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Standing Waves

  1. Jul 11, 2008 #1
    An aluminum wire of length .6 m, cross sectional area of .01 cm^2, density 2.60 g/cm^3, is tied to a steel wire of density 7.8 g/cm^3 and the same cross sectional area. The compound wire is joined to a pulley and then a block of mass 10 kg is tied at the end of the steel wire. Thisi is arranged so that the distance from the joint (where the 2 wires meet) to the pulley is .866m. Transverse waves are set up at a variable frequency with the pulley as a node. Find the frequency that generates a standing wave having the joint as one of the nodes.

    ---aluminum---x---steel-----------pulley
    |
    |
    |
    10kg mass


    So the length of the aluminum part is .6m and the steel part is .866m.
    So do I find the total mass of the wire, then divide that by the length (1.466m) to find the density. And then I use v = sqrt(tension/density) and then f = n*v/2*L to find the frequency, where L is the length of the wire?
    I don't see how both the .6m and the pulley can be a node? They aren't really multiples of one another, and so I'm wondering how I am to find n?
    Thanks much!
     
    Last edited: Jul 11, 2008
  2. jcsd
  3. Jul 11, 2008 #2
    The picture doesn't appear right; the mass is hanging on the pulley side. Thanks.
     
  4. Jul 11, 2008 #3

    alphysicist

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    Hi bodensee9,

    I don't believe that is right; here you need to treat the aluminum wire and the steel wire separately. But think about what quantities will be the same for the two wires.

    The two wires will have different linear mass densities; do you see how that will help?
     
  5. Jul 11, 2008 #4
    Hello:
    I think the tension is the same in both parts of the wire because they are connected? If I treat the two wires separately, wouldn't they have a different v? I am wondering though how the cross section area is relevant in that case ...
    Thanks!
     
  6. Jul 11, 2008 #5

    alphysicist

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    That's right: the tension is the same in both wires. You are also right that the velocity of the wave in the different wires will also be different; but what is the same for the wave in the steel and the wave in the aluminum?

    I think once you start calculating, you'll probably see where the cross-sectional area becomes important.
     
  7. Jul 11, 2008 #6
    Hello:

    Do the wavelengths have to be the same for both waves? Thanks.
     
  8. Jul 11, 2008 #7

    alphysicist

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    No, I don't think so. And there's no problem with the wavelength being different in each wire; since each wire has a node on each end (counting the pulley as one of the ends), it's almost like each wire is a separate problem. But what has to be the same, and why?
     
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