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Position of tranverse wave

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A transverse wave is described by y=0.02 sin(30x-400t). Determine the first positive x-coordinate (x>0) for which the transverse velocity of that point in the medium is at its maximum positive value when t=0. All variables are in SI units.

    a) 5.2 cm b) 10.4cm c) 15.7 cm d.20.9 e) none of the above.

    2. Relevant equations


    3. The attempt at a solution
    I found the derivative of the position equation to get the equation of velocity, at t=0 which is v = 0.6 cos(30x). I equated it to max velocity which is 0.6, therefore getting 1=cos(30x). arcos(1) is 2pi, which I later divided by 30 to get 0.209.
    Where did I make a mistake?
     
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    Double check your differentiation. Is this the correct dy/dt?

    Edit: p.s. arccos(1) is 0 but that's not relevant.
     
  4. Feb 18, 2015 #3
    I assumed we would be differentiating in terms of x, how do we know which variable we are differentiating in terms of?
     
  5. Feb 18, 2015 #4

    Nathanael

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    The definition of velocity is the derivative with respect to time.
     
  6. Feb 18, 2015 #5
    Oh right, my mistake. Also, even if I were to differentiate with respect to time, the equation obtained is y= 8 cos(30x-400t) and I plug in the value of t=0, getting the equation of y= 8 cos(30x). The position when the velocity is at maximum positive velocity, which is 8. Therefore you get the equation 8=8cos(30x). I will still get the same results, no?
     
  7. Feb 18, 2015 #6

    Nathanael

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    That's not quite the right equation. You're missing a factor of negative 1, aren't you?
     
  8. Feb 18, 2015 #7
    Oh right, now I see where I made a mistake! Thank you so much!
     
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