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To review the concepts that play roles in this problem, consult Multiple- Concept Example 4. Sometimes, when the wind blows across a long wire, a low- frequency "moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density \(=0.0140\) \(\mathrm{kg} / \mathrm{m}\) ) sustains a tension of 323 N because the wire is stretched between two poles that are \(7.60\) \(\mathrm{m}\) apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number \(n\) that could be responsible for the "moaning" sound?

Short Answer

Expert verified
The lowest harmonic number responsible is 2.

Step by step solution

01

Understand the Relationship Between Velocity and Frequency

The velocity of a wave on a string is determined by the tension in the string and the linear density (mass per unit length). This velocity equation is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the linear density. This velocity is also related to the frequency and wavelength by the equation \( v = f \cdot \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength.
02

Calculate the Velocity of the Wave on the Wire

Substitute \( T = 323 \text{ N} \) and \( \mu = 0.0140 \text{ kg/m} \) into the velocity formula to find the velocity of the wave:\[v = \sqrt{\frac{323}{0.0140}} \approx 151.6 \text{ m/s}.\]
03

Determine Wavelength for Standing Wave

For a string fixed at both ends, the wavelength \( \lambda \) is given by \( \lambda = \frac{2L}{n} \), where \( L \) is the length of the wire and \( n \) is the harmonic number. Here, \( L = 7.60 \text{ m} \). Substitute these values to express \( \lambda \) in terms of \( n \):\[\lambda = \frac{2 \times 7.60}{n} = \frac{15.2}{n} \text{ m.}\]
04

Express Frequency in Terms of Harmonic Number

The frequency \( f \) of the standing wave is tied to the velocity and wavelength by \( f = \frac{v}{\lambda} \). Substitute for \( \lambda \) from the previous step:\[ f = \frac{151.6}{\frac{15.2}{n}} = \frac{151.6 \cdot n}{15.2}. \]
05

Solve for the Lowest Harmonic n Corresponding to Audible Frequency

The minimum frequency of sound a human ear can detect is 20.0 Hz. Set the expression derived for frequency equal to 20 Hz to find the smallest harmonic number:\[\frac{151.6 \cdot n}{15.2} = 20. \]Solve for \( n \):\[151.6n = 20 \times 15.2\]\[151.6n = 304\]\[n = \frac{304}{151.6} \approx 2. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Harmonics
Harmonics are an essential concept when discussing standing waves, such as those found on strings or wires. They represent the fixed patterns of vibration that occur at specific frequencies. When a wave is reflected back along a medium such as a string under tension, constructive and destructive interference create a standing wave pattern. This pattern can be represented by different harmonic frequencies, each a whole number multiple of the fundamental frequency.
A standing wave is formed by nodes, where there's no movement, and antinodes, where the movement is largest. The first harmonic corresponds to the fundamental frequency, usually the lowest and longest wavelength. Higher harmonics, also called overtones, occur at frequencies that are integer multiples of the fundamental. In a string with fixed ends, harmonics directly relate to the number of antinodes along the string. For instance:
  • 1st harmonic: one antinode
  • 2nd harmonic: two antinodes
  • 3rd harmonic: three antinodes
In the problem, we need to find which harmonic number produces a `moaning' sound that is detectable by the human ear. Therefore, we solve for which harmonic's frequency first exceeds the minimum audible frequency of 20 Hz.
Wave Velocity
Wave velocity is a critical concept in understanding how waves travel through a medium. For mechanical waves traveling along a string or wire, the wave velocity (\(v\)) is determined by the tension (\(T\)) in the string and its linear density (\(\mu\), the mass per unit length). The relationship is expressed as:
  • \(v = \sqrt{\frac{T}{\mu}}\)
From the exercise, we know the tension in the wire is 323 N, and the linear density is 0.0140 kg/m. Using these values, we can calculate the velocity:\[v = \sqrt{\frac{323}{0.0140}} \approx 151.6 \text{ m/s}\]This velocity not only indicates how fast a wave can move along the wire, it also helps us derive the frequency and wavelength of the standing waves that form.
String Tension
String tension plays a vital role in determining both the velocity of the wave on a string as well as the properties of the resulting harmonics. Tension is the force applied to elongate the string, and more tension results in a tighter string, causing waves to travel faster. The fundamental relationship involving tension is encapsulated in the wave velocity equation:\[v = \sqrt{\frac{T}{\mu}}\]Where \(T\) is the tension, and \(\mu\) is the linear mass density. In scenarios like guitar strings or wires subjected to wind, higher tension typically raises the pitch of the sound by increasing the wave velocity.
In our exercise, the tension of 323 N results in the calculated wave velocity and subsequently affects the harmonics and audible sounds produced. Therefore, managing tension can significantly influence sound characteristics in musical and technical applications.
Audible Frequency
Audible frequency refers to the range of sound frequencies that the average human ear can detect. Typically, this range lies between 20 Hz and 20,000 Hz. In the context of waves, this concept is important when determining the perceptibility of sound produced by standing waves.
When a wave `moans', it's creating a frequency within the audible range of human hearing. In the problem, we are required to ensure the harmonic frequency is at least 20 Hz, the lower threshold of human audibility. The lowest harmonic number in this case ensures the frequency of the wave is detectable by human ears.
Solving for the smallest harmonic ensures that the frequency computed from the equation:\[f = \frac{151.6 \cdot n}{15.2} = \frac{15.6n}{15.2}\text{ Hz}\]becomes equal to or exceeds 20 Hz. Calculating this, we find the minimum harmonic number \(n\) that results in an audible frequency is approximately 2.

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Most popular questions from this chapter

A thin 1.2 -m aluminum rod sustains a longitudinal standing wave with vibration antinodes at each end of the rod. There are no other antinodes. The density and Young's modulus of aluminum are, respectively, \(2700\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(6.9 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) What is the frequency of the rod's vibration?

A string is fixed at both ends and is vibrating at \(130\) \(\mathrm{Hz},\) which is its third harmonic frequency. The linear density of the string is \(5.6 \times 10^{-3} \mathrm{kg} / \mathrm{m}\), and it is under a tension of \(3.3\) \(\mathrm{N}\). Determine the length of the string.

Two tubes of gas are identical and are open only at one end. One tube contains neon (Ne) and the other krypton (Kr). Both are monatomic gases, have the same temperature, and may be assumed to be ideal gases. The fundamental frequency of the tube containing neon is 481 Hz. Concepts: (i) For a gas-filled tube open only at one end, the fundamental frequency \((n=1)\) is \(f_{1}=v /(4 L),\) where \(v\) is the speed of sound and \(L\) is the length of the tube. How is the speed related to the properties of the gas? (ii) All of the factors that affect the speed of sound in this problem are the same except for the atomic masses, which are given by 20.180 u for neon, and 83.80 u for krypton. Is the speed of sound in krypton greater than, smaller than, or equal to the speed of sound in neon? Why? (iii) Is the fundamental frequency of the tube containing krypton greater than, less than, or equal to the fundamental frequency of the tube containing neon? Explain. Calculations: What is the fundamental frequency of the tube containing krypton?

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262\) \(\mathrm{Hz}\). The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\). What is the length of the pipe?

You and your team are exploring an antiquated research facility in the mountains of southern Argentina that had been abandoned in the 1960 s. You come to a giant locked door that has no visible handles or actuators, but you find a hand-held device nearby that has two buttons, labeled "Open" and "Close." It looks like some kind of crude remote control, but when you push the buttons they just click and nothing happens. You open the device's top cover and inspect it. The internal mechanism resembles an old acoustic remote control called the "Space Command \(600 "\) that your parents had for their ancient TV. Pushing a button on the remote actuated a small hammer on the inside that struck the end of an aluminum rod, about 2 or \(3 \mathrm{cm}\) in length. The rod vibrated and emitted an ultrasonic sound wave that actuated an electrical circuit in the TV that was sensitive to that frequency. The TV remote had three buttons, and therefore three rods that vibrated at different frequencies. The first frequency turned the \(\mathrm{TV}\) on and off, the second made the tuning dial click to the next station, and the third made the dial turn in the opposite direction. You pull off the cover of the device and find that it has places for two \(1 / 4\) -inch diameter rods, but both are missing. However, written on the inside of the cover of the device is the following: "Open \(=95.50 \mathrm{kHz}\) " and "Close \(=102.50 \mathrm{kHz}\) " Your team members search and eventually find a long piece of \(1 / 4\) -inch diameter aluminum rod. (a) To what lengths must you cut the rod in order to get the remote to work properly (i.e., so that the fundamental frequencies of the rods match those utilized by the remote)? (b) Suppose you had instead found a titanium rod. What lengths would be required in that case? (Young's moduli are \(Y_{\mathrm{Al}}=6.9 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\) and \(Y_{\mathrm{Ti}}=1.2 \times 10^{11} \mathrm{N} / \mathrm{m}^{2} ;\) the mass densities are \(\rho_{\mathrm{Al}}=2700 \mathrm{kg} / \mathrm{m}^{3}\) and \(\left.\rho_{\mathrm{Ti}}=4500 \mathrm{kg} / \mathrm{m}^{3} .\right)\)

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