A piano tuner stretches a steel piano wire with a tension of \(800\;{\rm{N}}\). The steel wire is \(0.400\;{\rm{m}}\) long and has a mass of \(3.00\;{\rm{g}}\).

(a) What is the frequency of its fundamental mode of vibration?

(b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000\;{\rm{Hz}}\)?

The given data can be listed below as,

• The tension in the wire is, \(F = 800\;{\rm{N}}\).
• The length of wire is, \(L = 0.400\;{\rm{m}}\).
• The mass of wire is, \(m = 3.00\;{\rm{g}} = 3.0 \times {10^{ - 3}}\;{\rm{kg}}\).

Initially, calculate the wave speed with the expression's help of the tension related to the speed v, substitute the value of v in the frequency and speed relationship equation and find the frequency. The following is the expression is the tension related to the speed v:

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

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

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

Hence,

\(v=\sqrt{\frac{{FL}}{m}}\) …(1)

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 frequency of its fundamental mode of vibration is expressed as,

\({f_1}=\frac{v}{{2L}}\) …(2)

Here \({f_1}\) is the frequency of its fundamental mode of vibration

Substitute \(800\;{\rm{N}}\) for \(F\), \(0.400\;{\rm{m}}\) for \(L\) and \(3.0 \times {10^{ - 3}}\;{\rm{kg}}\) for \(m\) in the equation (1)

\(\begin{array}{c}v = \sqrt {\frac{{\left( {800\;{\rm{N}}} \right) \times \left( {0.400\;{\rm{m}}} \right)}}{{3.0 \times {{10}^{ - 3}}\;{\rm{kg}}}}} \\v = 326.6\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\end{array}\)

The speed of the wave is, \(v = 326.6\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

Substitute \(326.6\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) for \(v\) and \(0.400\;{\rm{m}}\) for \(L\) in the equation (2)\(\begin{array}{c}{f_1} = \frac{{326.6\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}}}{{2 \times 0.400\;{\rm{m}}}}\\{f_1} = 408.2\;{\rm{Hz}}\end{array}\)

Hence the frequency of its fundamental mode of vibration is, \(408.2\;{\rm{Hz}}\).