-1260t is the angle being fed into the cosine function, so it should advance by 2pi (representing one cycle) every time x advances by lambda. If we hold t constant, what should lambda be so that this is true? You can use the same reasoning to find the period. Speed would then be wavelength/period.

    It's like 45,90,etc. if you plot sine or cosine with the angle on the x axis. In this case, bc+x is the angle being fed into the cosine function. You can definitely plot y=acos(bx+c)+d as a function of this angle, although for this question, I don't see why you would.

    PS. Anyways, try to use y=acos(wx-kt). It's much more common and intuitive than y=acos(bx+c)+d, since a, w, and k both have well-defined meanings.
     
  6. Ok so i've worked out λ to be 1.49m through 2∏/b. Then the frequency by 1/1.49 to get 0.23Hz. However to get the velocity I was told to use v= -Vmax sin(-1260t) and the formula for the maximum transverse velocity was Vmax= 2∏A/λ. Is 2∏A/λ the correct formula to use however? I know for maximum velocity it is, but is it also the formula for maximum transverse velocity?
     
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