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A horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's mass?

Short Answer

Expert verified
The length of the wire is \(18.84 \mathrm{~cm}\) and the mass of the wire is \(1.69 \mathrm{~g}\).

Step by step solution

01

Length of the Wire

A standing wave is formed by the superposition of two waves moving in opposite directions. In the second overtone (or the third harmonic), there are three loops (or three half-wavelengths) in the wave. One node-to-node distance equals one half-wavelength. Therefore, the length of the wire \(L\) is three times the node-to-node distance \(d\), or \(L = 3d\). Substituting the given \(d = 6.28 \mathrm{~cm}\) into the equation gives \(L = 3*6.28 \mathrm{~cm} = 18.84 \mathrm{~cm}\).
02

Wave Speed Calculation

The wave speed \(v\) can be calculated from the formula \(v = \sqrt{T/(\mu)}\), where \(T\) is the tension in the wire and \(\mu\) is the linear density (mass/length). We can rewrite this equation in terms of the desired variable \(\mu\), yielding \(\mu = T/v^2\). But first, we need to calculate the wave speed.
03

Period of Vibration

The period of vibration \(T\) is twice the time it takes for a point to travel from maximum upward to maximum downward displacement. This travel time is provided as \(8.40 \mathrm{~ms}\). Therefore, \(T = 2*8.40 \mathrm{~ms} = 16.8 \mathrm{~ms}\).
04

Calculate Wave Speed

The wave speed can now be calculated from the equation \(v = \lambda / T\), where \(\lambda\) is the wavelength and \(T\) is the period. In this case, \(\lambda = 2d\), because each full wave (or node-to-node travel) equates to a full wavelength \(\lambda\). Substituting these values into the equation gives \(v = (2*6.28 \mathrm{~cm}) / (16.8 \mathrm{~ms}) = 0.746 \mathrm{~m/s}\).
05

Calculate Mass Density

Now you can substitute \(T = 5.00 \mathrm{~N}\) and \(v = 0.746 \mathrm{~m/s}\) into the equation \(\mu = T/v^2\) from step 2 to get \(\mu = 5.00 \mathrm{~N} / (0.746 \mathrm{~m/s})^2 = 8.99 \mathrm{~kg/m}\).
06

Calculate Mass of the Wire

Finally, the mass \(M\) of the wire can be calculated from the equation \(M = \mu * L\), where \(\mu = 8.99 \mathrm{~kg/m}\) is the mass per unit length and \(L = 18.84 \mathrm{~cm} = 0.1884 \mathrm{~m}\) is the length of the wire. Substituting these values into the equation gives \(M = 8.99 \mathrm{~kg/m} * 0.1884 \mathrm{~m} = 1.69 \mathrm{~g}\).

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Key Concepts

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

Wave Speed
Wave speed is fundamental in understanding wave phenomena. In the case of standing waves, it illustrates how quickly disturbances travel through a medium, like a wire. For a wave traveling in a wire, this speed is affected by the wire's tension and its mass per unit length (linear density, \(\mu\)).

The wave speed \(v\) can be determined by the equation:
  • \(v = \sqrt{T/\mu}\)
Here, \(T\) represents the tension in the wire, which in our case is \(5.00\, \mathrm{N}\). The linear density \(\mu\) is the mass of the wire per unit length.

In order to fully understand wave speed, consider its relationship with wavelength \(\lambda\) and period \(T\):
  • \(v = \lambda / T\)
This relation highlights that wave speed can also be expressed as the product of the frequency and wavelength of the wave.
Tension
Tension is a pulling force exerted by a string, cable, or another form of extended object under stress. In the context of standing waves on a wire, tension significantly impacts wave speed.

The formula linking tension \(T\) to wave speed \(v\) and linear density \(\mu\) is:
  • \(v = \sqrt{T/\mu}\)
Therefore, the higher the tension in the wire, the faster the wave travels. In practical applications, this means that by increasing the tension in the wire, we can increase the speed of the wave. For our wire, this has been set to \(5.00\, \mathrm{N}\) to create the second overtone's specific standing wave pattern.

Understanding tension's role helps in predicting and controlling wave behavior in various scenarios, whether musically in string instruments or technically in transmission wires.
Overtones
An overtone is a wave frequency higher than the fundamental frequency of a waveform, and it is crucial in the study of harmonics. For standing waves, overtones can be visualized as additional nodes and antinodes along the medium. In the problem provided, the wire vibrates at its second overtone, also known as the third harmonic. This means that there are three loops or antinodes present on the wire.

Each harmonic can be associated with a specific number of half-wavelengths that fit into the length of the wire:
  • In the second overtone scenario, the wire encompasses three half-wavelengths, where one half-wavelength corresponds to a single node-to-node distance.
Understanding overtones helps explain how different sound qualities are produced in musical instruments and how standing waves can be visualized in physics.
Linear Density
Linear density \(\mu\) refers to the mass per unit length of a wire or string. It's a critical factor in determining how a wave will behave on a medium like a string or wire.

The formula to find linear density in the context of a wave on a wire is derived from:
  • \(\mu = T/v^2\)
Where \(v\) is the wave speed, which we calculated as \(0.746 \, \mathrm{m/s}\). By substituting the tension \(T\) and wave speed \(v\) into the equation, you can find the wire's linear density. For our scenario, this results in \(\mu = 8.99 \, \mathrm{kg/m}\).

Understanding linear density is essential for anticipating how waves might behave as they travel through different media. It affects how quickly waves can travel through the medium and the efficiency of transmission in practical applications.

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Most popular questions from this chapter

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