Question

How far from the mouthpiece of the flute in Example 12–11 should the hole be that must be uncovered to play \({\bf{F}}\) above middle \({\bf{C}}\) at \({\bf{349}}\;{\bf{Hz}}\)?

Short answer

The length of the mouthpiece of the flute is \(0.491\;{\rm{m}}\).

Step-by-step solution

Concept of fundamental frequency.

The fundamental frequency is also known as the first harmonic of the instrument.

The fundamental frequency of a string that stretched between two fixed supports is given as,

\(f = \frac{v}{{2l}}\)

Here, f is the frequency, v is the speed of sound, and l is the length of the pipe.

Calculation of speed of sound at \({\bf{20^\circ C}}\).

The expression for the speed of the sound at temperature T is,

\(v = \left( {331 + 0.60\left( T \right)} \right)\;{\rm{m/s}}\)

The expression for the speed of the sound at temperature \(20^\circ {\rm{C}}\) is,

\(\begin{array}{c}v = \left( {331 + 0.60\left( {20{\rm{^\circ C}}} \right)} \right)\;{\rm{m/s}}\\ = 343\;{\rm{m/s}}\end{array}\)

Calculation of the length of the mouthpiece of the flute.

The expression of the frequency of a string between two fixed supports is,

\(\begin{array}{c}f = \frac{v}{{2l}}\\l = \frac{v}{{2f}}\end{array}\)

Here \(l\) denotes the distance between the mouthpiece (antinode) and the flute tube's first open side hole (antinode).

Substitute the values in the above equation,

\(\begin{array}{c}l = \frac{{\left( {343\;{\rm{m/s}}} \right)}}{{2\left( {349\;{\rm{Hz}}} \right)}}\\ = 0.491\;{\rm{m}}\end{array}\)

Thus, the length of the mouthpiece of the flute is \(0.491\;{\rm{m}}\).