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Wire problem

  1. Sep 30, 2005 #1
    A wire with mass (m) is stretched so that its ends are tied down at points a distance (s) apart. The wire vibrates in its fundamental mode with frequency (f) and with an amplitude at the antinodes of (A).

    [] How would I compute for the tension of the wire?

    since f = (1/L)sqrt(Tension/omega) and in this problem L = s

    so Tension = (sf)^2/omega?

    Would that be right?

    [] What is the speed of propagation of transverse waves in the wire?

    Would that just be Velocity = fs ?

    since f = v/L and L = s in this problem


    Thank you for your help
     
  2. jcsd
  3. Sep 30, 2005 #2

    quasar987

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    It should be written somewhere in your textbook that the velocity at which the transverse waves propagate is

    [tex]v = \sqrt{\frac{T}{\mu}}[/tex]

    Use this along with

    [tex]v = \lambda f[/tex]

    to find the answer to both question.

    Note: [itex]\lambda \neq s[/itex]. [itex]\lambda[/itex] is the spatial period (wavelenght) period of the sine wave corresponding to the given mode. In the fundamental mode, [itex]\lambda[/itex] is greater than s. I let you figure ou how much bigger.


    Edit: You might also be wondering v is the speed of what in the case of standing waves such as the fundamental mode. The fact is that standing waves can be seen as the sum of 2 identical sine waves voyaging in opposite direction. So there is no problem using the concept of "speed of a transverse wave" in solving a problem concerning normal modes/stationary waves, since they are just the sum of two traveling waves voyaging at speed v. :smile:
     
    Last edited: Sep 30, 2005
  4. Sep 30, 2005 #3
    For the first part, solving for T , I get Tension = (lambdaf)^2/(m/s)

    Second part, Im still stuck with (lambda)f


    Does this mean I have to solve for lambda in using terms such as the s?
     
    Last edited: Sep 30, 2005
  5. Sep 30, 2005 #4

    quasar987

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    You must always answer in terms of the quantities you're given! Here, m, s, f, A.

    Of course you can always express lambda in terms of the rope lenghts. The "conversion formula" ought to be somewhere in your book. If it's not written in terms of lambda, maybe it is written in terms of k. Since k = lambda/2pi, you can easily recover the formula in terms of lambda.

    But here you don't even need the general formula. Just by looking at the shape of the fundamental mode and using the info I gave in first post, you should be able to find the wavelenght.

    And I believe your answer for T is incorect. Did you make use of [itex]v = \sqrt{T/\mu}[/itex] ?!?
     
  6. Sep 30, 2005 #5
    Whoops, I meant Tension = (m/s)((lambda)f)^2

    From my notes, I have k= 2pi/(lambda) = 2pif/v

    Then I tried

    Since v = omega/k

    (m/s)(omega/k)^2 = Tension

    Since omega = 2pi/f

    (m/s)(2pif/k)^2 = Tension

    now I am stuck with k

    I cant seem to find a way that allows k or lambda to be expressed with the quantities I am given; m, s, f, A.



    Another formula I found in the book was f = (1/2L)sqrt(Tension/(m/s))
    Now if L = s

    then Tension = (m/s)(2sf)^2 Would that be the answer to my first question?


    Thank you for your help
     
  7. Sep 30, 2005 #6

    quasar987

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    With those 2 expressions of the tension you found independantly, aren't you tempted to take a guess at what lambda is in terms of s?

    Then just by drawing the modes, find what lambda is for the second normal mode. Then for the third and fourth normal mode. By then, you should be able to find the general formula of lambda in terms of s and of the mode number n.
     
    Last edited: Sep 30, 2005
  8. Oct 1, 2005 #7
    humm so the general formula for lambda should be ns right?

    so that would make Tension = (m/s)(nsf)^2 ?

    however, in this problem is it possible to find the value of n?
     
  9. Oct 1, 2005 #8

    quasar987

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    Fundamental mode is a synonym of n = 1.

    Btw that is not the formula for lambda. If you look at the fundamental mode, you see that if you imagine the sine wave to continue past the walls where the rope is fixed, you see that it would complete a full cycle at x = 2s. So [itex]\lambda_1 = 2s[/itex]. In the 2nd natural mode, the sine wave fits exactly within the walls. So [itex]\lambda_2 = s[/itex]. For the third mode, you see that the sine wave as complete a full period at 2/3 of the rope lenght. So [itex]\lambda_3 = 2L/3[/itex]. Etc, [itex]\lambda_4 = s/2[/itex], [itex]\lambda_5 = 2s/5[/itex], ...

    Do you see the patern?
     
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