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1. Homework Statement
An aluminum wire of length l_{1} = 60cm, cross section area 1.00 x 10^{2}cm^{2}, and density 2.60g/cm^{3}, is joined to a steel wire of density 7.8g/cm^{3} and same cross sectional area. The compound wire loaded with a block of mass m = 10kg is arranged as shown so that the distance l_{2}from 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(F_{T}/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
F_{T} = μv^{2}
F_{T} = mg = 98N
μ = m/l = ρV/l = ρA where A represents the cross sectional area.
so μ_{1} = 2600*0.0001 = 0.26
μ_{2} = 7800*0.0001 = 0.78
v_{1} = sqrt(98/0.26) = 19.4m/s
v_{2} = sqrt(98/0.78) = 11.2m/s
so f = nv/2l
where n = 2 for the second harmonic
f_{1} = 2*0.26/(2*0.6) = 0.43Hz
f_{2} = 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!
An aluminum wire of length l_{1} = 60cm, cross section area 1.00 x 10^{2}cm^{2}, and density 2.60g/cm^{3}, is joined to a steel wire of density 7.8g/cm^{3} and same cross sectional area. The compound wire loaded with a block of mass m = 10kg is arranged as shown so that the distance l_{2}from 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(F_{T}/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
F_{T} = μv^{2}
F_{T} = mg = 98N
μ = m/l = ρV/l = ρA where A represents the cross sectional area.
so μ_{1} = 2600*0.0001 = 0.26
μ_{2} = 7800*0.0001 = 0.78
v_{1} = sqrt(98/0.26) = 19.4m/s
v_{2} = sqrt(98/0.78) = 11.2m/s
so f = nv/2l
where n = 2 for the second harmonic
f_{1} = 2*0.26/(2*0.6) = 0.43Hz
f_{2} = 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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