What is the lowest-frequency standing wave

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To determine the lowest-frequency standing wave with a node at the junction of a copper and aluminum wire, the tension and linear mass density (μ) of both materials must be considered. The user initially calculated the wavelength using the total length of the wire and derived the frequency from the wave speed equation. However, confusion arose regarding which material's density to use for calculating μ, as using aluminum's density did not yield the correct frequency. The correct approach involves calculating the linear mass density for both wires based on their respective densities and lengths, then finding the effective μ for the combined system. Ultimately, the solution requires careful consideration of the properties of both metals to accurately determine the standing wave characteristics.
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Homework Statement


A 22-cm-long, 1.0-mm-diameter copper wire is joined smoothly to a 60-cm-long, 1.0-mm-diameter aluminum wire. The resulting wire is stretched with 20 N of tension between fixed supports 82 cm apart. The densities of copper and aluminum are 8920kg/m^3 and , 2700 kg/m^3 respectively.

a) What is the lowest-frequency standing wave for which there is a node at the junction between the two metals?

b) At that frequency, how many antinodes are on the aluminum wire?

Homework Equations


v = sqrt(T/mu)
v = lambda*f


The Attempt at a Solution


lambda = 2L
L = 0.82m
lambda = 1.64m

v = lambda*f
f = v / lambda
f = v / (1.64m)

v = sqrt(T/mu)
v = sqrt(20/mu)

d = mass / volume

mu = mass / length

ok...now the problem is, i dont' know which denstity (copper or aluminum) to use to figure out the mu. i tried using aluminum's since it's lower and hence will give the lower frequency, but it did not give the right answer. please help! thanks
 
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