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Chapter 17: Problem 29

A string has a linear density of \(8.5 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is under a tension of \(280 \mathrm{~N}\). The string is \(1.8 \mathrm{~m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

Short Answer

Expert verified
The wave speed is approximately 181.44 m/s, the wavelength is 3.6 m, and the frequency is about 50.4 Hz.

Step by step solution

01

Calculate the Speed of the Wave on the String

The speed of a wave traveling on a string can be found using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear density. Here, \( T = 280 \ \text{N} \) and \( \mu = 8.5 \times 10^{-3} \ \text{kg/m} \). Substitute these values into the formula:\[v = \sqrt{\frac{280}{8.5 \times 10^{-3}}} = \sqrt{32941.18} \approx 181.44 \ \text{m/s}.\]
02

Determine the Wavelength of the Standing Wave

The length of the string and the mode number determine the wavelength. For a string fixed at both ends, the wavelength depends on the number of antinodes (or nodes) in the standing wave pattern. If the pattern shows a half-wavelength fitting into the string length, the wavelength is twice the length of the string. Here, \( \lambda = 2 \times 1.8 \text{ m} = 3.6 \text{ m}. \)
03

Calculate the Frequency of the Wave

Frequency can be found using the relationship \( f = \frac{v}{\lambda} \). Using the values obtained previously, \( v = 181.44 \ \text{m/s} \) and \( \lambda = 3.6 \ \text{m} \), substitute these values into the formula:\[f = \frac{181.44}{3.6} \approx 50.4 \ \text{Hz}.\]

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Key Concepts

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

Standing Wave
Standing waves are a fascinating phenomenon that occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. On a string fixed at both ends, standing waves appear as patterns that seem to stand still. They are characterized by nodes and antinodes. Nodes are points where the string does not move, while antinodes are points where the string moves with maximum amplitude.

For a string to form a standing wave, it must vibrate at certain frequencies, which are called harmonic frequencies. These frequencies depend on the length of the string and the speed of the wave traveling along it. Only specific wavelengths that fit into the length of the string create standing waves, leading to specific frequencies that produce vibrant patterns.
Wavelength
The wavelength is a crucial parameter in understanding waves. It's defined as the distance over which the wave's shape repeats. In standing waves on a string, the wavelength can be determined by examining the number of nodes and antinodes.
  • For a string fixed at both ends, only those wavelengths that create whole numbers of half-wavelengths fit into the string's length will produce standing waves.
  • In our specific problem, if the string's length fits one-half of the wavelength, then the entire wavelength is twice the string's length.
  • So for a string that is 1.8 meters long, the wavelength would be 3.6 meters if there's a single antinode in the middle of the string.
This understanding of wavelength helps in predicting the pattern of the standing wave.
Wave Frequency
Wave frequency refers to how often the particles of a medium oscillate as a wave passes through it. It's typically measured in hertz (Hz), which is the number of cycles per second.
  • When it comes to standing waves on strings, frequency is pivotal because it dictates the wave pattern that forms.
  • It is derived from the wave speed and the wavelength using the formula: \( f = \frac{v}{\lambda} \).
  • In our exercise, with a wave speed of approximately 181.44 m/s and a wavelength of 3.6 meters, the wave frequency is calculated to be approximately 50.4 Hz.
This relationship highlights how the speed and the wavelength directly influence the frequency of a wave.
Tension in String
Tension in the string is a force that is exerted along the length of the string and plays a significant role in wave properties. It heavily influences the speed of waves traveling through the string.
  • The greater the tension, the faster the wave can travel.
  • In mathematical terms, this relationship is captured by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) represents tension and \( \mu \) stands for the linear mass density of the string.
  • For instance, in our example where the tension in the string is 280 N and the linear density is \(8.5 \times 10^{-3} \text{kg/m}\), the wave speed is calculated as approximately 181.44 m/s.
This formula emphasizes tension as a controlling factor in how quickly signals travel down a string.

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Most popular questions from this chapter

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{~cm} .\) The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

The fundamental frequency of a vibrating system is \(400 \mathrm{~Hz}\). For each of the following systems, give the three lowest frequencies (excluding the fundamental) at which standing waves can occur: (a) a string fixed at both ends, (b) a cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end open.

The A string on a string bass is tuned to vibrate at a fundamental frequency of 55.0 \(\mathrm{Hz}\). If the tension in the string were increased by a factor of four, what would be the new fundamental freauency?

(a) When sound emerges from a loudspeaker, is the diffraction angle determined by the wavelength, the diameter of the speaker, or a combination of these two factors? (b) How is the wavelength of a sound related to its frequency? Explain your answers. The following two lists give diameters and sound frequencies for three loudspeakers. Pair each diameter with a frequency, so that the diffraction angle is the same for each of the speakers. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Find the common diffraction angle. $$ \begin{array}{|l|c|} \hline \text { Diameter, } D & \text { Frequency, } f \\ \hline 0.050 \mathrm{~m} & 6.0 \mathrm{Khz} \\ \hline 0.10 \mathrm{~m} & 4.0 \mathrm{kHz} \\ \hline 0.15 \mathrm{~m} & 12.0 \mathrm{kHz} \\ \hline \end{array} $$

A row of seats is parallel to a stage at a distance of \(8.7 \mathrm{~m}\) from it. At the center and front of the stage is a diffraction horn loudspeaker. This speaker sends out its sound through an opening that is like a small doorway with a width \(D\) of \(7.5 \mathrm{~cm} .\) The speaker is playing a tone that has a frequency of \(1.0 \times 10^{4} \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the distance between two seats, located near the center of the row, at which the tone cannot be heard?
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