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A string that is fixed at both ends has a length of \(2.50\) \(\mathrm{m}\). When the string vibrates at a frequency of \(85.0\) \(\mathrm{Hz},\) a standing wave with five loops is formed. (a) What is the wavelength of the waves that travel on the string? (b) What is the speed of the waves? (c) What is the fundamental frequency of the string?

Short Answer

Expert verified
(a) 1.0 m; (b) 85.0 m/s; (c) 17.0 Hz.

Step by step solution

01

Identify the Relationship between Length and Wavelength

For a string with fixed ends, a standing wave forms loops or segments. Each loop corresponds to half a wavelength (\(\lambda/2\)). Given that there are 5 loops in the 2.50 m string, the total length equals 5 times half the wavelength, or \(5 \times \frac{\lambda}{2} = 2.50\).
02

Calculate the Wavelength

Solve the equation \(5 \times \frac{\lambda}{2} = 2.50\) to find the wavelength (\(\lambda\)). Multiply both sides by 2: \(5\lambda = 5\). Then, divide by 5: \(\lambda = 1.0\, \text{m}\).
03

Determine the Speed of the Waves

The wave speed \(v\) is calculated using the formula \(v = f \lambda\), where \(f = 85.0\, \text{Hz}\) is the frequency. Substitute the values: \(v = 85.0 \times 1.0 = 85.0\, \text{m/s}\).
04

Calculate the Fundamental Frequency

The fundamental frequency \(f_1\) is the lowest frequency that forms a single loop (half of \(\lambda\)), covering the string's length \(L = 2.50\, \text{m}\). Use \(f_1 = \frac{v}{2L}\). Since \(v = 85.0\, \text{m/s}\), substitute to get \(f_1 = \frac{85.0}{2 \times 2.50} = 17.0\, \text{Hz}\).

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Key Concepts

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

Wavelength Calculation
Calculating the wavelength of a wave can be straightforward once we understand the concept of standing waves. A standing wave is a pattern that forms when two waves of the same frequency travel in opposite directions and superimpose on each other. In a string with fixed ends, standing waves create distinct loops or segments along the string.
To find the wavelength of waves forming on such a string, consider each loop as half a wavelength (\(\lambda/2\)). If there are five loops present, this means the entire length of the string is equal to five times half the wavelength.
Thus, the formula can be established as:
  • Total length = Number of loops \(\times (\lambda/2)\)
In this problem, the string is 2.50 meters long and contains five loops, so the relationship becomes: \(5 \times (\lambda/2) = 2.50\) meters.
By solving for \(\lambda\), you multiply both sides by 2 to isolate the wavelength: \(5\lambda = 5\), and then solve for \(\lambda\) to find \(\lambda = 1.0\) meter.
Wave Speed
Wave speed is key to understanding how energy traverses through a medium, such as a string. The speed at which waves travel on a string is dependent on both the wavelength and frequency of the waves involved. These quantities are related through a simple equation.
The formula to calculate wave speed \(v\) is:
  • \(v = f \lambda\)
where \(f\) represents the frequency and \(\lambda\) signifies the wavelength.
In practical terms, it means that to find the wave speed, one multiplies the frequency at which the wave oscillates by its wavelength. In the given scenario, with a frequency of 85 Hz and a wavelength of 1.0 meter, substituting these values yields:
  • \(v = 85.0\, \text{Hz} \times 1.0\, \text{m} = 85.0\, \text{m/s}\)
This calculation demonstrates that waves travel along the string at a speed of 85.0 meters per second.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a wave will resonate, producing a standing wave with the least number of loops possible. For a string fixed at both ends, this results in a pattern consisting of only one loop or segment.
The fundamental frequency \(f_1\) can be calculated using the wave speed and the string's length. It follows the equation:
  • \(f_1 = \frac{v}{2L}\)
Here, \(v\) is the wave speed and \(L\) is the length of the string. With the known wave speed of 85.0 m/s and a string length of 2.50 meters, substituting these into the equation gives:
  • \(f_1 = \frac{85.0\, \text{m/s}}{2 \times 2.50\, \text{m}} = 17.0\, \text{Hz}\)
This shows that the foundational frequency is 17.0 Hz, corresponding to the simplest form of standing waves on the string with just one central loop.

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Most popular questions from this chapter

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 \(\mathrm{E}\) string to be used to produce the note \(\mathrm{D},\) 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?

A \(440.0-\mathrm{Hz}\) tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of \(3 \mathrm{Hz}\) is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?

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?

A sound wave is traveling in seawater, where the adiabatic bulk modulus and density are \(2.31 \times 10^{9}\) \(\mathrm{Pa}\) and \(1025\) \(\mathrm{kg} / \mathrm{m}^{3},\) respectively. The wavelength of the sound is \(3.35\) \(\mathrm{m} .\) A tuning fork is struck under water and vibrates at \(440.0\) \(\mathrm{Hz}\). What would be the beat frequency heard by an underwater swimmer?

The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of \(343\) \(\mathrm{m} / \mathrm{s}\) for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ?

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