# Compound Wire - Standing Wave

#### Stealth849

1. Homework Statement

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. Homework 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!

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#### Simon Bridge

Homework Helper
When it asks for the lowest frequency of excitation, I expect that means the second harmonic
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.

#### Stealth849

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

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#### Simon Bridge

Homework Helper
Well done.
The other approach was to solve the wave equation directly :)

#### Stealth849

Thanks!

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?

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#### Simon Bridge

Homework Helper
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?

#### Stealth849

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?

Homework Helper

#### Stealth849

Deriving the 1D case from Hook's Law used to be a college-level exercise...
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...?