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Chapter 15: Problem 44

One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

Short Answer

Expert verified
(a) The tension is approximately 169.95 N. (b) The fundamental frequency is about 149.89 Hz.

Step by step solution

01

Understand the Physical Context

The string vibrates in the second overtone, which is the third harmonic. In a vibrating string, the frequency of the nth harmonic is given by \( f_n = \frac{n}{2L} \cdot \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( L \) is the length of the string, and \( \mu \) is the linear mass density of the string. We are given the wavelength of sound produced and need the tension for the second overtone, which corresponds to the third harmonic.
02

Calculate Linear Mass Density

Convert the mass of the string from grams to kilograms: \( 8.75 \text{ g} = 0.00875 \text{ kg} \). The length is given in centimeters, which we need in meters: \( 75.0 \text{ cm} = 0.75 \text{ m} \). The linear mass density \( \mu \) is calculated as \( \mu = \frac{\text{mass}}{\text{length}} = \frac{0.00875 \text{ kg}}{0.75 \text{ m}} = 0.01167 \text{ kg/m} \).
03

Relate Wavelength to Frequency

The second overtone (or third harmonic) has a frequency \( f_3 \) such that the speed of sound \( v = \lambda f_3 \), where \( \lambda = 0.765 \text{ m} \) is the wavelength. Thus, \( f_3 = \frac{344 \text{ m/s}}{0.765 \text{ m}} \approx 449.67 \text{ Hz} \).
04

Calculate Tension in the String

Using the frequency relation for the third harmonic \( f_3 = \frac{3}{2L} \sqrt{\frac{T}{\mu}} \) and solving for tension \( T \), we get: \[ T = \left( \frac{2Lf_3}{3} \right)^2 \cdot \mu = \left( \frac{2 \cdot 0.75 \cdot 449.67}{3} \right)^2 \cdot 0.01167 \approx 169.95 \text{ N} \].
05

Calculate Fundamental Frequency

For the fundamental mode, the first harmonic frequency is given by \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). Using the tension found, \[ f_1 = \frac{1}{2 \cdot 0.75} \sqrt{\frac{169.95}{0.01167}} \approx 149.89 \text{ Hz} \].

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Key Concepts

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

Wave Mechanics
Wave mechanics is a crucial part of understanding how vibrations transform into sound. When a string vibrates, it creates standing waves that lead to the production of musical notes. These waves are described by the wave equation, linking the wave speed, frequency, and wavelength. In our scenario, the string vibrates, creating a sound wave that travels through air at a velocity of 344 meters per second.
  • The speed of a wave on a string depends on the string's tension and density.
  • Understanding wave mechanics allows us to relate wave properties and solve problems regarding stringed instruments.
Grasping the concept of standing waves is essential, as it highlights points called nodes where the wave does not move, and antinodes where the wave has maximum amplitude. The number of nodes and antinodes relate to harmonics, discussed next.
Harmonics
Harmonics are integral to understanding musical sound and resonance. In musical terms, harmonics are the "overtones" that occur at integer multiples of the fundamental frequency. Each harmonic corresponds to a mode of vibration of the string.
  • The first harmonic, or fundamental frequency, is the lowest possible frequency of vibration.
  • The second overtone, often the third harmonic, means the string vibrates in three segments.
When musicians tune their instruments, they often rely on harmonics to ensure that the correct frequencies are being produced. Harmonics add depth and richness to the sound, which is why they are so important in musical acoustics.
String Tension
String tension is a vital concept that dictates the speed at which waves travel across a string. The tenser a string, the faster the wave speed. For a fixed length and mass, this means the frequency of the sound produced changes.
The relationship between tension, length, and mass density is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where \( v \) is the wave speed, \( T \) is the tension, and \( \mu \) is the linear mass density. Adjusting the tension changes the string's pitch, which is why tuning adjustments are made by tightening or loosening the string. Higher tension results in higher pitch, while lower tension produces a lower pitch.
Frequency Calculation
Frequency calculation is essential for determining the pitch a string will produce. It connects the physical properties of the string to the sound it generates. To calculate frequency, the equation used for string harmonics is helpful: \[ f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \] where \( f_n \) represents the frequency of the n-th harmonic, \( L \) is the length, \( T \) is the tension, and \( \mu \) is the mass density.
  • This formula shows how increasing tension or decreasing length raises the frequency.
  • Knowing how to manipulate these factors allows musicians to precisely control instrument pitch.
Using this understanding, musicians can predict how changes in instrument setup will affect the sound output, leading to a harmoniously tuned instrument.

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Most popular questions from this chapter

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

A thin, 75.0-cm wire has a mass of 16.5 g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 625 vibrations per second? (b) How fast would this wave travel?

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t) = (5.60 \, \mathrm{cm}) \mathrm{sin} [(0.0340 \mathrm{rad/cm})x] \mathrm{sin} [(150.0 \, \mathrm{rad/s})t]\), where the origin is at the left end of the string, the x-axis is along the string, and the \(y\)-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x,t)\) for this string if it were vibrating in its eighth harmonic?

A certain transverse wave is described by $$y(x, t) = (6.50 \, \mathrm{mm}) \mathrm{cos} \, 2\pi \Big( \frac{x}{28.0 \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \Big)$$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.
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