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

a)
`f_1=5.587Hz`

b)
`f_(n+1)-f_n=5.587Hz`

Step-by-step 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=(n)/(2L) \sqrt{(T)/(\mu)`

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

`f_1=(1)/(2(347m)) \sqrt{(65.4*10^6N)/(4.35kg/m)}=5.587Hz`

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

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