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Tuning an Instrument. A musician tunes the C.string of her instrument to a fundamental frequency of 65.4 \(\mathrm{Hz}\) . The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g}\) . (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

Short Answer

Expert verified
Tension for 65.4 Hz is 65.9 N; a 22.1% increase in tension is needed for 73.4 Hz.

Step by step solution

01

Understand the relationship between frequency, tension, and string characteristics

The fundamental frequency of a stretched string is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( f \) is the frequency, \( L \) is the length, \( T \) is the tension, and \( \mu \) is the mass per unit length of the string.
02

Calculate the mass per unit length (\( \mu \))

First, convert the mass of the string to kilograms: \( 14.4 \text{ g} = 0.0144 \text{ kg} \). The mass per unit length is then given by \( \mu = \frac{m}{L} = \frac{0.0144 \text{ kg}}{0.600 \text{ m}} = 0.024 \text{ kg/m} \).
03

Solve for tension (\( T \)) using the frequency equation

Rearrange the frequency formula to solve for tension: \( T = (2Lf)^2 \mu \). Substitute the known values: \( L = 0.600 \text{ m} \), \( f = 65.4 \text{ Hz} \), and \( \mu = 0.024 \text{ kg/m} \). Then calculate: \[T = (2 \times 0.600 \times 65.4)^2 \times 0.024 = 65.9 \text{ N}.\]
04

Calculate the new tension for an updated frequency of 73.4 Hz

Use the rearranged formula for tension with the new frequency: \( f = 73.4 \text{ Hz} \). Calculate the new tension \[T' = (2 \times 0.600 \times 73.4)^2 \times 0.024 = 80.5 \text{ N}.\]
05

Find the percentage increase in tension

The percentage increase is calculated using the initial and new tension values: \[\text{Increase} = \frac{T' - T}{T} \times 100 = \frac{80.5 - 65.9}{65.9} \times 100 \approx 22.1\%.\]

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Key Concepts

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

Frequency
In wave physics, frequency refers to how often a wave repeats itself in a given unit of time. For vibrating strings, like those in musical instruments, frequency determines the pitch of the sound. The frequency is measured in Hertz (Hz), which represents cycles per second.

A string vibrates at a certain fundamental frequency, governed by its length, tension, and mass per unit length. When musicians tune their instruments, they adjust the string tension to achieve the desired frequency, thus ensuring the correct pitch. As seen in our exercise, adjusting the tension modifies the frequency from 65.4 Hz to 73.4 Hz, illustrating how frequency and tension influence the sounds produced by vibrating strings.
Tension
Tension in the context of a vibrating string is the force applied to stretch the string. This force significantly affects how the string vibrates and, consequently, the frequency or pitch of the sound it produces.

Higher tension typically results in higher frequency, producing a higher pitch, while lower tension results in a lower frequency. In the exercise, initially setting the string's tension to 65.9 N achieves the fundamental frequency of 65.4 Hz. Increasing this tension to 80.5 N subsequently raises the frequency to 73.4 Hz.

### Importance of Tension
  • Tension determines sound quality and volume.
  • Excessive tension might snap the string, while insufficient tension may lead to a flat note.
Vibrating String
A vibrating string is the source of sound in many musical instruments. As the string vibrates, it disturbs the surrounding air, creating sound waves that we hear as music.

The vibration occurs in patterns or modes, with the fundamental mode providing the characteristic pitch. The fundamental frequency is primarily determined by the string’s length, tension, and mass per unit length.

### Characteristics of Vibrating Strings
  • Longer strings vibrate at lower frequencies.
  • Thicker strings have a lower pitch.
  • Proper tuning adjusts tension to reach desired frequencies.
Mass per Unit Length
The mass per unit length, denoted by the symbol \( \mu \), is a crucial factor in determining the frequency of a vibrating string. This value represents the string’s density across its length, influencing how it vibrates.

If the string’s mass per unit length is high, it results in a lower frequency, producing a lower pitch. In the exercise, the value calculated as \( \mu = 0.024 \text{ kg/m} \) impacts the frequency and is key in determining the necessary tension for tuning.

### Why Mass per Unit Length Matters
  • A dense string requires more tension for higher pitches.
  • Lighter strings respond more swiftly to tension changes.

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Most popular questions from this chapter

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g}\) . The wave speed is 30.0 \(\mathrm{m} / \mathrm{s}\) , and the wavelength is 0.200 \(\mathrm{m}\) (a) If the wave is to have an aver- age power of 50.0 \(\mathrm{W}\) , what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 \(\mathrm{W} / \mathrm{m}^{2}\) at a distance of 4.3 \(\mathrm{m}\) from the source. (a) What is the intensity at a distance of 3.1 \(\mathrm{m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

Visible Light. Light is a wave, bnt not a mechanical wave. The quantities that oscillate are electric and magnetic fields. Light visible to humans has wavelengths between 400 \(\mathrm{nm}\) (violet) and \(700 \mathrm{nm}(\mathrm{red}),\) and all light travels through vacuum at speed \(c=3.00 \times 10^{8} \mathrm{m} / \mathrm{s}\) (a) What are the limits of the frequency and period of visible light? (b) Could you time a single light vibration with a stopwatch?

Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

Waves of Arbitrary Shape. (a) Explain why any wave described by a function of the form \(y(x, t)=f(x-v t)\) moves in the \(+x\) -direction with speed \(v\) . (b) Show that \(y(x, t)=f(x-v t)\) satisfies the wave equation, no matter what the functional form of \(f\) . To do this, write \(y(x, t)=f(u),\) where \(u=x-v t\) . Then, to take partial derivatives of \(y(x, t),\) use the chain rule: $$ \begin{array}{l}{\frac{\partial y(x, t)}{\partial t}=\frac{d f(u)}{d u} \frac{\partial u}{\partial t}=\frac{d f(u)}{d u}(-v)} \\ {\frac{\partial y(x, t)}{\partial x}=\frac{d f(u)}{d u} \frac{\partial u}{\partial x}=\frac{d f(u)}{d u}}\end{array} $$ (c) A wave pulse is described by the function \(y(x, t)=D e^{-(B x-C)^{2}}\) where \(B, C,\) and \(D\) are all positive constants. What is the speed of this wave?

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