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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 thin, 75.0-cm wire has a mass of 16.5 g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 625 vibrations per second? (b) How fast would this wave travel?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to have a wavelength of 0.750 m?

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the \(+x\)-axis and is fixed at \(x = 0\). (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

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 km from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in m/s and in km/h? Does your answer help you understand why the waves caused such devastation?

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