Problem 48 - Fundamentals of Physics

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 \(347 \mathrm{~m}\), a linear density of \(3.35 \mathrm{~kg} / \mathrm{m}\), and a tension of \(65.2 \mathrm{MN}\), what are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes?

Step-by-Step Solution

Answer

Fundamental frequency is 6.36 Hz; frequency difference is also 6.36 Hz.

Step 1: Understand the Problem

We are tasked with finding the frequency of the fundamental mode (\(f_1\)) of oscillation of a transmission line and the frequency difference (\(\Delta f\) ). The given parameters include the length (\(L = 347 \, \text{m}\) ), linear density (\(\mu = 3.35 \, \text{kg/m}\) ), and tension (\(T = 65.2 \, \text{MN} = 65.2 \times 10^6 \, \text{N}\) ).

Step 2: Calculate Wave Speed

The speed (\(v\) ) of waves on a string is given by the formula \[ v = \sqrt{\frac{T}{\mu}} \].Plug in the values:\[ v = \sqrt{\frac{65.2 \times 10^6}{3.35}} \].Calculate \(v\) to find the wave speed.

Step 3: Wave Speed Calculation Result

Performing the calculation, we get\[ v = \sqrt{\frac{65.2 \times 10^6}{3.35}} \approx \sqrt{19477611.94} \approx 4412.23 \text{ m/s} \].This is the speed of the wave on the transmission line.

Step 4: Calculate Fundamental Frequency

The fundamental frequency (\(f_1\)) is calculated using the formula \[ f_1 = \frac{v}{2L} \].Substitute \(v = 4412.23 \, \text{m/s}\) and \(L = 347 \, \text{m}\) into the formula:\[ f_1 = \frac{4412.23}{2 \times 347} \].

Step 5: Fundamental Frequency Calculation Result

Perform the calculation:\[ f_1 = \frac{4412.23}{694} \approx 6.36 \text{ Hz} \].This is the frequency of the fundamental mode.

Step 6: Calculate Frequency Difference

The frequency difference between successive modes \(\Delta f\) is given by \[ \Delta f = \frac{v}{2L} \].Since this is the same formula we used to find \(f_1\), the frequency difference between successive modes is \(\Delta f = f_1 \).

Step 7: Frequency Difference Calculation Result

Since \(\Delta f = f_1 = 6.36 \, \text{Hz} \), the frequency difference between successive modes is also \(6.36 \, \text{Hz}\).

Key Concepts

Transmission Line Oscillations

Transmission line oscillations, also known as galloping, occur when there are vibrational movements in electric power lines. These oscillations can be triggered by specific weather conditions, such as ice formation on the lines. The ice alters the line's diameter, creating vortex patterns when wind blows across it. These vortexes can cause the line to oscillate dramatically if the wind's frequency matches one of the line's resonant frequencies. This is a type of physical resonance where external vibrations align with a structure's natural frequencies, potentially leading to large oscillations. These resonant frequencies are closely spaced, especially in long lines, making it easier for various wind speeds to induce potentially destructive modes of vibration.

Wave Speed Calculation

Understanding wave speed on a transmission line is vital for analyzing oscillations. In physics, the wave speed (\(v\)) on a string or line under tension is determined by the formula:\[v = \sqrt{\frac{T}{\mu}}\]where \(T\) is the tension in the line, and \(\mu\) is the linear density. Linear density is the mass per unit length of the string. For example, if a tension of \(65.2 \times 10^6 \, \text{N}\) and a linear density of \(3.35 \, \text{kg/m}\) are given, we can calculate:

• Substitute the values into the formula.
• Calculate the quotient inside the square root.
• Solve for \(v\) to find approximately\(4412.23 \, \text{m/s}\).

This speed tells us how fast oscillations travel through the line.

Fundamental Frequency

The fundamental frequency (\(f_1\)) is the lowest frequency at which a system oscillates. For transmission lines undergoing resonant oscillations, \(f_1\) can be calculated using:\[f_1 = \frac{v}{2L} \]where \(v\) is the wave speed, and \(L\) is the line length. For the given example with \(v = 4412.23 \, \text{m/s}\) and length \(L = 347 \, \text{m}\):

• Plug in the values.
• Perform the division to find\(f_1 = 6.36 \, \text{Hz}\).

This fundamental frequency represents the lowest resonant mode of vibration for the line, where half a wavelength fits along the line's length.

Frequency Difference

Frequency difference (\(\Delta f\)) between successive modes of oscillation on a transmission line is crucial in understanding resonance. This concept explains how closely spaced these resonant frequencies are. For a line, \(\Delta f\) is the frequency increment between one resonant mode and the next. It can be determined using the same formula for the fundamental frequency:\[\Delta f = \frac{v}{2L} \]Here, since \(\Delta f = f_1\), it shows that each additional mode of vibration occurs at an interval equal to the fundamental frequency. Given \(\Delta f = 6.36 \, \text{Hz}\) in this scenario, this constant spacing means that subsequent modes are also 6.36 Hz apart from each other, ensuring a systematic frequency structure for the resonant modes.