Suggested languages for you:

Americas

Europe

Q48P

Expert-verified
Found in: Page 475

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 3.35 kg/m , and a tension of 65.2 MN . What are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes?

1. The frequency of the fundamental mode is 6.36 Hz
2. The frequency difference between successive modes is 6.36 Hz
See the step by step solution

## Step 1: Given data

Length of the transmission line is L = 347 m .

Linear density of the transmission line is $\mu =3.35\mathrm{kg}/\mathrm{m}$ .

Tension of the transmission line is T = 65.2 MN .

## Step 2: Understanding the concept of frequency

We can find the frequency of the fundamental mode using the formula for it. Taking the difference between two successive frequencies obtained from the above formula we can the frequency difference between the successive modes.

Formula:

The frequency of nth mode, $f=\frac{nv}{2L}......\left(1\right)$

The speed of the wave, $v=\sqrt{\left(\frac{T}{\mu }\right)}........\left(2\right)$

## Step 3(a): Calculation of the frequency of the fundamental node

For fundamental mode is n = 1 substituted in equation (1), thus we get the frequency formulas as:

$f=\frac{v}{2L}......\left(3\right)$

Again, using equation (2) and the given values, we het the speed of the wave as given:

$v={\left(\frac{65.2×{10}^{6}}{3.35}\right)}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=4.4116×{10}^{3}\mathrm{m}/\mathrm{s}$

Hence, substituting the value of equation (3) in equation (1), we get the frequency of the fundamental node as:

$f=\frac{4.4116×{10}^{3}}{2×347}\phantom{\rule{0ex}{0ex}}=6.36\mathrm{Hz}$

Hence, the required value of fundamental frequency is 6.36 Hz

## Step 4(b): Calculation of the frequency difference

The frequency difference between two successive modes using equation (1) is given as:

$∆f={f}_{n}-{f}_{n-1}\phantom{\rule{0ex}{0ex}}=n\frac{v}{2L}-\left(n-1\right)\frac{v}{2L}\phantom{\rule{0ex}{0ex}}=\frac{v}{2L}\phantom{\rule{0ex}{0ex}}=\frac{4.4116×{10}^{3}}{2×347}\phantom{\rule{0ex}{0ex}}=6.36\mathrm{Hz}$

Hence, the value of frequency difference between two successive modes is 6.36 Hz