How Do Wave Velocities and Frequencies Change in a Compound Wire System?

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In a compound wire system consisting of aluminium and steel, the velocities of transverse waves are determined by their respective densities and the applied tension. The frequency of vibration must remain constant across both materials, allowing for the relationship between the number of half-wavelengths in each section to be established. The lowest frequency standing wave condition requires that the number of half-wavelengths in each wire be minimized while maintaining the node at the joint. Calculations indicate that the relationship between the half-wavelengths can be expressed as n2 = 2.5n1, leading to integer values for n1 and n2. The differing velocities result in varying wavelengths, but the frequency remains uniform throughout the system.
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Homework Statement



(a) An aluminium wire of length L1, cross-sectional area A and density ρ1 is connected to
a steel wire with the same cross-sectional area and density ρ2. This compound wire,
loaded with a block of mass m, is arranged as shown below so that the distance between
the joint and the supporting pulley is L2. Transverse waves are set up on the wire by using
an external source of variable frequency.(i) What are the velocities of the waves on the aluminium, v1, and steel wires, v2?

(ii) If we require that the joint is a node find the frequency of the wave on each part of the
wire in terms of the number of half wavelengths on that part of the wire, n1 on the
aluminium and n2 on the steel wires.

(iii) Given that L1 = 60.0 cm, L2 = 86.6 cm, ρ1 = 2.60 g cm-3, ρ2 = 7.86 g cm-3, A = 1:00 x
10-2 cm2 and m = 10.0 kg how many half-wavelengths are there on each part of the wire
for the lowest frequency standing wave such that the joint is a node?

(iv) What is the frequency of this vibration?

[Hint: In (a) (i) and i) the expressions should be in terms of in terms of the area A,
densities ρ1 and ρ2, the lengths L1 and L2, the number of half-wavelengths n1 and n2,mass
of the block m, and g the acceleration due to gravity.]

Homework Equations


\mu=A*\rho1
where \mu is the linear mass density

V=\sqrt{(F)/\mu}
where V is velocity
f1=\frac{Vn}{2L}

The Attempt at a Solution



i) using the first equation for velocity I get: V1=\sqrt{(mg)/A\rho1} and the same for V2

ii)I then plug into the second and get:
f1=\frac{n1}{2L1}\sqrt{\frac{mg}{A\rho}} with the same for f2

iii) This is where I run into problems:

For aluminium: f_low= f1=f1=\frac{V}{2L} because the largest wavelength is going to be 2L
therefore I get f1=\frac{V}{2L}=f1=\frac{Vn}{2L}
so n1=n2=1. Irrespective of the numbers they give us. This isn't right is it?

Homework Statement


Homework Equations


The Attempt at a Solution

 
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gboff21 said:
iii) This is where I run into problems:

For aluminium: f_low= f1=f1=\frac{V}{2L} because the largest wavelength is going to be 2L
therefore I get f1=\frac{V}{2L}=f1=\frac{Vn}{2L}
so n1=n2=1. Irrespective of the numbers they give us. This isn't right is it?
No, it isn't.

Hint: for something to be vibrating at some frequency, what is true about the frequency of vibration at different parts of the object?
 
Hi Redbelly

I know this thread is a year old but I'm working on the same problem at the moment. Does your hint mean that the frequency of vibration is the same throughout the wire regardless of the fact that it is a compound wire? So you can set f1=f2 and work out n1 in terms of n2 but how can you then work out what n1 and n2 is?

Thanks
 
dbatten said:
Hi Redbelly

I know this thread is a year old but I'm working on the same problem at the moment. Does your hint mean that the frequency of vibration is the same throughout the wire regardless of the fact that it is a compound wire? So you can set f1=f2 and work out n1 in terms of n2...
Yes, correct.
... but how can you then work out what n1 and n2 is?
They want the lowest possible (nonzero) frequency, so n1 and n2 must be as small as possible, but not zero.
 
Redbelly98 said:
Yes, correct.

They want the lowest possible (nonzero) frequency, so n1 and n2 must be as small as possible, but not zero.

Ok so I worked out that n1 is roughly 2.5n2 using the data given so can I just let n1 = 2 and n2 equal 5?
 
dbatten said:
Ok so I worked out that n1 is roughly 2.5n2 using the data given so can I just let n1 = 2 and n2 equal 5?
I think you have it backwards somewhere but yes, that is the idea.
 
Redbelly98 said:
I think you have it backwards somewhere but yes, that is the idea.

Oh yes I got it the wrong way around. n2 = 2.5n1. Thanks for your help
 
On a related note, just to make sure I have the concept: Each part of the 'string' would have a different wavelength due to the differing velocities v=SQRT(mg/Aρ), with ρ being different for each, right? Only the frequency would necessarily be constant?
 
NotACrook said:
On a related note, just to make sure I have the concept: Each part of the 'string' would have a different wavelength due to the differing velocities v=SQRT(mg/Aρ), with ρ being different for each, right? Only the frequency would necessarily be constant?
Yes, correct.
 
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