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Two wires are made of the same material but the second wire has twice the diameter and twice the length of the first wire. When the two wires are stretched and the tension in the second wire is also twice the tension in the first wire, the fundamental frequency of the first wire is 480 Hz. What is the fundamental frequency of the second wire? Answer in units of Hz.

The speed of a wave in a wire is given by:

v=sqrt(T/mu) where T is the tension and mu is the mass per unit length.

The wire with twice the diameter will have 4 times the mass per unit length since doubling the diameter will quadruple the volume and therefore the mass of the wire. Therefore:

mu2a=4*mu1

As is given the tension in the 2nd wire is twice the tension in the first wire as well:

T2=2*T1

This means that the wave speed in the second wire will be:

v2=sqrt(2*T/4*mu)=1/sqrt(2)*sqrt(T/mu)=1/sqrt(2)*v1

Since the second wire has twice the length, the corresponding wavelength in the second wire must be twice as great:

L2=2*L1

Finally, therefore, the frequency in the first wire will be:

f1=v1/L1

While the frequency in the 2nd wire will be:

f2=v2/L2=1/sqrt(2)*v1/2*L1=1/[2*sqrt(2)]*v1/L11=1/[2*sqrt(2)]*f1=0.354*f1=0.354*480=170Hz

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.