Standing Waves

  • Thread starter bodensee9
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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
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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!
 
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  • #2
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The picture doesn't appear right; the mass is hanging on the pulley side. Thanks.
 
  • #3
alphysicist
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Hi bodensee9,

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.
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.

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!
The two wires will have different linear mass densities; do you see how that will help?
 
  • #4
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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!
 
  • #5
alphysicist
Homework Helper
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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!
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.
 
  • #6
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Hello:

Do the wavelengths have to be the same for both waves? Thanks.
 
  • #7
alphysicist
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Hello:

Do the wavelengths have to be the same for both waves? Thanks.
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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