Length of a Steel Wire (Standing Waves)

[smashd]

Question: A copper wire and steel wire with identical diameters are placed under identical tensions. The frequency of the third resonant mode for the copper wire is found to be the same as the frequency of the fourth resonant mode for the steel wire. If the length of the copper wire is 3.44 m and the density of steel is 7.85 g/cc, find the length of the steel wire.

Known Values: $~ ~ ~ ~ ~ L_c = 3.44 ~ m ~ ~ ~ ~ ~ , ~ ~ ~ ~ ~ ρ_s = 7.85 ~ g/cc$

Relevant Equations: $~ ~ ~ ~ ~ ƒ_n = \frac{v}{λ_n} ~ ~ ~ ~ ~ , ~ ~ ~ ~ ~ λ_n = \frac{2L}{n} ~ ~ ~ ~ ~ ~ , ~ ~ ~ ~ v = \sqrt{\frac{T}{μ}}$

Attempt at a Solution:

ƒ_{3-c} = ƒ_{4-s}
\frac{v_c}{λ_{3-c}} = \frac{v_s}{λ_{4-s}}
\frac{\sqrt{\frac{T}{μ_c}}}{λ_{3-c}} = \frac{\sqrt{\frac{T}{μ_s}}}{λ_{4-s}}

μ_c = \frac{m_c}{L_c} ~ ~ ~ , ~ ~ ~ ρ_c = \frac{m_c}{V_c} ~ ~ ~ ⇒ ~ ~ ~ m_c = ρ_c ~ V_c = ρ_c ~ π ~ r^2 ~ L_c

μ_c = ρ_c ~ π ~ r^2​

μ_c = \frac{m_s}{L_s} ~ ~ ~ , ~ ~ ~ ρ_s = \frac{m_s}{V_s} ~ ~ ~ ⇒ ~ ~ ~ m_s = ρ_s ~ V_s = ρ_s ~ π ~ r^2 ~ L_s

μ_s = ρ_s ~ π ~ r^2​

\frac{\sqrt{\frac{T}{ρ_c ~ π ~ r^2}}}{\frac{2}{3} ~ L_c} = \frac{\sqrt{\frac{T}{ρ_s ~ π ~ r^2}}}{\frac{1}{2} ~ L_s}
\frac{\frac{T}{ρ_c ~ π ~ r^2}}{\frac{4}{9} ~ L_{c}^2} = \frac{\frac{T}{ρ_s ~ π ~ r^2}}{\frac{1}{4} ~ L_{s}^2}
\frac{T}{\frac{4}{9} ~ ρ_c ~ π ~ r^2 ~ L_{c}^2} = \frac{T}{\frac{1}{4} ~ ρ_s ~ π ~ r^2 ~ L_{s}^2}
\frac{4}{9} ~ ρ_c ~ L_{c}^2 = \frac{1}{4} ~ ρ_s ~ L_{s}^2
∴ ~ L_{s} = \frac{4}{3} ~ L_{c} ~ \sqrt{\frac{ρ_c}{ρ_s}}

My problem is that I don't know what $ρ_c$ is. Any ideas how to solve this? Am I even going about this the right way?

 

[smashd]

Meh... The density of copper was something that's just supposed to be looked up. I thought I was supposed to solve for it another way. It's 8.96 g/cc.

Never mind, problem solved!