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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 45.2 N.

Step by step solution

01

Identify Given Information

We are given that the length of the violin string is \( L = 30.0 \; \mathrm{cm} = 0.30 \; \mathrm{m} \). The linear density of the string is \( \mu = 0.650 \; \mathrm{g/m} = 0.650 \times 10^{-3} \; \mathrm{kg/m} \). The string resonates at frequencies \( f_1 = 880 \; \mathrm{Hz} \) and \( f_2 = 1320 \; \mathrm{Hz} \).
02

Determine Harmonics

Given frequencies \(880 \; \mathrm{Hz}\) and \(1320 \; \mathrm{Hz}\) correspond to the string's natural frequencies. Since these values form a simple ratio \( \frac{1320}{880} = \frac{3}{2} \), \(880 \; \mathrm{Hz}\) could be the second harmonic \((n=2)\), which means \(1320 \; \mathrm{Hz}\) is the third \((n=3)\) harmonic.
03

Use the Formula for Harmonic Frequencies

The frequency of the \(n^{th}\) harmonic on a string fixed at both ends is given by \[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]Solve for \(T\) using the second harmonic frequency \(f_2 = 880 \; \mathrm{Hz}\) (since it corresponds to \(n=2\)): \[ 880 = \frac{2}{2 \cdot 0.30} \sqrt{\frac{T}{0.650 \times 10^{-3}}} \]
04

Substitute Values and Solve for Tension

Substitute \(f_2 = 880 \; \mathrm{Hz}\), \( L = 0.30 \; \mathrm{m} \) and \( \mu = 0.650 \times 10^{-3} \; \mathrm{kg/m} \) into the equation:\[ 880 = \frac{1}{0.30} \sqrt{\frac{T}{0.650 \times 10^{-3}}} \]Square both sides to remove the square root: \[ 774400 = \left(\frac{1}{0.30}\right)^2 \frac{T}{0.650 \times 10^{-3}} \]Solve for \(T\):\[ T = 774400 \times 0.09 \times 0.650 \times 10^{-3} \approx 45.2 \; \mathrm{N} \]
05

Confirm the Calculation with Multiple Harmonics

As a check, verify the calculated tension using \(f_3 = 1320 \; \mathrm{Hz}\) and verify that it fits the equation using \(n = 3\):\[ 1320 = \frac{3}{2 \cdot 0.30} \sqrt{\frac{T}{0.650 \times 10^{-3}}} \]This should yield the same value of \(T = 45.2 \; \mathrm{N}\), confirming our earlier calculation.

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Key Concepts

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

String Tension
String tension is a vital concept when dealing with musical instruments like violins. It refers to the force exerted along the length of the string. This force is crucial in determining how the string vibrates and, consequently, the sound it produces. When you pluck a violin string or slide a bow across it, the string vibrates in waves.
Understanding string tension is important because it affects the frequency of these waves. The tighter the string, the faster it vibrates, giving a higher pitch. Conversely, a looser string vibrates more slowly, producing a lower pitch.
To find the tension in a string experimentally, one can use the harmonics, which are the natural frequencies at which the string vibrates. By knowing the harmonics, as was done in the exercise by identifying frequencies such as 880 Hz and 1320 Hz, we can backtrack to calculate the tension with the help of a formula. The formula includes the string's length, frequency, and linear density.
Linear Density
Linear density is a way of expressing how much mass is in a given length of a material, in this context, a violin string. It is written as mass per unit length, typically in kilograms per meter (kg/m). Linear density matters greatly in musical strings, as it influences how easily the string can oscillate.
In simple terms, the denser the string, the harder it is to get it moving at high frequencies because more mass means more inertia. In the provided problem, the linear density is given as 0.650 g/m, which translates to 0.650 x 10^{-3} kg/m.
This value was important in calculating the string's tension because the oscillation of the string at given frequencies (880 Hz and 1320 Hz) allowed us to relate the density to tension using the harmonic frequency formula. - Greater linear density generally leads to lower frequencies for given tension and length. - Lower density means less resistance to wave motion and potentially higher frequencies given the same conditions.
Frequency and Resonance
The concepts of frequency and resonance are pivotal in understanding how strings on instruments produce sound. Frequency refers to the number of wave oscillations per second and is measured in Hertz (Hz). When you change the frequency by tightening or loosening a string or by changing its length, you change the sound's pitch.
Resonance occurs when a string vibrates at a natural frequency, where the waves reinforce each other, leading to a larger amplitude. In the exercise, the string vibrated at "resonant" frequencies of 880 Hz and 1320 Hz, corresponding to certain harmonics.
Harmonics are specific frequencies at which the string naturally prefers to vibrate, determined by its physical properties and boundary conditions. A string under resonance exhibits strong oscillations because energy efficiently transfers from the source of vibration to the string.
  • First harmonic or fundamental frequency is the lowest frequency at which a system naturally vibrates.
  • Overtones or higher harmonics are integer multiples of that fundamental frequency
Recognizing these harmonics allows musicians to tune instruments to achieve a desired sound.
Physics Problem Solving
Solving physics problems like the one with the violin string involves systematic and logical reasoning. Such problems often break down into identifiable steps. In this exercise, the key was to determine the tension in a string based on observed resonances. Here's a detail-oriented approach: 1. **Extract Given Information**: Identify what is known, such as string length, given resonant frequencies, and linear density. 2. **Understand Fundamental Principles**: Apply appropriate physics concepts, such as string harmonics. 3. **Set Up the Correct Equations**: Use established formulas, e.g., harmonic frequency equations. 4. **Perform Calculations**: Substitute the known values into the formula to derive the unknowns, such as string tension. 5. **Verify and Check Work**: Whenever possible, confirm findings using alternative approaches or additional data.
Applying this method ensures efficiency and accuracy when exploring the physics of string instruments. Physics problem solving is as much about the process as it is about getting the right answer.

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Most popular questions from this chapter

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