One string of a certain musical instrument is \(75.0\;{\rm{cm}}\) long and has a mass of \(8.75\;{\rm{g}}\). It is being played in a room where the speed of sound is \(344\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\). (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength \(0.765\;{\rm{m}}\)? (Assume that the breaking stress of the wire is very large and isn’t exceeded.)

(b) What frequency sound does this string produce in its fundamental mode of vibration?

The given data can be listed below as,

• The length of instrument is, \(L = 75.0\;{\rm{cm}} = 75.0 \times {10^{ - 2}}\;{\rm{m}}\).
• The mass of the instrument is, \(m = 8.75\;{\rm{g}} = 8.75 \times {10^{ - 3}}\;{\rm{kg}}\).
• The speed of sound is, \({v_s} = 344\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).
• The wavelength of sound is, \({\lambda _s} = 0.765\;{\rm{m}}\)

To begin with, determine the string's linear mass density using its mass and length. Next, determine the wavelength of the second overtone using the standing wave theory.

The frequency of the sound wave is then determined using the sound's velocity and wavelength.

The speed of a wave through the string can then be determined using the sound wave's frequency and wavelength. The tension in the string is then determined using the velocity and linear mass density. The following is the expression is the tension related to the speed v:

\(v = \sqrt {\frac{F}{\mu }} \)

\(F = \mu {v^2}\) …(1)

The mass\(m\)per length\(L\)is equivalent to the specific mass of wire (\(\mu \)). So,

\(\mu = \frac{m}{L}\)

Here\(F\)is the tension in the string,\(m\)is the mass of the wire,\(L\)is the length of wire and\(v\)is the speed of wave.

The following expression is used to calculate linear mass density:

\(\mu = \frac{m}{L}\) …(2)

Substitute \(8.75 \times {10^{ - 3}}\;{\rm{kg}}\) for \(m\) and \(75.0 \times {10^{ - 2}}\;{\rm{m}}\) for \(L\) in the equation (2).

\(\begin{array}{c}\mu = \frac{{8.75 \times {{10}^{ - 3}}\;{\rm{kg}}}}{{75.0 \times {{10}^{ - 2}}\;{\rm{m}}}}\\\mu = 0.0117\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{m}}}} \right. \\} {\rm{m}}}\end{array}\)

The value of linear mass density is, \(\mu = 0.0117\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{m}}}} \right. \\} {\rm{m}}}\).

The standing wave wavelength for the second overtone can be expressed as follows:

\(\lambda = \frac{{2L}}{3}\) …(3)

Substitute \(75.0 \times {10^{ - 2}}\;{\rm{m}}\) for \(L\) in the equation (3).

\(\begin{array}{c}\lambda = \frac{{2 \times 75.0 \times {{10}^{ - 2}}\;{\rm{m}}}}{3}\\\lambda = 0.5\;{\rm{m}}\end{array}\)

The wavelength is, \(\lambda = 0.5\;{\rm{m}}\).

The sound wave's frequency can be expressed as follows:

\({f_s} = \frac{{{v_s}}}{{{\lambda _s}}}\) …(4)

Here \({v_s}\) is the speed of sound and \({\lambda _s}\) is the wavelength of sound.

Substitute \(0.765\;{\rm{m}}\) for \({\lambda _s}\) and \(344\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}}\right\\} {\rm{s}}}\) for \({v_s}\) in the equation (4).

\(\begin{array}{c}{f_s} = \frac{{344\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}}}{{0.765\;{\rm{m}}}}\\{f_s} = 449.67\;{\rm{Hz}}\end{array}\)

The sound wave's frequency is, \({f_s} = 449.67\;{\rm{Hz}}\).

The wave's speed in the string can be expressed as follows:

\(v = {f_s}\lambda \) …(5)

Substitute \(0.5\;{\rm{m}}\) for \(\lambda \) and \(449.67\;{\rm{Hz}}\) for \({f_s}\) in the equation (5).

\(\begin{array}{c}v = \left( {449.67\;{\rm{Hz}}} \right) \times \left( {0.5\;{\rm{m}}} \right)\\v = 224.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\end{array}\)

The wave's speed is, \(v = 224.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

Substitute \(224.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) for \(v\) and \(0.0117\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{m}}}} \right. \\} {\rm{m}}}\) for \(\mu \) in the equation (1).

\(\begin{array}{c}F = \left( {0.0117\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{m}}}} \right. \\} {\rm{m}}}} \right) \times {\left( {224.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)^2}\\F = 591.45\;{\rm{N}}\end{array}\)

Hence the tension in the instrument is, \(591.45\;{\rm{N}}\).