Q45E
Question
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?
Step-by-Step Solution
Verified(a) The distance from bridge is, \(0.45\;{\rm{m}}\).
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, \({f_w} = 587\;{\rm{Hz}}\) .
The difference between two similar points (adjacent crests) in two consecutive waves of a waveform signal travelling in space or over a wire.
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,
\(v = f\lambda \) …(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,\(\lambda = \frac{v}{{{f_w}}}\) …(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, \(0.45\;{\rm{m}}\).