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Chapter 17: Problem 51

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?

Short Answer

Expert verified
The new fundamental frequency is 110.0 Hz.

Step by step solution

01

Understanding the Relationships

The fundamental frequency of a vibrating string is determined by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), 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. Here, we focus on the relationship between frequency \( f \) and tension \( T \).
02

Setting up the relationship

Since the length \( L \) and the linear density \( \mu \) of the string do not change, the frequency relationship simplifies to \( f \propto \sqrt{T} \). This implies any change in tension \( T \) results in a change in frequency in proportion to the square root of the change in tension.
03

Calculate the New Frequency

If the tension \( T \) is increased by a factor of four, the new tension is \( 4T \). The new frequency \( f' \) is then given by: \[ f' = \sqrt{4} \cdot f = 2f \]. Since the original frequency is \( 55.0 \) Hz, the new frequency becomes \( f' = 2 \times 55.0 \) Hz.

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Key Concepts

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

Tension in Vibrating Strings
Vibrating strings, such as those on a musical instrument, require tension to produce sound. The tension is the force that stretches the string, keeping it taut. When a string is plucked, the tension enables it to vibrate and produce sound. By altering the tension, we can change the characteristics of the sound produced.
Imagine a rubber band; if you stretch it, the tension increases, and if you twang it, the sound becomes higher-pitched. Similarly, the tension in a musical instrument string affects its vibration and the pitch it produces. Higher tension results in strings that vibrate more quickly, leading to a higher pitch.
Frequency and Tension Relationship
In the world of physics and music, the frequency of a vibrating string is closely related to its tension. The fundamental frequency is the lowest frequency at which a string vibrates and depends largely on the tension of the string.
For a fixed string length and density, if we increase the tension, we increase the vibrational frequency, leading to a higher pitch. This relationship can be explained by the formula: \[f \propto \sqrt{T}\] where \( f \) is the frequency and \( T \) is the tension. This shows that frequency is proportional to the square root of tension. So, if the tension is quadrupled, the frequency becomes twice as large.
Vibrating String Formulas
Understanding vibrating string formulas is essential for grasping how stringed instruments work. The fundamental frequency of a string is given by:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( f \) is the frequency, \( L \) is the string length, \( T \) is the tension, and \( \mu \) is the linear density. In this formula:
  • The length \( L \) is how long the string is when measured straight from end to end.
  • The tension \( T \) is how tight the string is pulled.
  • The linear density \( \mu \) is the mass of the string per unit length.
The formula helps to show how vibration frequency is tied to physical properties of the string and is used to design and tune musical instruments.
Square Root Dependency in Physics
Square root dependencies are common in physics, explaining a non-linear relationship between two variables. It occurs when one quantity depends on the square root of another, showing diminishing returns with increasing values of the dependent variable.
In the case of vibrating strings, frequency depends on the square root of tension. This means that doubling the tension does not double the frequency; instead, it increases the frequency by a factor of the square root of two. This root reflects how properties like mass and elasticity balance with measurements such as length, density, and tension to define vibrational behavior.

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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 \(n^{\text {th }}\) natural frequency of the helium- filled tube to the \(n^{\text {th }}\) natural frequency of the air-filled tube.

A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of \(425 \mathrm{~Hz},\) while the piece open only at one end has a fundamental frequency of \(675 \mathrm{~Hz}\). What is the fundamental frequency of the original tube?

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.)

When a guitar string is sounded along with a 440 -Hz tuning fork, a beat frequency of \(5 \mathrm{~Hz}\) is heard. When the same string is sounded along with a 436 -Hz tuning fork, the beat frequency is \(9 \mathrm{~Hz}\). What is the frequency of the string?

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \mathrm{X} 10^{5} \mathrm{~Pa}\). The tube has a length of \(0.75 \mathrm{~m}\). Mercury (mass density \(=13\) \(\left.600 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?
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