What is the lowest-frequency standing wave

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SUMMARY

The discussion focuses on calculating the lowest-frequency standing wave in a composite wire made of copper and aluminum. The wire consists of a 22-cm copper segment and a 60-cm aluminum segment, stretched under a tension of 20 N between fixed supports 82 cm apart. The correct approach involves determining the linear mass density (μ) for both materials, where the density of copper is 8920 kg/m³ and aluminum is 2700 kg/m³. The lowest frequency standing wave occurs at the junction of the two metals, requiring the use of the appropriate μ for each segment to find the correct frequency and number of antinodes in the aluminum wire.

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  • Knowledge of tension and its effect on wave speed
  • Basic algebra for solving equations involving frequency and wavelength
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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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