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neoking77
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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