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Frequency in two different wires

  • #1
[SOLVED] Frequency in two different wires

Homework Statement



An aluminum wire. of length L1=60.0cm, cross-section area 1.00 x 10^-2 cm^2, and density 2.60g/cm^3, is joined to a steel wire, of density 7.80g/cm^3 and the same cross-section area. The compound wire, loaded with a block of mass m-10.0kg, is arranged so that the distance L2=86.6cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency?

Homework Equations

and

The Attempt at a Solution



f= v/(wavelength) = (n x v)/(2L)
v= (F/(linear density))

I know the F is 10.0kg x 9.80m/s^2 = 98.0N
...for the velocities do you do them saperate or together
...how do you get the wavelength
 

Answers and Replies

  • #2
Day and hours have pasted and I have gotten not, just need to know if anyone have some suggestions on the approach to get me started.
 
  • #3
Please Help!
 
  • #4
I am lost with the question why is no one helping?
 
  • #5
Doc Al
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The wave speed is different on each stretch of wire. Once you figure out the wave speeds, find the wavelengths using [itex]v = f \lambda[/itex].

(Look up the formula for wave speed; yours is not quite right.)
 
  • #6
I don't know the frequency or wavelength

v= (F/(linear density))^(.5)

v= ([10.0 x 9.8]/[2.60 x 1.00 x 10^-2])^(.5) = 61.4
v= ([10.0 x 9.8]/[7.80 x 1.00 x 10^-2])^(.5) = 35.4

I can get that but then I am lost
 
  • #7
Doc Al
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I don't know the frequency or wavelength
The frequency is what you're asked to find. Since the frequency is constant over both segments of wire, what can you say about the ratio of the wavelength?

Then you just have to play around until you find a set of wavelengths that fit the conditions for a standing wave with a node at one end and at the joint. (Which end is attached to the source?)
 
  • #8
I don't know which end is attached to the source, what do I do with the nodes because part b ask for them?
 
  • #9
I am lost with what was previously said can anyone help
 
  • #10
Doc Al
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Assume that you have nodes at each end and at the joint. Hint: You need to have an integral number of "loops" in each wire section to have a standing wave matching those conditions. How do the number of loops in each section compare? (Find their ratio.) Then figure out the smallest values that will give you an integral number of loops in both sections. Use that to find the frequency.
 
  • #11
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Can't make sense

I have found how many nodes there are(I used the idea that L2 = 1.44 x L1, so maybe this is where I went wrong.) and how long the wavelength is. I am not sure I follow what is wrong with LandOfStandar's speed formula, but that may be my problem. Since I know the number of wavelengths, I should be able to use this to tell me frequency. I have different wave speeds in L1 and L2 which is what makes me confused.

my textbook says that v = wavelength x frequency

From the sound of Doc Al's posts this is incorrect?

If the two speeds are different then how can then this says to have the same frequency, the length of wavelengths must be different. If this is the case then I'm not sure how to relate the number of 'loops' that make up each length.
 
  • #12
Doc Al
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my textbook says that v = wavelength x frequency

From the sound of Doc Al's posts this is incorrect?
No, that's perfectly correct.

If the two speeds are different then how can then this says to have the same frequency, the length of wavelengths must be different. If this is the case then I'm not sure how to relate the number of 'loops' that make up each length.
Yes, the wavelengths (and thus the size of the loops) are different in each section. How are they related?
 

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