/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The A string on a string bass vi... [FREE SOLUTION] | 91Ó°ÊÓ

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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 Hz.

Step by step solution

01

Understanding the Frequency of a String

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 mass density.
02

Increasing Tension and Its Effect on Frequency

When the tension \( T \) is increased by a factor of four, the new tension \( T' = 4T \). We need to see how the frequency changes with this new tension, using the relation \( f' = \frac{1}{2L} \sqrt{\frac{4T}{\mu}} \).
03

Simplifying the New Frequency Formula

Substitute the new tension into the frequency formula: \( f' = \frac{1}{2L} \sqrt{\frac{4T}{\mu}} = \frac{1}{2L} \cdot 2 \sqrt{\frac{T}{\mu}} = 2 \times \frac{1}{2L} \sqrt{\frac{T}{\mu}} = 2f \).
04

Calculating the New Frequency

The original frequency \( f = 55\, \text{Hz} \). Since \( f' = 2f \), the new frequency \( f' = 2 \times 55 = 110\, \text{Hz} \).

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Key Concepts

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

Fundamental Frequency
The fundamental frequency of a string is the lowest frequency at which it naturally vibrates. This occurs when the string oscillates in a single segment, often resulting in the string's purest sound. The formula to calculate the fundamental frequency of a vibrating string 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 applied to the string,
  • \( \mu \) is the linear mass density.
The equation indicates that the fundamental frequency is inversely proportional to the length and depends on the square root of the tension and mass density.
Hence, even a small change in tension or mass density can lead to significant frequency changes, affecting the pitch of the string.
Tension in Strings
Tension is a critical factor in the physics of string instruments. It refers to the tightening force applied along a string, which affects its vibration and resulting sound.
In the frequency formula, tension is represented by \( T \).
When a string's tension is increased, it impacts its vibrational properties in the following ways:
  • The frequency increases, making the pitch higher.
  • The vibrations become quicker, leading to a sharper sound.
When tension is multiplied by a factor, such as four, the effect on frequency can be observed mathematically. Using the example from the exercise, when tension increases fourfold, the new frequency \( f' = \frac{1}{2L} \sqrt{\frac{4T}{\mu}} \) simplifies to \( 2f \).
This illustrates that quadrupling the tension results in doubling the frequency.
Mass Density
The mass density, also known as linear mass density, is a measure of the mass of the string per unit length, symbolized by \( \mu \). It plays a crucial role in determining the vibrational characteristics of the string. Mass density is influenced by:
  • The material composition of the string.
  • The thickness of the string—thicker strings typically have a higher mass density.
In the formula for fundamental frequency, the mass density \( \mu \) appears in the denominator within the square root. Thus, a higher mass density results in a lower frequency, assuming tension and length remain constant. This means that for strings of the same length and tension, a thicker string will produce a lower pitch than a thinner one.
Understanding mass density is crucial for instrument design and playing, as it helps musicians choose the right strings to achieve desired sound qualities.

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Most popular questions from this chapter

A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and 98 Hz. The frequency of \(42\) \(\mathrm{Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s} .\) How deep is the well?

Two ultrasonic sound waves combine and form a beat frequency that is in the range of human hearing for a healthy young person. The frequency of one of the ultrasonic waves is \(70\) \(\mathrm{kHz}\). What are (a) the smallest possible and (b) the largest possible value for the frequency of the other ultrasonic wave?

A sound wave with a frequency of 15 kHz emerges through a circular opening that has a diameter of \(0.20 \mathrm{m} .\) Concepts: (i) The diffraction angle for a wave emerging through a circular opening is given by \(\sin \theta=1.22 \lambda / D,\) where \(\lambda\) is the wavelength of the sound and \(D\) is the diameter of the opening. What is meant by the diffraction angle? (ii) How is the wavelength related to the frequency of the sound? (iii) Is the wavelength of the sound in air greater than, smaller than, or equal to the wavelength in water? Why? (Note: The speed of sound in air is \(343 \mathrm{m} / \mathrm{s}\) and the speed of sound in water is \(1482 \mathrm{m} / \mathrm{s} .\) ) (iv) Is the diffraction angle of the sound in air greater than, smaller than, or equal to the diffraction angle in water? Explain. Calculations: Find the diffraction angle \(\theta\) when the sound travels (a) in air and (b) in water.

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of 450 Hz. The next higher harmonic has a frequency of \(750\) \(\mathrm{Hz}\). The speed of sound in air is \(343\) \(\mathrm{m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450\) \(\mathrm{Hz} ?\) (b) What is the length of the 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?

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