What is the lowest-frequency standing wave for which there is a node at the junc

Question

What is the lowest-frequency standing wave for which there is a node at the junction between the two metals? At that frequency, how many antinodes are on the aluminum wire?

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 and , respectively. What is the lowest-frequency standing wave for which there is a node at the junction between the two metals? At that frequency, how many antinodes are on the aluminum wire?

Explanation / Answer

length of Cu wire = l1 = 22 cm     Diameter of cu Wire d1 = 1.0 mm, density 1=8920 kg/m3

length of Al wire = l2 = 60 cm, Diameter of Al wire d2 = 1.0 mm, density 2 = 2700 kg/m3

Tension in the wire = T = 20N

Mass per unit length of cu = m1 = density* vol of unit lenth of Cu wire=1*d1*10^-6*

Mass per unit length of Al = m2 = 2*d2*10^-6*

Say, for the lowest frequency no. of loops in Cu is x and no. of loops in Al is y.

Now,

We have,   frequency n = (x/2l1)*(T/m1) = (y/2l2)*(T/m2)

=> x/y = (l1/l2)*(m1/m2)

=> x/y = (22/60)*(1*d1*10^-6*/2*d2*10^-6*)

=> x/y = (22/60)*(8920/2700)

=> x/y = 2/3

So at lowest frequency Al has 3 antinodes.

lowest frequency n = (3/2*60*10^-2)*sqrt(20/2700*1*10^-6)

                            = 243 Hz

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