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Instaneous Power in a standing Wave

  1. Feb 28, 2008 #1

    TFM

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

    The instantaneous rate at which a wave transmits energy along a string (instantaneous power) is

    [tex] P(x,t) = =F\frac{\partial y(x,t)}{\partial x} \frac{\partial y(x,t)}{\partial t}[/tex]

    where F is the tension.

    Show that for all values of x, the average power [tex]P_a_v[/tex] carried by the standing wave is zero. (Equation [tex] P_a_v = \frac{1}{2}\sqrt{\mu F} \omega^2 A^2[/tex] does not apply here. Can you see why?)

    2. Relevant equations

    [tex] P(x,t) = -F sin(A_S_W(\omega t)kA_S_Wcos(kx))(A_S_Wsin(kx)\omega cos(\omega t)) [/tex]

    ^ Power equation calculated from part 1

    3. The attempt at a solution

    The answer section is in the form of a writing box (I am using Mastering Physics), Not the normal maths box.

    What is the best way to show for all values of x the power is zero?

    TFM
     
  2. jcsd
  3. Feb 28, 2008 #2

    TFM

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    Does anyone have any idea what the best way is to approach this question?

    Would be grearly appreciated,

    TFM
     
  4. Feb 29, 2008 #3

    TFM

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    This is the annoying part of Mastering Physics - I can't get onto the next question until I finish this one. What I would normally do for a "Show that for all values of x..." question, I would just stick in thousands of values for x.

    As you can see, any ideas would be very greatly appreciated.

    TFM
     
  5. Feb 29, 2008 #4

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    What is the best way to go about this question, should I just use a lot o different values, or is there a way to change the equation into an average one?

    ANy help would be greatly appreciated,

    TFM
     
  6. Feb 29, 2008 #5

    Doc Al

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    Staff: Mentor

    Given the instantaneous power, find the time average. What's the average value of [itex]\sin\omega t[/itex] or [itex]\cos\omega t[/itex] over a complete cycle?
     
  7. Feb 29, 2008 #6

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    Since the maximam value is 1, the minimum is -1, the average will be 0.

    Would you insert this intop the orginal power equation?

    TFM
     
  8. Feb 29, 2008 #7

    Doc Al

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    I don't quite understand your instantaneous power equation. Can you check for typos? If you can write it as [itex]P(x,t) = f(x)\cos\omega t[/itex], then you can immediately see that the average is zero since the average of sin or cos is zero (they are negative as much as positive).
     
  9. Feb 29, 2008 #8

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    If cos(average) = 0, then it will make the average power 0 becasue all the terms are multiplied together, so the 0 from cos(average) and sin(average) will make the whole equation = 0.

    The equation is right, because I calculated it for the first part oif the question, on Mastering Physics, and it came up as it correct.

    TFM

    edit:

    It will make the formula:

    [tex] p_a_v_g = -F*(0) [/tex]
     
    Last edited: Feb 29, 2008
  10. Feb 29, 2008 #9

    Doc Al

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    You cannot independently average different factors. You must average the entire function of t as a whole. You can show that [itex]\sin\omega t \cos\omega t[/itex] can be written as a simple sine function and thus its average must be zero.

    Your equation in post #1 as you wrote it here:
    doesn't quite make sense to me. (You've got some parentheses mixed up or something.)
     
  11. Feb 29, 2008 #10

    TFM

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    The equation from Mastering Physics:

    [tex] P(x,t) = -F (sin(\omega t)kA_S_Wcos(kx))(A_S_Wsin(kx) \omega cos(\omega t)) [/tex]

    TFM
     
  12. Feb 29, 2008 #11

    Doc Al

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    That's better.

    Now you can write that equation as [itex]P(x,t) = f(x) \sin\omega t \cos\omega t[/itex]. Then show that the sine*cosine factor is equivalent to a simple sine function and thus averages to zero.
     
  13. Feb 29, 2008 #12

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    Rearranging the equation, then:

    [tex] P(x,t) = sin(\omega t)cos(\omega t)[-F(kA_S_Wcos(kx))(\omega A_S_Wsin (kx))] [/tex]

    do I need to use doubke angle functions to show that the sine*cosine factor is equivalent to a simple sine function?

    TFM
     
  14. Feb 29, 2008 #13

    Doc Al

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    That's what I would do. :wink:
     
  15. Feb 29, 2008 #14

    TFM

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    Should I use:

    [tex] sin2\theta = 2sin\theta cos\theta [/tex]

    rearranging:

    [tex] \frac{sin2\theta }{2} = sin\theta cos\theta [/tex]

    Inserting into the equation, giving:

    [tex] P(x,t) = \frac{sin2\theta }{2} F_x [/tex]

    where [tex] F_x [/tex] is the other half of the equation

    TFM
     
  16. Feb 29, 2008 #15

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    I meant

    [tex] P(x,t) = \frac{sin2\theta }{2} F_x [/tex]

    TFM

    (I tried editting, but it didn't change)
     
  17. Feb 29, 2008 #16

    Doc Al

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    Looks good, but keep it as [itex]\sin 2\omega t[/itex].
     
  18. Feb 29, 2008 #17

    TFM

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    [tex] P(x,t) = \frac{sin2\omega t}{2} F_x [/tex]

    What would be the best thing to do now?

    TFM
     
  19. Feb 29, 2008 #18

    Doc Al

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    Declare victory! Assuming that you accept that the average value of sin(t) = 0, you're done.
     
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