Q40E - University Physics with Modern Physics

Question

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 \(10,000\;{\rm{Hz}}\)?

Step-by-Step Solution

Answer

(a) The frequency of its fundamental mode of vibration is, \(408.2\;{\rm{Hz}}\)

Step 1: Identification of the given data

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, \(m = 3.00\;{\rm{g}} = 3.0 \times {10^{ - 3}}\;{\rm{kg}}\).

Step 2: Concept

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.

Step 3: Determination of the frequency of its fundamental mode of vibration

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, \(408.2\;{\rm{Hz}}\).