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Frequency in two different wires

  1. Feb 18, 2008 #1
    [SOLVED] Frequency in two different wires

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

    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?

    2. Relevant equations and 3. 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
     
  2. jcsd
  3. Feb 19, 2008 #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.
     
  4. Feb 20, 2008 #3
    Please Help!
     
  5. Feb 21, 2008 #4
    I am lost with the question why is no one helping?
     
  6. Feb 21, 2008 #5

    Doc Al

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    Staff: Mentor

    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.)
     
  7. Feb 21, 2008 #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
     
  8. Feb 21, 2008 #7

    Doc Al

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    Staff: Mentor

    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?)
     
  9. Feb 21, 2008 #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?
     
  10. Feb 21, 2008 #9
    I am lost with what was previously said can anyone help
     
  11. Feb 21, 2008 #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.
     
  12. Apr 14, 2008 #11
    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.
     
  13. Apr 14, 2008 #12

    Doc Al

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    Staff: Mentor

    No, that's perfectly correct.

    Yes, the wavelengths (and thus the size of the loops) are different in each section. How are they related?
     
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