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Transverse wave particle speeds

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A transverse traveling wave on a cord is represented by D = 0.38sin(5.1x+30t) where D and x are in meters and t is in seconds. Determine maximum and minimum speeds of particles of the cord.

    2. Relevant equations

    First derivative

    3. The attempt at a solution

    I took the first derivative with respect to x

    dx/dD = (5.1) .38cos(5.1x+10t)

    cosine's max value is at 1 so i set cos(5.1x+30t) = 1

    then I solved : (5.1)(.38)(1) = 1.94 m/s
    Im not sure if this is right. All help will be appreciated.
     
  2. jcsd
  3. Nov 14, 2008 #2

    Hootenanny

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    Firstly let's clear up some notation.

    [tex]\frac{dD}{dx}[/tex]

    represents the derivative of D with respect to x.

    Secondly, why are you taking the derivative with respect to x?
     
  4. Nov 14, 2008 #3
    yes sorry that was honestly a typo. but thank you.

    I was taking the derivative with respect to x because of speed's definition. so i thought by taking it with respect to x instead of time, i wud be getting the rate of change of distance...
     
  5. Nov 14, 2008 #4

    Hootenanny

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    But the definition of speed is the derivative of displacement with respect to time? Why would you want to find the derivative of amplitude with respect to displacement?
     
  6. Nov 14, 2008 #5
    ok thank you for clearing that up. it makes sense now. so apart from the confusion in the derivation, the way i solved the problem ,is that still the route i will take to solve it?
     
  7. Nov 14, 2008 #6

    Hootenanny

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    Yes, differentiating D with respect to time will give you the velocity of a particle on the cord at position x and time t.

    A word of warning when finding the minimum speed: note that the minimum value of cosine does not correspond to the speed.
     
  8. Nov 14, 2008 #7
    Wouldnt the minimum speed be 0?
     
  9. Nov 14, 2008 #8
    and when i take the derivative with respect to t, i calculate the v as 30(.38) I am getting 11.4 m/s. However, the homework program is not accepting it as the correct answer. Any more suggestions?
     
  10. Nov 14, 2008 #9

    Hootenanny

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    Yes, that was the point I was making. The minimum value of cosine is -1, but this corresponds to the maximum rather than minimum speed, but it looks like you've already got it figured out.
     
  11. Nov 14, 2008 #10

    Hootenanny

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    Care to elaborate on your attempt? It's difficult to see where you're going wrong without seeing your solution.
     
  12. Nov 14, 2008 #11
    sure. D = 0.38sin(5.1x+30t)
    so dt/dx = (30)(0.38)cos(5.1x+30t)
    and again since we obtain cosine's max value at 1 i am setting cos(5.1x+30t) = 1
    hence ===> (30)(.38)(1) = 11.4 m/s

    I guess the derivative has a glitch in it somewhere..please let me know
     
  13. Nov 14, 2008 #12

    Hootenanny

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    Except for your notation (dt/dx => dD/dt), your solution looks fine to me.
     
  14. Nov 14, 2008 #13
    oops, thank you again. My only bet is that the homework may have been programmed incorrectly.
    Thanks for the help mate.
     
  15. Nov 14, 2008 #14

    Hootenanny

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    No worries.
     
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