(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

v = sqrt(T/mu)

v = lambda*f

3. 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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# Homework Help: Wave Question

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