Question

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind. The air pressure variations in the vortexes tend to cause the line to oscillate (gallop), especially if the frequency of the variations matches a resonant frequency of the line. In long lines, the resonant frequencies are so close that almost any wind speed can set up a resonant mode vigorous enough to pull down support towers or cause the line to short out with an adjacent line. If a transmission line has a length of 347 m, a linear density of 4.35 kg/m, and a tension of 65.4 MN, what are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes

Answer 1

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length L, linear density $\mu$ under a tension T is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the fundamental mode (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The frequency difference between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$