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Homework Help: Waves question, finding new frequency, please help

  1. Mar 30, 2010 #1

    Ush

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

    A stretched wire vibrates in its fundamental mode at a frequency of 375 Hz. What would be the fundamental frequency if the wire were one third as long, its diameter were tripled, and its tension were increased five-fold?


    2. Relevant equations

    V = F*w = sqrt(T/u)

    mass of cylinder = Volume x Density
    m = πr2Lp

    where
    F = Frequency
    w = wavelength,
    T = tension
    u = linear mass density
    m = mass
    L = length
    p = density

    3. The attempt at a solution

    m = πr2Lp
    m/L = πr2p
    1. u = πr2p

    F*w = sqrt(T/u)
    F*2L = sqrt(T/u)
    F24L2 = T/u
    2. F2 = T/(u4L2)

    sub in u from 1. into 2.

    F2 = T/(πr2p4L2)

    Question says..
    T x 5
    L x 1/3
    D x 3 => r x 6

    F22 / F12 = [5T/(π(6r)2p4(1/3L)2)] / [T/(πr2p4L2)]

    F22 / F12 = 5 / 62(1/3)2

    F2 / F1 = sqrt(5/4)
    F2 = sqrt(5/4)*F1

    ...This gave me the wrong answer,
    Could someone please tell me where I went wrong? or where I missed something?
    Thank you
     
  2. jcsd
  3. Mar 30, 2010 #2

    Doc Al

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    Staff: Mentor

    You're almost right. Keep it simple. Express the fundamental frequency in terms of the speed and length. Then you can figure out how the frequency changes by understanding how the speed and length changes. (By what factor does each variable change?)
     
  4. Mar 30, 2010 #3

    Ush

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    thank you for the reply,

    but I don't understand how I can express it in terms of velocity, don't I need to change V into sqrt(T/pπr2) because we're changing diameter?
     
  5. Mar 31, 2010 #4

    Ush

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    Hm, I tried that, but it doesn't seem to give me the correct answer, =/ it gave me a oddly high frequency of 7546.5Hz
     
  6. Mar 31, 2010 #5

    Ush

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    I believe there may be something wrong with your reasoning for calculating linear mass density?
     
  7. Mar 31, 2010 #6
    Yeah, I fixed it in my example. I was probably looking at a center of mass problem earlier and went with mass per unit area instead of mass per unit length. Hopefully the problem is cleared up now.
     
  8. Mar 31, 2010 #7

    Doc Al

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    My suggestion was to do it in steps. Much easier (to me) and less chance for error.

    The fundamental frequency in terms of speed and length is given by: f = v/(2L).

    How does the speed change? How does the length change?
     
  9. Mar 31, 2010 #8

    Ush

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    AtticusFinch,
    your method gave me a frequency of 279.508Hz, unfortunately this is the wrong answer,

    This may be because Volume of a cylinder doesn't vary with D3; It varies with D2, or r2, depending on the volume equation used..; but even when I tried the same reasoning, with D2, it still gave me the wrong answer, (484.12Hz)

    ---

    Doc Al,
    I know the length is reduced to a 1/3 of original
    Fnew = Vnew / (2/3)L

    I don't understand how I can calculate how speed changes? Other then trying => speed varies with sqrt(T/u) ,Which isn't really working at the moment =/
     
  10. Mar 31, 2010 #9

    Doc Al

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    Express that as:
    L' = L/3

    Why isn't it working? Use v = sqrt(T/μ). How does T change? How does μ change?
     
  11. Mar 31, 2010 #10

    Ush

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    can I express it as,
    Wold = Wnew/3 ?

    Tension changes by a factor of 5,

    the equation for mass of a cylinder is
    m = πr2pL

    therefore u (m/L) is u = πr2p

    Diameter changes by a factor of 3
    and since volume changes by r2
    wouldn't that mean u changes by a factor of 36?

    or am I complicating it again?
     
  12. Mar 31, 2010 #11

    Doc Al

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    Staff: Mentor

    Sure. But let's keep L' = L/3.

    OK. T' = 5T.

    How did you get 36? Since r' = 3r, r² increases by a factor of 3².

    (Throughout this I assume that the new wire is made from the same material as the original wire, thus it has the same density.)
     
  13. Mar 31, 2010 #12

    Ush

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    But 1 diameter = 2 radius?
    so shouldn't an increase in diameter by 3, increase radius by 6?
    and 62 => 36?
     
  14. Mar 31, 2010 #13

    Doc Al

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    Nope. Both the radius and diameter are proportional to each other. If the area is proportional to r² (as in A = πr²), then it's also proportional to d² (as in A = π(d/2)² = (π/4)d²).
     
  15. Mar 31, 2010 #14

    Ush

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    hm, then u would increase by a factor of 9 correct?

    also, do we have to consider change in length when we're finding u?

    since u = m/L
    and L decreases to 1/3Lo

    would that make unew = 3 uold

    -> b/c
    mold = πd2pL/4

    mnew = π(3d)2p(1/3)L/4

    3u = π(3d)2p /4
    3u = π9d2p /4

    divide by 3 again goes back to

    u = π3d2p /4

    so.. u increases by a factor of 3? :S
     
  16. Mar 31, 2010 #15

    Doc Al

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    Right.

    No. μ is the mass per unit length, so it does not depend on the length.

    That would be true if you kept the same mass and just squished it into 1/3 of its original length. But that's not what we're doing here. We're just using a thicker piece of wire.
     
  17. Mar 31, 2010 #16

    Ush

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    ah, okay

    so that would mean..

    T' = 5T
    u' = 9u
    L' = L/3

    Fold*2L = sqrt(T/u)

    Fnew*(2/3)L = sqrt(5T/9u)

    Fnew / Fold = sqrt5
    Fnew = sqrt5*Fold ?

    is this correct?
     
  18. Mar 31, 2010 #17

    Doc Al

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    Looks good to me.
     
  19. Mar 31, 2010 #18

    Ush

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    thank you so much =]
     
  20. Mar 31, 2010 #19
    Whoops yeah I saw that I was thinking in terms of a sphere not a wire. But you got it now so I just took out the example :)
     
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