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Q48P

Expert-verifiedFound in: Page 475

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**

- The frequency of the fundamental mode is 6.36 Hz
- The frequency difference between successive modes is 6.36 Hz

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 .

**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, $v=\sqrt{\left(\frac{T}{\mu}\right)}........\left(2\right)$

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:

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

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

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

Hence, the required value of fundamental frequency is 6.36 Hz

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

$\u2206f={f}_{n}-{f}_{n-1}\phantom{\rule{0ex}{0ex}}=n\frac{v}{2L}-\left(n-1\right)\frac{v}{2L}\phantom{\rule{0ex}{0ex}}=\frac{v}{2L}\phantom{\rule{0ex}{0ex}}=\frac{4.4116\times {10}^{3}}{2\times 347}\phantom{\rule{0ex}{0ex}}=6.36\mathrm{Hz}$

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

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