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Chapter 12: Q33P (page 328)

Question: (II) An unfingered guitar string is 0.68 m long and is tuned to play E above middle C (330 Hz). (a) How far from the end of this string must a fret (and your finger) be placed to play A above middle C (440 Hz)? (b) What is the wavelength on the string of this 440-Hz wave? (c) What are the frequency and wavelength of the sound wave produced in air at 22°C by this fingered string?

Short Answer

Expert verified
  1. The string should fret at a distance of \(0.17{\rm{ m}}\).
  2. The wavelength of the string is \(1.02{\rm{ m}}\).
  3. The frequency and the wavelength of the sound is \(440{\rm{ Hz}}\) and \(0.783{\rm{ m}}\).

Step by step solution

01

Step-1:-Relation of frequency and wavelength of a string

The frequency of the string is the division of the speed of the wave by the wavelength. Also, the wavelength is equivalent to 2 times the string's length.

02

Step-2:-Given data

The length of the un-fingered string is \({l_{\rm{E}}} = 0.68{\rm{ m}}\).

The frequency of the un-fingered string is \({f_{\rm{E}}} = 330{\rm{ Hz}}\).

The frequency A above middle C is \({f_{\rm{A}}} = 440{\rm{ Hz}}\).

03

Step-3:-Calculation of the length

The relation between length of the string and frequency is given by,

\(\begin{array}{c}{f_{\rm{A}}}{l_{\rm{A}}} = {f_{\rm{E}}}{l_{\rm{E}}}\\{l_{\rm{A}}} = \frac{{{f_{\rm{E}}}{l_{\rm{E}}}}}{{{f_{\rm{A}}}}}\end{array}\)

Here \({l_{\rm{A}}}\) is the length at which the fret be placed.

Substitute the values in the above relation.

\(\begin{array}{c}{l_{\rm{A}}} = \frac{{330{\rm{ Hz}}}}{{440{\rm{ Hz}}}} \times 0.68{\rm{ m}}\\{l_{\rm{A}}} = {\rm{0}}{\rm{.51 m}}\end{array}\)

The length at which the string should fret is calculated as,

\(\begin{array}{l}L = {l_{\rm{E}}} - {l_{\rm{A}}}\\L = 0.68{\rm{ m}} - 0.51{\rm{ m}}\\{\rm{L}} = 0.17{\rm{ m}}\end{array}\)

Thus, the required length is \(0.17{\rm{ m}}\).

04

Step-4:-Calculation of the wavelength

The wavelength is calculated as;

\({\lambda _{\rm{A}}} = 2 \times {l_{\rm{A}}}\)

Here \({\lambda _{\rm{A}}}\) is the wavelength A above middle C.

Substitute the values in the above relation.

\(\begin{array}{c}{\lambda _{\rm{A}}} = 2 \times 0.51{\rm{ m}}\\{\lambda _{\rm{A}}} = 1.{\rm{02 m}}\end{array}\)

Thus, the wavelength of the string with frequency \(440{\rm{ Hz}}\) is \(1.02{\rm{ m}}\).

05

Step-5:-Calculation of the frequency and wavelength of sound wave

The frequency of the sound wave is same as the frequency of the string that is \(440{\rm{ Hz}}\).

The speed of sound at \(22\;^\circ {\rm{C}}\) is \(v = 344.9\;{\rm{m/s}}\).

The wavelength of the sound wave is calculated as;

\(\lambda = \frac{v}{{{f_{\rm{A}}}}}\)

Substitute the values in the above relation.

\(\begin{array}{c}\lambda = \frac{{344.9{\rm{ m/s}}}}{{440{\rm{ Hz}}}}\\\lambda = 0.783{\rm{ m}}\end{array}\)

Thus, the frequency of the sound wave is \(440{\rm{ Hz}}\) and the wavelength of the sound is \(0.783{\rm{ m}}\).

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Most popular questions from this chapter

Question: (II) Two sound waves have equal displacement amplitudes, but one has 2.2 times the frequency of the other. What is the ratio of their intensities?

Question: Manufacturers typically offer a particular guitar string in a choice of diameters so that players can tune their instruments with a preferred string tension. For example, a nylon high-E string is available in a low- and high-tension model with diameter 0.699 mm and 0.724 mm, respectively. Assuming the density \(\rho \) of nylon is the same for each model, compare (as a ratio) the tension in a tuned high-and-low tension string.

(I)(a) Calculate the wavelengths in air at\(20^\circ {\rm{C}}\)for sounds in the maximum range of human hearing, 20 Hz to 20,000 Hz.

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Question: The “alpenhorn” (Fig. 12–42) was once used to send signals from one Alpine village to another. Since lower frequency sounds are less susceptible to intensity loss, long horns were used to create deep sounds. When played as a musical instrument, the alpenhorn must be blown in such a way that only one of the overtones is resonating. The most popular alpenhorn is about \({\bf{3}}{\bf{.4}}\,{\bf{m}}\) long, and it is called the \({{\bf{F}}^{\bf{\# }}}\) horn. What is the fundamental frequency of this horn, and which overtone is close to\({{\bf{F}}^{\bf{\# }}}\)? (See Table12–3.) Model as a tube open at both ends.

Determine the fundamental and first overtone frequencies when you are in a \({\bf{9}}{\bf{.0 - m}}\) long hallway with all doors closed. Model the hallway as a tube closed at both ends.
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