Problem 44
Question
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?
Step-by-Step Solution
Verified Answer
(a) The tension is approximately 169.95 N. (b) The fundamental frequency is about 149.89 Hz.
1Step 1: 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.
2Step 2: 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 \( \mu = \frac{\text{mass}}{\text{length}} = \frac{0.00875 \text{ kg}}{0.75 \text{ m}} = 0.01167 \text{ kg/m} \).
3Step 3: 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, \( f_3 = \frac{344 \text{ m/s}}{0.765 \text{ m}} \approx 449.67 \text{ Hz} \).
4Step 4: 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: \[ T = \left( \frac{2Lf_3}{3} \right)^2 \cdot \mu = \left( \frac{2 \cdot 0.75 \cdot 449.67}{3} \right)^2 \cdot 0.01167 \approx 169.95 \text{ N} \].
5Step 5: 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, \[ f_1 = \frac{1}{2 \cdot 0.75} \sqrt{\frac{169.95}{0.01167}} \approx 149.89 \text{ Hz} \].
Key Concepts
Wave MechanicsHarmonicsString TensionFrequency Calculation
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.
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.
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.
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.
Other exercises in this chapter
Problem 41
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{si
View solution Problem 42
The wave function of a standing wave is \(y(x, t) = 4.44 \, \mathrm{mm} \, \mathrm{sin}[(32.5 \, \mathrm{rad/m})x] \mathrm{sin}[(754 \, \mathrm{rad/s})t]\). For
View solution Problem 46
(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wave
View solution Problem 47
(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wave
View solution