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A piano tuner stretches a steel piano wire with a tension of 800 \(\mathrm{N}\) . The steel wire is 0.400 \(\mathrm{m}\) long and has a mass of 3.00 \(\mathrm{g}\) . (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{Hz} ?\)

Short Answer

Expert verified
(a) 407.5 Hz; (b) 24th harmonic

Step by step solution

01

Understand the Problem

We are given a steel piano wire with a tension of 800 N, a length of 0.400 m, and a mass of 3.00 g. We need to find its fundamental frequency and determine the highest harmonic audible to a person who can hear up to 10,000 Hz.
02

Calculate Linear Mass Density

The linear mass density \( \mu \) of the wire is its mass per unit length. Convert the mass from grams to kilograms: \( 3.00 \, ext{g} = 0.003 \, ext{kg} \.\) Calculate \( \mu: \)\[ \\mu = \\frac{0.003 \text{ kg}}{0.400 \text{ m}} \= 0.0075 \text{ kg/m} \\]
03

Calculate the Wave Speed on the Wire

The wave speed \( v \) on the wire is given by the formula \( v = \sqrt{\frac{T}{\mu}} \,\) where \( T \) is the tension in the wire. Substitute the given values:\[ \v = \\sqrt{\frac{800 \text{ N}}{0.0075 \text{ kg/m}}} \\approx 326 \text{ m/s} \\]
04

Find the Fundamental Frequency

The fundamental frequency \ ( u_1 \ ) is determined by the equation \( u_1 = \\frac{v}{2L} \,\) where \( L \) is the length of the wire. Substituting the values:\[ \u_1 = \\frac{326 \text{ m/s}}{2 \\times 0.400 \text{ m}} \= 407.5 \text{ Hz} \\]
05

Calculate the Highest Audible Harmonic

The number of a harmonic is determined when \( n \\times \u_1 \= 10000 \\text{ Hz} \,\) where \( n \) is the harmonic number. Solve for \( n \):\[ = \\frac{10000 \\text{ Hz}}{407.5 \\text{ Hz}} \\approx 24.54 \\]Since \( n \) must be an integer, the highest harmonic is 24.

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Key Concepts

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

Harmonics
Harmonics refer to the multiple frequencies at which a physical system, such as a piano wire, can vibrate naturally. These are whole number multiples of the fundamental frequency, which is the lowest frequency produced by the vibration. For example, if the fundamental frequency is 408 Hz, the second harmonic would be 816 Hz, the third harmonic 1224 Hz, and so on.

Understanding harmonics is crucial in music since they affect the timbre and sound quality of instruments. Each harmonic contributes to the overall tone heard when a string vibrates. As we increase the harmonic number, the pitch becomes higher, but it also becomes quieter and less prominent than the lower harmonics. Musicians and audio engineers often use these concepts to tune instruments or design room acoustics for optimal sound quality.

In this exercise, calculating harmonics allows us to determine which frequencies are audible when a note is played. For this specific piano wire, we calculated that the highest harmonic an average person can hear is the 24th, as it approximates 10,000 Hz, based on the given listener's range.
Wave Speed
Wave speed is a measure of how quickly a wave travels through a medium. For a vibrating wire or string, the wave speed determines how fast the vibrations move along it. This speed is determined by the tension in the wire and its linear mass density. Using the formula \[v = \sqrt{\frac{T}{\mu}}\]where:
  • \( v \) is the wave speed
  • \( T \) is the tension in the wire
  • \( \mu \) is the linear mass density
Plugging in the numbers from our problem gives us a wave speed of approximately 326 m/s for the wire.

This concept is not only foundational in physics but also in fields dealing with waves, like acoustics and electromagnetics. By adjusting the tension or the density, one can alter the wave speed, achieving different musical notes or signals in various applications.
Linear Mass Density
Linear mass density, represented by the symbol \( \mu \), is a measure of the mass of a wire per unit of its length. It's calculated by dividing the mass by the total length of the wire:\[\mu = \frac{\text{mass}}{\text{length}}\]For example, in our given problem, a 3 g wire stretched over 0.4 m gives a linear mass density of 0.0075 kg/m.

This concept is crucial in determining the wave speed through the wire, as it directly influences how the force (or tension) is distributed along it. In musical instruments, linear mass density affects not only the pitches produced but also their quality and richness. Heavier strings produce lower pitches and vice versa. Understanding \( \mu \) helps in designing strings and wires for specific applications, such as musical instruments where precise tones are required.
Fundamental Frequency
The fundamental frequency, also known as the first harmonic, is the lowest frequency at which a system like a vibrating wire can naturally oscillate. It is calculated using the equation:\[ u_1 = \frac{v}{2L}\]where:
  • \( u_1 \) is the fundamental frequency
  • \( v \) is the wave speed
  • \( L \) is the length of the wire
In our scenario, we calculated a fundamental frequency of 407.5 Hz for the wire, indicating this is the main tone it produces when struck.

The fundamental frequency is intrinsic to the identity of any musical note. It forms the basis for the harmonic series, affecting the pitch, clarity, and timbre of sounds produced. In tuning and sound engineering, achieving the desired fundamental frequency is critical for accurate sound reproduction and instrument setup. Knowing how to calculate \( u_1 \) allows musicians to tune their instruments appropriately and helps physicists and engineers to design and test systems where wave behavior is essential.

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Most popular questions from this chapter

Guitar String. One of the 63.5 -cm-long strings of an ordinary guitar is tuned to produce the note \(\mathbf{B}_{3}\) (frequency 245 \(\mathrm{Hz} )\) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is 344 \(\mathrm{m} / \mathrm{s}\) , find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

Tsunami! On December \(26,2004,\) a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some \(200,000\) people. Satellites observing these waves from space measured 800 \(\mathrm{kn}\) from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) ? Does your answer help you understand why the waves caused such devastation?

(a) Show that for a wave on a string, the kinetic energy per unit length of string is $$ u_{k}(x, t)=\frac{1}{2} \mu v_{y}^{2}(x, t)=\frac{1}{2} \mu\left(\frac{\partial y(x, t)}{\partial t}\right)^{2} $$ where \(\mu\) is the mass per unit length. (b) Calculate \(u_{\mathrm{k}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7) .\) (c) There is also elastic potential energy in the string, associated with the work required to deform and stretch the string. Consider a short segment of string at position \(x\) that has unstretched length \(\Delta x,\) as in \(\mathrm{Fg} .15 .13 .\) Ignoring the (small) curvature of the segment, its slope is \(\partial y(x, t) / \partial x\) Assume that the displacement of the string from equilibrium is small, so that \(\partial y / \partial x\) has a magnitude much less than unity. Show that the stretched length of the segment is approximately $$ \Delta x\left[1+\frac{1}{2}\left(\frac{\partial y(x, t)}{\partial x}\right)^{2}\right] $$ (Hint: Use the relationship \(\sqrt{1+u} \approx 1+\frac{1}{2} u,\) valid for \(|u| \ll 1 . )\) (d) The potential energy stored in the segment equalsthe work done by the string tension \(F\) (which acts along the string) to stretch the segment from its unstretched length \(\Delta x\) to the length calculated in part (c). Calculate this work and show that the potential energy per unit length of string is $$ u_{p}(x, t)=\frac{1}{2} F\left(\frac{\partial y(x, t)}{\partial x}\right)^{2} $$ (e) Calculate \(u_{\mathrm{o}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7)\) (f) Show that \(u_{k}(x, t)=u_{p}(x, t)\) for all \(x\) and \(t .(g)\) Show \(y(x, t)\) \(u_{k}(x, t),\) and \(u_{p}(x, t)\) as functions of \(x\) for \(t=0\) in one graph with all three functions on the same axes. Explain why \(u_{k}\) and \(u_{p}\) are maximum where \(y\) is zero, and vice versa. (h) Show that the instantaneous power in the wave, given by Eq. \((15.22),\) is equal to the total energy per unit length multiplied by the wave speed \(v\) Explain why this result is reasonable.

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?

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