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The A string on a string bass vibrates at a fundamental frequency of \(55.0 \mathrm{Hz}\). If the string's tension were increased by a factor of four, what would be the new fundamental frequency?

Short Answer

Expert verified
The new fundamental frequency is 110.0 Hz.

Step by step solution

01

Understand the Relationship of Frequency and Tension

The fundamental frequency \( f \) of a vibrating string is related to its tension \( T \) by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the string and \( \mu \) is the linear mass density. This tells us that frequency is proportional to the square root of tension.
02

Determine the New Tension's Effect

Since the tension is increased by a factor of four, the new tension \( T' \) can be expressed as \( T' = 4T \). The relationship for frequency becomes \( f' = \frac{1}{2L} \sqrt{\frac{4T}{\mu}} = \frac{1}{2L} \cdot 2 \cdot \sqrt{\frac{T}{\mu}} = 2 \cdot f \).
03

Calculate the New Frequency

Using the initial fundamental frequency \( f = 55.0 \mathrm{Hz} \), the new fundamental frequency will be \( f' = 2 \times 55.0 \mathrm{Hz} = 110.0 \mathrm{Hz} \).

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Key Concepts

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

Vibrating String
A vibrating string is a fundamental concept in understanding sound and music. When a string vibrates, it creates waves along its length that ultimately produce sound. These vibrations are caused by plucking, striking, or bowing the string, depending on the instrument. The speed at which waves travel along the string determines the pitch of the sound. This speed is influenced by various factors including the tension of the string, its mass, and length.

In instruments like guitars, violins, or basses, the frequency of the vibration (how often the string moves back and forth in one second) determines the pitch. A higher frequency results in a higher pitch. The fundamental frequency is the lowest frequency of vibration for a string and the one that defines its pitch. It is calculated based on the properties of the string, which we will explore further in terms of tension and mass density.
  • Plucking, striking, or bowing the string initiates vibration.
  • Vibrations create waves along the string that produce sound.
  • Frequency of vibrations determines the pitch.
  • Fundamental frequency is the basic pitch of the vibrating string.
Tension of a String
The tension of a string significantly influences its vibration and the sound it produces. When you increase the tension on a string, such as by tightening a tuning peg on a guitar or bass, you increase the speed at which vibrations travel through the string. This, in turn, increases the frequency of the vibration, leading to a higher pitch.

The relationship between tension and frequency is captured by a key principle: if you increase the tension, the frequency increases. This is because frequency is proportional to the square root of the tension according to the mathematical expression:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where:
  • \(f\) is the frequency of the vibrating string.
  • \(L\) is the length of the string.
  • \(T\) is the tension applied to the string.
  • \(\mu\) is the linear mass density of the string.
Understanding this relationship helps musicians and technicians to fine-tune their instruments to reach desired frequencies and pitches.
Linear Mass Density
Linear mass density is a critical factor in determining how a string vibrates. It refers to the mass per unit length of the string, usually denoted by \(\mu\). Different materials and thicknesses of strings will have different linear mass densities. For instance, a thicker string or one made of a denser material will typically have a higher mass per unit length.

The linear mass density impacts the speed at which vibrations travel along the string. A string with greater mass density will vibrate slower than a string with a lower mass density when under the same tension, resulting in a lower frequency and thus a lower pitch. This is because a heavier string requires more force to vibrate.
  • \(\mu\) represents mass per unit length of the string.
  • Higher linear mass density means a lower vibration frequency.
  • Material and thickness affect the density of the string.
Square Root Relationship
The square root relationship plays a crucial role in understanding the physics of vibrating strings. It is evident from the formula for calculating the frequency of a vibrating string:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]This formula illustrates that the frequency of a string is proportional to the square root of the tension divided by its linear mass density.

What this means in simple terms is that:
  • Doubling the tension of the string does not double the frequency; instead, it increases by a factor of the square root of two.
  • Similarly, if you increase the linear mass density, the frequency decreases, dictated by the square root relationship.
This relationship is important because it allows for precise calculation of changes in pitch when adjusting tension or changing strings, helping musicians maintain accurate tunings.

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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 highpitched 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 tube with a cap on one end, but open at the other end, has a fundamental frequency of \(130.8 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\) (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and mass per unit length, but they differ in length by \(0.57 \mathrm{cm} .\) The waves on the shorter string propagate with a speed of \(41.8 \mathrm{m} / \mathrm{s},\) and the fundamental frequency of the shorter string is \(225 \mathrm{Hz}\). Determine the beat frequency produced by the two standing waves.

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.

Two speakers, one directly behind the other, are each generating a 245-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343 \mathrm{m} / \mathrm{s}\).

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