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Chapter 17: Problem 25

(a) Find the speed of waves on a violin string of mass \(860 \mathrm{mg}\) and length \(22.0 \mathrm{~cm}\) if the fundamental frequency is \(920 \mathrm{~Hz}\). (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

Short Answer

Expert verified
(a) 404.8 m/s, (b) 641.2 N, (c) 0.44 m, (d) 0.373 m

Step by step solution

01

Convert Units

First, we need to convert the mass and length of the string into kilograms and meters, respectively. The mass is given as \(860 \text{ mg}\), which can be converted to kilograms as follows: \(860 \text{ mg} = 0.860 \text{ g} = 0.000860 \text{ kg}\). The length of the string is \(22.0 \text{ cm} = 0.22 \text{ m}\).
02

Calculate Linear Mass Density

The linear mass density (\( \mu \)) of the string is the mass per unit length. It is calculated using \( \mu = \frac{\text{mass}}{\text{length}}\). Therefore, \( \mu = \frac{0.000860 \text{ kg}}{0.22 \text{ m}} \approx 0.003909 \text{ kg/m}\).
03

Calculate Wave Speed on String

The wave speed (\( v \)) on the string can be calculated using the formula for the speed of a wave on a string: \( v = f \lambda \). For the fundamental frequency, \( \lambda = 2L \) where \( L \) is the length of the string. Hence, \( \lambda = 2 \times 0.22 \text{ m} = 0.44 \text{ m} \). Using \( v = \lambda f \), we have \( v = 0.44 \text{ m} \times 920 \text{ Hz} = 404.8 \text{ m/s}\).
04

Calculate Tension in the String

The tension (\( T \)) in the string is determined using the wave speed formula \( v = \sqrt{\frac{T}{\mu}} \). Solving for \( T \), we get \( T = \mu v^2 \). Thus, \( T = 0.003909 \text{ kg/m} \times (404.8 \text{ m/s})^2 \approx 641.2 \text{ N} \).
05

Wavelength of Sound Waves in Air

The wavelength of sound waves emitted by the string can be calculated using the speed of sound in air, approximately \(343 \text{ m/s}\). Using \( \lambda_{sound} = \frac{v_{sound}}{f} = \frac{343 \text{ m/s}}{920 \text{ Hz}} \approx 0.373 \text{ m} \).

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Key Concepts

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

Linear Mass Density
The linear mass density of a string is an important concept when dealing with waves on a string, such as those on a musical instrument like a violin. Basically, it tells us how heavy the string is per unit length. You can find this value by dividing the total mass of the string by its length.
  • Convert units first, if necessary, to keep the consistency in measurements. For example, if the mass is in milligrams and the length is in centimeters, convert them to kilograms and meters respectively.
  • Use the formula: \( \mu = \frac{\text{mass}}{\text{length}} \) to calculate the linear mass density.
  • Understanding linear mass density helps us figure out other properties, like wave speed and tension.
Knowing the linear mass density is essential for sound quality in musical instruments as it influences the vibration and tension of the strings.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system like a string can naturally vibrate. It鈥檚 often called the first harmonic. For musical instruments, this frequency determines the pitch of the sound we hear.
  • The fundamental frequency occurs when the string vibrates in its simplest form, often with two nodes and one antinode.
  • To find the fundamental frequency of a string, use the formula: \( f = \frac{v}{\lambda} \), where \( v \) is the speed of the wave and \( \lambda \) is the wavelength.
  • Changing the tension or the linear mass density of the string also affects the fundamental frequency, hence the pitch of the note produced.
Understanding this allows musicians to tune their instruments precisely to produce the desired notes harmonically.
Wave Wavelength
Wavelength is a key concept in understanding waves on strings. It refers to the distance between consecutive points of a wave in phase, such as from crest to crest.
  • For a string attached at both ends, like a violin string, the wavelength of the fundamental frequency is twice the length of the string (\( \lambda = 2L \)).
  • This concept helps visualize how waves travel across the string and resonate to produce sound.
  • For more complex wave patterns, called higher harmonics, the wavelength becomes fractions of this fundamental wavelength.
Visualizing the concept of wavelength helps in understanding how musical notes are generated and altered by changing the length or tension of the string.
Sound Wave Wavelength
Sound waves emitted from vibrating strings travel through the air and are perceived as sound by our ears. The wavelength of these sound waves can differ from the wavelength of the waves on the string.
  • The speed of sound in air is typically around \( 343 \text{ m/s} \) under standard conditions, and this can change based on temperature and pressure conditions.
  • To calculate the sound wave wavelength, use \( \lambda_{sound} = \frac{v_{sound}}{f} \), where \( f \) is the frequency of the sound.
  • The sound wave's wavelength helps us understand how the sound will travel and interact within a space, which affects acoustics and sound quality.
Being aware of sound wave wavelength is valuable in fields like music, acoustics, and audio engineering, as it assists in creating environments with optimal sound properties.

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Most popular questions from this chapter

The crest of a Parasaurolophus dinosaur skull is shaped somewhat like a trombone and contains a nasal passage in the form of a long, bent tube open at both ends. The dinosaur may have used the passage to produce sound by setting up the fundamental mode in it. (a) If the nasal passage in a certain Parasaurolophus fossil is \(1.8 \mathrm{~m}\) long, what frequency would have been produced? (b) If that dinosaur could be recreated (as in Jurassic Park), would a person with a hearing range of \(60 \mathrm{~Hz}\) to \(20 \mathrm{kHz}\) be able to hear that fundamental mode and, if so, would the sound be high or low frequency? Fossil skulls that contain shorter nasal passages are thought to be those of the female Parasaurolophus. (c) Would that make the female's fundamental frequency higher or lower than the male's?

A violin string \(15.0 \mathrm{~cm}\) long and fixed at both ends oscillates in its \(n=1\) mode. The speed of waves on the string is \(280 \mathrm{~m} / \mathrm{s}\), and the speed of sound in air is \(348 \mathrm{~m} / \mathrm{s}\). What are the (a) frequency and (b) wavelength of the emitted sound wave?

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(425 \mathrm{~Hz}\). The fifth harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe \(A\). How long are (a) pipe \(A\) and (b) pipe \(B ?\)

A sound wave in a fluid medium is reflected at a barrier so that a standing wave is formed. The distance between nodes is \(4.9 \mathrm{~cm}\), and the speed of propagation is \(1250 \mathrm{~m} / \mathrm{s}\). Find the frequency of the sound wave.

Pipe \(A\), which is \(1.80 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of \(B\) happens to match the frequency of \(A\). An \(x\) axis extends along the interior of \(B\), with \(x=0\) at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of \(x\) locating those nodes? (d) What is the fundamental frequency of \(B\) ?
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