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Chapter 17: Problem 49

A violin string \(30.0 \mathrm{~cm}\) long with linear density \(0.650 \mathrm{~g} / \mathrm{m}\) is placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880 and \(1320 \mathrm{~Hz}\) as the frequency of the oscillator is varied over the range \(500-1500 \mathrm{~Hz}\). What is the tension in the string?

Short Answer

Expert verified
The tension in the string is approximately 181.21 N.

Step by step solution

01

Identifying Vibrating Modes

The string is set into oscillation at 880 Hz and 1320 Hz. These frequencies are harmonic frequencies of the string, indicating the string is vibrating at its fundamental (first harmonic) and second harmonic frequencies. Thus, the fundamental frequency \( f_1 = 880 \) Hz and the second harmonic frequency \( f_2 = 1320 \) Hz.
02

Using Frequency Formula

For a vibrating string, the frequency of the \( n^{th} \) harmonic is given by \( f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the string, \( T \) is the tension, \( \mu \) is the linear density, and \( n \) is the harmonic number. For the fundamental frequency \( f_1 \), \( n = 1 \). Thus, \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \).
03

Solving for Tension

Using the equation \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), substitute \( f_1 = 880 \text{ Hz}, L = 0.30 \text{ m}, \text{and} \ \mu = 0.650 \text{ g/m} \ \text{(converted to 0.00065 kg/m)} \). The equation becomes:\[ 880 = \frac{1}{2 \times 0.30} \sqrt{\frac{T}{0.00065}} \]Rearrange to solve for \( T \):\[ T = (2L \cdot f_1)^2 \cdot \mu = (2 \times 0.30 \times 880)^2 \times 0.00065 \]
04

Calculating Tension

Compute the tension using the formula derived in the previous step:\[ T = (2 \times 0.30 \times 880)^2 \times 0.00065 \]Calculate inside the square first:\[ 2 \times 0.30 \times 880 = 528 \]Then square it:\[ 528^2 = 278784 \]Finally, multiply by \( 0.00065 \):\[ T = 278784 \times 0.00065 = 181.21 \text{ N} \]
05

Verification

Verify that using the second harmonic (1320 Hz) gives the same tension. Use the formula \( f_2 = \frac{2}{2L} \sqrt{\frac{T}{\mu}} \), which simplifies to the earlier equation since \( f_2/f_1 = 3/2 \) does not alter the tension calculation. This confirms the initial result.

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Key Concepts

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

Harmonic Frequencies
When a string vibrates, it produces sound waves at certain frequencies known as harmonic frequencies. These are natural vibrating frequencies of the string determined by its physical properties. In this problem, the violin string oscillates at 880 Hz and 1320 Hz.
These correspond to the first and second harmonics:
  • The first harmonic, often called the fundamental frequency, is 880 Hz.
  • The second harmonic is 1320 Hz, which is 1.5 times the fundamental frequency, indicating a simple mathematical relationship between the harmonics.
Understanding harmonic frequencies is essential since they determine the musical notes produced by string instruments. When a string vibrates, it generally vibrates in multiple harmonics at once. These harmonics combine to create the complexity and richness of the sound.
String Tension Calculation
String tension affects the pitch of the note it produces. To calculate tension, we utilize the formula for harmonic frequency of a vibrating string:
\[f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}\]In this formula:
  • \( f_n \) is the frequency of the nth harmonic,
  • \( L \) is the length of the string,
  • \( T \) is the tension in the string,
  • \( \mu \) is the linear density (mass per unit length),
  • \( n \) is the harmonic number.
For the fundamental frequency \( f_1 = 880 \) Hz, we set \( n = 1 \). Solving the equation for tension \( T \), and substituting the given values yields a tension of 181.21 N.
Calculating string tension helps in tuning musical instruments accurately, ensuring they produce the correct notes when played.
Linear Density
Linear density \( \mu \) is a measure of the mass per unit length of the string, influencing its vibration. Given in grams per meter (g/m), it should be converted to kilograms per meter (kg/m) for calculations.
The violin string in the problem has a linear density of 0.650 g/m, which converts to:
  • 0.00065 kg/m.
Linear density impacts how tightly a string must be tensioned to achieve a desired pitch. A higher linear density requires more tension to vibrate at the same frequency. Thus, linear density is crucial for understanding the relationship between string properties and the sound it produces.
Vibrating Modes
Vibrating modes describe the distinct patterns of standing waves that form on a string when it vibrates. These are determined by fixed points (nodes) and points of maximum displacement (antinodes) on the string.
In our exercise, the string vibrates at its fundamental mode and the second harmonic:
  • The fundamental mode has one antinode at the midpoint and nodes at each end.
  • The second mode has an additional node and antinode, altering its vibration pattern.
Each mode corresponds to a specific harmonic frequency. Nodes and antinodes result from wave interference, leading to standing wave patterns that define the vibration behavior of stringed instruments.

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Most popular questions from this chapter

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