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Multiple-Concept Example 4 deals with the same concepts as this problem. A 41-cm length of wire has a mass of 6.0 g. It is stretched between two fixed supports and is under a tension of \(160 \mathrm{N}\). What is the fundamental frequency of this wire?

Short Answer

Expert verified
The fundamental frequency of the wire is approximately 127.05 Hz.

Step by step solution

01

Identify Given Values

We are given a wire with a length \( L = 41 \) cm, which is \( 0.41 \) m when converted. The mass of the wire is \( m = 6.0 \) g or \( 0.006 \) kg. The tension in the wire is \( T = 160 \) N. Our task is to find the fundamental frequency of the vibrating wire.
02

Calculate Linear Mass Density

To find the fundamental frequency, we first need to compute the linear mass density \( \mu \), which is the mass per unit length of the wire. We use the formula: \[ \mu = \frac{m}{L} = \frac{0.006 \, \text{kg}}{0.41 \, \text{m}} = 0.01463 \, \text{kg/m} \]
03

Use the Formula for Fundamental Frequency

The fundamental frequency \( f \) of a string is given by the equation: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear mass density.
04

Substitute Values into the Frequency Formula

Substitute the values for \( L \), \( T \), and \( \mu \) into the fundamental frequency formula: \[ f = \frac{1}{2 \times 0.41} \sqrt{\frac{160}{0.01463}} \] Calculating inside the square root first, we get: \( \sqrt{\frac{160}{0.01463}} \approx 104.18 \). Then, \[ f = \frac{1}{0.82} \times 104.18 \approx 127.05 \, \text{Hz} \]
05

Round and Conclude

Round the frequency to two decimal places. The fundamental frequency of the wire is approximately \( 127.05 \) Hz.

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Key Concepts

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

Tension in Strings
Tension in strings is a key concept in understanding how vibrating bodies produce sound. In physics, tension refers to the pulling force applied to a string or wire, stretching it between two points. This force is fundamental as it directly influences the wave speed along the string, and consequently, the sound it produces.
When a string is under tension, it stores potential energy. This energy is converted into kinetic energy as the string vibrates, producing sound waves. The greater the tension, the higher the potential energy, which allows the string to sustain vibrations at greater frequencies.
  • Higher tension increases wave speed.
  • Increased tension leads to a higher frequency of sound.
  • Tension must be properly balanced to avoid string breakage or unwanted sound distortion.
Understanding and calculating tension correctly is crucial in determining other properties like wave speed or fundamental frequency. For students and musicians alike, mastering this concept is essential for tuning string instruments accurately.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a string or wire vibrates when set in motion. It's also known as the first harmonic and is a key determinant of the sound's pitch.
A wire's fundamental frequency can be affected by several factors, including its tension, length, and linear mass density. Mathematically, it is expressed as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where:
  • \( f \) = fundamental frequency
  • \( L \) = length of the string
  • \( T \) = tension in the string
  • \( \mu \) = linear mass density
By manipulating the tension or the length, musicians can change the fundamental frequency to produce notes of varying pitches. This concept not only aids in understanding musical acoustics but also in solving physics problems related to vibrations and waves.
Linear Mass Density
Linear mass density is a measure of the mass of a wire per unit length, represented by the symbol \( \mu \). It plays a critical role in determining how a wire vibrates and what sound it produces.
The formula to calculate linear mass density is straightforward:
\[ \mu = \frac{m}{L} \]where:
  • \( m \) = mass of the wire
  • \( L \) = length of the wire
A high linear mass density means the string is heavier per unit length, which generally results in a lower fundamental frequency. Conversely, a lighter string will likely have a higher pitch.
In the context of physics problem-solving, finding \( \mu \) first is often a crucial step toward computing other variables such as tension or frequency. Recognizing this relationship allows for a comprehensive understanding of the behavior of vibrating strings.
Vibrating Wire
The phenomenon of a vibrating wire explains how sounds are produced, particularly in musical instruments like guitars or pianos. When tension is applied and the wire is plucked or struck, it vibrates back and forth around a fixed position, producing sound waves.
Factors like tension, length, and linear mass density determine how the wire vibrates. When these vibrational modes occur, they generate harmonics, each with a frequency that is a whole-number multiple of the fundamental frequency.
  • Vibrations create nodes and antinodes along the length of the wire.
  • The fundamental frequency is the simplest standing wave pattern.
  • Harmonics add richness to the sound, creating complex tones.
By studying the properties of vibrating wires, we gain insight not only into musical acoustics but also into engineering, where similar principles are applied in the design of devices like sensors and transducers.

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Most popular questions from this chapter

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?

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 \(323 \mathrm{N}\) 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 \(20.0 \mathrm{Hz}\). What is the lowest harmonic number \(n\) that could be responsible for the "moaning" sound?

The E string on an electric bass guitar has a length of \(0.628 \mathrm{m}\) and, when producing the note \(\mathrm{E}\), vibrates at a fundamental frequency of \(41.2 \mathrm{Hz} .\) Players sometimes add to their instruments a device called a "D-tuner." This device allows the E string to be used to produce the note D, which has a fundamental frequency of \(36.7 \mathrm{Hz}\). The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the E string?

Two ideal gases have the same temperature and the same value for \(\gamma\) (the ratio of the specific heat capacities at constant pressure and constant volume). A molecule of gas A has a mass of \(7.31 \times 10^{-26} \mathrm{kg}\) and a molecule of gas \(\mathrm{B}\) has a mass of \(1.06 \times 10^{-25} \mathrm{kg} .\) When gas \(\mathrm{A}\) (speed of sound \(=259 \mathrm{m} / \mathrm{s}\) ) fills a tube that is open at both ends, the first overtone frequency of the tube is \(386 \mathrm{Hz}\). Gas \(\mathrm{B}\) fills another tube open at both ends, and this tube also has a first overtone frequency of \(386 \mathrm{Hz}\) What is the length of the tube filled with gas B?

One method for measuring the speed of sound uses standing waves. A cylindrical tube is open at both ends, and one end admits sound from a tuning fork. A movable plunger is inserted into the other end at a distance \(L\) from the end of the tube where the tuning fork is. For a fixed frequency, the plunger is moved until the smallest value of \(L\) is measured that allows a standing wave to be formed. Suppose that the tuning fork produces a \(485-\mathrm{Hz}\) tone, and that the smallest value observed for \(L\) is \(0.264 \mathrm{m} .\) What is the speed of sound in the gas in the tube?

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