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Chapter 2: Q45E (page 500)

The portion of the string of a certain musical instrument between the bridge and upper end of the finger board (that part of the string that is free to vibrate) is \(60.0\;{\rm{cm}}\) long, and this length of the string has mass \(2.00\;{\rm{g}}\). The string sounds an \({A_4}\) note (\(440\;{\rm{Hz}}\)) when played.

(a) Where must the player put a finger (what distance x from the bridge) to play a \({D_5}\) note (\(587\;{\rm{Hz}}\))? (See Fig. E15.45.) For both the \({A_4}\) and \({D_5}\) notes, the string vibrates in its fundamental mode.

(b) Without retuning, is it possible to play a \({G_4}\) note (\(392\;{\rm{Hz}}\)) on this string? Why or why not?

Short Answer

Expert verified

(a) The distance from bridge is, \(0.45\;{\rm{m}}\).

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The length of string is, \(L = 60.0\;{\rm{cm}} = 60.0 \times {10^{ - 2}}\;{\rm{m}}\).
  • The mass of string is, \(m = 2.00\;{\rm{g}} = 2.0 \times {10^{ - 3}}\;{\rm{kg}}\).
  • The frequency of sounds is, \(f = 440\;{\rm{Hz}}\).
  • The frequency of wave is, \({f_w} = 587\;{\rm{Hz}}\) .
02

Significance of the wavelength

The difference between two similar points (adjacent crests) in two consecutive waves of a waveform signal travelling in space or over a wire.

03

Determination of the distance

For the fundament or first harmonic, the wavelength is expressed as,

\(\begin{array}{c}L = \frac{\lambda }{2}\\\lambda = 2L\end{array}\)

Substitute the value of \(L\) in the above equation.

\(\begin{array}{c}\lambda = 2 \times 60.0 \times {10^{ - 2}}\;{\rm{m}}\\\lambda = 1.2\;{\rm{m}}\end{array}\)

The wavelength is, \(\lambda = 1.2\;{\rm{m}}\).

The relation between speed, frequency and wavelength is expressed as,

\(v = f\lambda \) 鈥�(1)

Here \(f\) is the frequency of sound.

Substitute \(440\;{\rm{Hz}}\) for \(f\) and \(1.2\;{\rm{m}}\) for \(\lambda \) in the equation (1).

\(\begin{array}{l}v = \left( {440\;{\rm{Hz}}} \right) \times \left( {1.2\;{\rm{m}}} \right)\\v = 528\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\end{array}\)

The velocity is, \(v = 528\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

The string and velocity will be same for note \({D_5}\). So,

The relation between speed, frequency and wavelength is expressed as,\(\lambda = \frac{v}{{{f_w}}}\) 鈥�(2)

Substitute \(528\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) for \(v\) and \(587\;{\rm{Hz}}\) for \({f_w}\) in the equation (2).\(\begin{array}{l}\lambda = \frac{{528\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}}}{{587\;{\rm{Hz}}}}\\\lambda = 0.899\;{\rm{m}}\end{array}\)

The wavelength is, \(0.899\;{\rm{m}}\).

For the fundament or first harmonic, the length \(L = x\) of the string and its is expressed as,

\(\begin{array}{l}L = \frac{\lambda }{2}\\x = \frac{\lambda }{2}\end{array}\)

Substitute the value of \(\lambda \) in the above equation.

\(\begin{array}{l}x = \frac{{0.899\;{\rm{m}}}}{2}\\x = 0.45\;{\rm{m}}\end{array}\)

Hence the distance from the bridge is, \(0.45\;{\rm{m}}\).

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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 \(9.00\;{\rm{m/s}}\). 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?

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Transverse waves on a string have wave speed 8 m/s, amplitude 0.07 m, and wavelength 0.32 m . The waves travel in the -x-direction, and at t=0 the x=0 end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at x=36 m and time t=0.15 s. (d) How much time must elapse from the instant in part (c) until the particle at x=0.36 m next has maximum upward displacement?
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