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Compound Wire - Standing Wave

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    An aluminum wire of length l1 = 60cm, cross section area 1.00 x 10-2cm2, and density 2.60g/cm3, is joined to a steel wire of density 7.8g/cm3 and same cross sectional area. The compound wire loaded with a block of mass m = 10kg is arranged as shown so that the distance l2from the joint to the supporting pulley is 86.6cm. Transverse waves are set up in the wire using an external source of variable frequency. A node is located at the pulley.

    a) find the lowest frequency of excitation for which a standing wave is observed to have a node at the joint.

    b) How many nodes are observed at this frequency?

    2. Relevant equations

    f = v/λ = nv/2l

    ρ = m/V

    v = sqrt(FT/μ)

    3. The attempt at a solution
    Not too sure where to begin on this one. When it asks for the lowest frequency of excitation, I expect that means the second harmonic, because that would leave a node at the joint with the lowest frequency.

    So knowing the densities of the wires, I can get v where

    FT = μv2
    FT = mg = 98N

    μ = m/l = ρV/l = ρA where A represents the cross sectional area.

    so μ1 = 2600*0.0001 = 0.26

    μ2 = 7800*0.0001 = 0.78

    v1 = sqrt(98/0.26) = 19.4m/s
    v2 = sqrt(98/0.78) = 11.2m/s

    so f = nv/2l

    where n = 2 for the second harmonic

    f1 = 2*0.26/(2*0.6) = 0.43Hz

    f2 = 2*0.78/(2*0.866) = 0.90Hz

    No idea if I'm approaching this correctly, due to the idea of reflection of waves at the joint.. Say a wave does reflect at the joint and a portion of it is sent back to the source, there would be a lot more waves thus being a much higher harmonic - not the second.

    Any and all help appreciated. Thanks!

    Attached Files:

    Last edited: Jan 24, 2013
  2. jcsd
  3. Jan 23, 2013 #2

    Simon Bridge

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    er ... that would be the case if the wave-speed is the same in each section of wire.

    Think in terms of the modes in each section of wire, if the sections were separate.
  4. Jan 23, 2013 #3
    Okay.. So if I know that

    nv/2l = f

    and am treating each section separately, except for the fact that frequency must be the same for each section, could I say that

    nv1/2l1 = mv2/2l2

    essentially setting frequency equal to frequency, but being able to solve for the number of nodes, or at least a ratio of nodes...?

    as in

    n/m = l1*v2/l2*v1

    Thanks for the reply :)
    Last edited: Jan 23, 2013
  5. Jan 24, 2013 #4

    Simon Bridge

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    Well done.
    The other approach was to solve the wave equation directly :)
  6. Jan 24, 2013 #5

    I don't see how I could solve the wave equation directly as you said, however. Can you define what you mean by wave equation?

    Using what I mentioned previously...

    n/m = l1*v2/l2*v1
    n/m = 0.6*11.2/(0.866*19.4) which is roughly 2/5

    So this means that the aluminum section, l_1, has 2 nodes and the steel section has 5 nodes, yes?

    and frequency can be calculated by either side,

    f = nv/2l
    f = 19.4/0.6 = 32.3Hz
    f = 5*11.2/2*0.866 = 32.3Hz

    and the total number of nodes will be 2 + 5 = 7?
    Last edited: Jan 24, 2013
  7. Jan 24, 2013 #6

    Simon Bridge

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    Wave equation:
    $$\frac{\partial^2}{\partial x^2}y(x,y) = \frac{1}{c^2}\frac{\partial^2}{\partial t^2}y(x,t)$$ ... in your case, the wave-speed, c, depends on x so it will need some modification.

    Since ##c=f\lambda##, which will have the longer wavelength for the same frequency? The fast one or the slow one?
  8. Jan 24, 2013 #7
    Well, if c is proportional to to wavelength, the faster one should have the longer wavelength...no?

    How did you get to that wave equation? Or is that a standard model for a wave?
  9. Jan 24, 2013 #8

    Simon Bridge

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    That's how you check your answers make sense :)
    Deriving the 1D case from Hook's Law used to be a college-level exercise...
  10. Jan 24, 2013 #9
    I'm sure it still is... Looking forward to learning it eventually when college rolls around. :)

    Do you mind giving me a quick introduction to how that equation works...?
  11. Jan 24, 2013 #10

    Simon Bridge

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    It's a second order partial differential equation - if we consider y(x,t) to be a transverse displacement of a point on the string at position x along the string, then it says that the way a point on a string accelerates (the RHS) depends on how the string around that point is curved (the LHS).

    Any wave has to be consistent with that equation, and the external "boundary conditions" - like being fixed at both ends and having a node in a particular place.
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