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The approach to solving this problem is similar to that taken in Multiple- Concept Example 4. On a cello, the string with the largest linear density \(\left(1.56 \times 10^{-2} \mathrm{kg} / \mathrm{m}\right)\) is the C string. This string produces a fundamental frequency of \(65.4\) \(\mathrm{Hz}\) and has a length of \(0.800 \mathrm{m}\) between the two fixed ends. Find the tension in the string.

Short Answer

Expert verified
The tension in the C string is approximately 170.85 N.

Step by step solution

01

Understand the Formula

To solve this problem, we will use the formula for the fundamental frequency of a vibrating string which is \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( f \) is the frequency, \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear density of the string.
02

Rearrange the Formula

We need to find the tension \( T \). Rearranging the formula, we find\[ T = (2Lf)^2 \mu \].This allows us to calculate \( T \) given the values for \( L \), \( f \), and \( \mu \).
03

Insert Known Values

Substitute the given values into the rearranged formula:- \( L = 0.800 \) m- \( f = 65.4 \) Hz- \( \mu = 1.56 \times 10^{-2} \) kg/mThus, the equation becomes\[ T = (2 \times 0.800 \times 65.4)^2 \times 1.56 \times 10^{-2} \].
04

Calculate Tension

Solve for \( T \):\[ T = (1.600 \times 65.4)^2 \times 1.56 \times 10^{-2} = (104.64)^2 \times 1.56 \times 10^{-2} \].First, calculate \( 104.64^2 = 10951.29 \). Then multiply by \( 1.56 \times 10^{-2} \):\[ T = 10951.29 \times 0.0156 = 170.85 \text{ N} \].
05

Conclude the Solution

The calculation gives us the tension in the string. After performing all these operations, we find that the tension \( T \) in the cello string is approximately \( 170.85 \) N.

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Key Concepts

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

Fundamental Frequency
When a string vibrates, it produces a sound characterized by its fundamental frequency. This is the lowest frequency at which the string vibrates, creating the characteristic tone of the instrument. The fundamental frequency depends significantly on the string's length, tension, and linear density. In the given problem, the fundamental frequency of the cello's C string is 65.4 Hz. This value is crucial because it dictates how we perceive the pitch of the sound. Understanding how various factors affect the fundamental frequency helps musicians and technicians optimize the tuning and quality of instruments.

Key takeaways about fundamental frequency include:
  • A higher frequency results in a higher pitch.
  • The frequency can be manipulated by changing string tension, length, or density.
  • It is the "base note" that harmonics build upon in complex sounds.
Knowing these allows students and musicians to grasp how precise changes in the physical properties of a string can affect the sound produced.
Linear Density
Linear density (\( \mu \)) is a fundamental property of a vibrating string. It is defined as the mass per unit length of the string and is measured in kilograms per meter (kg/m). For the cello string problem, the linear density is given as \(1.56 \times 10^{-2}{\text{ kg/m}}\). This parameter is vital because it directly affects how the string responds when tension is applied.

The main points to understand about linear density are:
  • Thicker strings tend to have higher linear density.
  • It influences the propagation speed of waves along the string.
  • It's one part of the equation that determines frequency.
Selecting strings with different densities can change the instrument's tone and playability, making it an important consideration in the design and setup of stringed instruments.
Cello String Tension
Tension in a string is the force exerted along its length. For the C string on a cello, the tension was calculated to be approximately 170.85 N. This force is necessary for producing sound, as it affects the speed at which the waves travel along the string and thus influences the pitch of the sound. In practical terms, adjusting the string tension allows musicians to tune their instruments.

Important points about cello string tension include:
  • Higher tension increases the string's pitch.
  • Careful balancing is needed; too much tension can damage the instrument.
  • Affects playability – higher tension might make strings harder to press down.
Understanding string tension is crucial for musicians tuning their instruments and for luthiers designing and maintaining them.
Vibrating String Formula
The vibrating string formula is a core aspect of understanding how strings produce sound. The specific formula used here is:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( f \) is the fundamental frequency, \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear density. This formula helps predict how changes in physical properties of the string will alter its frequency.

This formula underscores key relationships:
  • Frequency increases with higher tension or reduced linear density.
  • It decreases with an increase in string length.
  • Solves for one variable if the others are known, useful for instrument design and tuning.
By using this formula, one can adjust string parameters to achieve the desired musical note, making it an essential tool for both music students and professional luthiers.

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Most popular questions from this chapter

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the "bends"), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a high-pitched quality, like that of Donald Duck's voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20^{\circ} \mathrm{C} .\) One is filled with air, in which the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s}\) To see the effect of helium on voice quality, calculate the ratio of the \(n\) th natural frequency of the helium-filled tube to the \(n\) th natural frequency of the air-filled tube.

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?

The fundamental frequencies of two air columns are the same. Column A is open at both ends, while column B is open at only one end. The length of column A is 0.70 m. What is the length of column B?

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 \(\left.\rho_{\mathrm{Ti}}=4500 \mathrm{kg} / \mathrm{m}^{3} .\right)\)

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of \(262\) \(\mathrm{Hz}\), while the other produces \(266\) \(\mathrm{Hz}\). When a tuning fork is sounded together with the 262 - Hz tone, a beat frequency of 1 Hz is produced. When the same tuning fork is sounded together with the \(266-\mathrm{Hz}\) tone, a beat frequency of \(3\) \(\mathrm{Hz}\) is produced. What is the frequency of the tuning fork?

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