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A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 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 Hz?

Short Answer

Expert verified
(a) 408.25 Hz; (b) 24th harmonic.

Step by step solution

01

Convert Units

The mass of the wire is given in grams and must be converted to kilograms for our calculations since the standard unit of mass in physics is the kilogram. Therefore, we convert 3.00 g to kg:\[ 3.00 \, \text{g} = 0.003 \, \text{kg} \]
02

Calculate Linear Mass Density

Next, we determine the linear mass density (\( \mu \)) of the wire, which is mass per unit length. It's given by the formula:\[ \mu = \frac{m}{L} \]Where:- \( m = 0.003 \, \text{kg} \) (mass of the wire)- \( L = 0.400 \, \text{m} \) (length of the wire)Substituting the values:\[ \mu = \frac{0.003}{0.400} = 0.0075 \, \text{kg/m} \]
03

Calculate Speed of Wave on Wire

The speed of the wave on the string \( v \) is calculated using the formula:\[ v = \sqrt{\frac{T}{\mu}} \]Where:- \( T = 800 \, \text{N} \) (tension)- \( \mu = 0.0075 \, \text{kg/m} \) (linear mass density)\[ v = \sqrt{\frac{800}{0.0075}} \approx 326.60 \, \text{m/s} \]
04

Calculate Fundamental Frequency

The fundamental frequency \( f \) of a stretched string is given by the formula:\[ f = \frac{v}{2L} \]Substitute the values we have:\[ f = \frac{326.60}{2 \times 0.400} \approx 408.25 \, \text{Hz} \]
05

Determine the Highest Audible Harmonic

To find the highest harmonic number \( n \) that can be heard (\( \leq 10,000 \, \text{Hz} \)), use the relation:\[ f_n = n \times f \leq 10,000 \]Using the fundamental frequency \( f = 408.25 \, \text{Hz} \):\[ n \leq \frac{10,000}{408.25} \approx 24.50 \]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 are fascinating phenomena observed with vibrating systems like strings and air columns. They occur when a wave can exist at multiple frequencies that are integer multiples of a fundamental frequency. The fundamental frequency is often the lowest frequency and the first harmonic. On a musical string instrument, this represents the base note you hear, while the higher harmonics contribute to the instrument's timbre, providing richness and depth to the sound.

Understanding harmonics is key because it helps us calculate higher frequencies that a vibrating object can produce. For a wire or string, each harmonic increases in frequency by whole number multiples. If the wire's fundamental frequency is 408.25 Hz, the next harmonic (the second harmonic) would be twice this frequency, and so on. This is particularly useful in music to help understand notes and their octaves. Learn about harmonics, and you'll better appreciate how musicians can produce a range of notes with just a few strings or instruments.
Linear Mass Density
Linear mass density is the mass of a wire or string divided evenly along its length. It's denoted by the Greek letter μ (mu) and has the units of kilograms per meter (kg/m). This measure is crucial when examining how a string vibrates, as it affects both the tension and the wave speed.

In the example exercise, we calculated the linear mass density of a piano wire by dividing the total mass (0.003 kg) by its length (0.400 m), resulting in 0.0075 kg/m. This density is important because it directly influences the string's wave speed and, subsequently, the frequencies of the harmonics it can produce. In musical terms, the linear mass density affects which notes are heard and their pitch. Thicker or denser strings typically have lower pitches, while lighter, less dense strings tend to have higher pitches.
Wave Speed
Wave speed on a string or wire is a pivotal concept when considering vibrations and sound. It can be calculated using the tension in the string and its linear mass density. The formula to determine wave speed is:
  • \( v = \sqrt{\frac{T}{\mu}} \)
  • where \( v \) is the wave speed, \( T \) is the string tension, and \( \mu \) is the linear mass density.

In our exercise, with a tension of 800 N and a linear mass density of 0.0075 kg/m, we computed the wave speed as approximately 326.60 m/s. This speed guides how quickly waves or vibrations travel along the string. Faster wave speeds lead to higher frequency sounds. Understanding wave speed is essential for tuning string instruments, designing musical instruments, and any applications involving waves and vibrations, including engineering and multimedia fields. Wave speed, in the context of sound, influences how far and how fast sound can travel, impacting both musical acoustics and communication systems.

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Most popular questions from this chapter

A strong string of mass 3.00 g and length 2.20 m is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is 9.00 m/s. The tension in the string is 330 N. (a) What is the amplitude of the standing wave at its antinode? (b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\)? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling \(down\) the string?

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string's mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

A simple harmonic oscillator at the point \(x = 0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave. Assume that the oscillator has its maximum upward displacement at time \(t = 0\). (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time \(t = 0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 m to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverse velocity of a particle 1.35 m to the right of the origin at time \(t = 0.0625\) s.

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