Question

An organ is in tune at \({\bf{22}}{\bf{.}}{{\bf{0}}^{\bf{o}}}{\bf{C}}\). By what percent will the frequency be off at \({\bf{1}}{{\bf{1}}^{\bf{o}}}{\bf{C}}\)?

Short answer

The percentage frequency be off at \(11{\rm{^\circ C}}\) is \( - 1.917\% \).

Step-by-step solution

Relationship between the speed of sound, frequency, and temperature.

The speed of sound is directly related to the change in the temperature, and this change in temperature is directly related to the change in frequency.\(v \propto T \propto f\).

Calculation of change in frequency

The expression for the frequency at\(22{\rm{^\circ C}}\)is given as,

\({f_{22{\rm{^\circ C}}}} = \frac{{{v_{22{\rm{^\circ C}}}}}}{\lambda }\)

Here,\({v_{22^\circ C}}\)is the speed of the sound at\(22{\rm{^\circ C}}\)and\(\lambda \)is the resonant wavelength inside the pipe.

The expression for the frequency at\(11{\rm{^\circ C}}\)is given as,

\({f_{{\rm{11^\circ C}}}} = \frac{{{v_{{\rm{11^\circ C}}}}}}{\lambda }\)

Here,\({v_{11^\circ C}}\)is the speed of the sound at\(11{\rm{^\circ C}}\).

The expression for the change in frequency is given as,

\(\Delta f = {f_{11{\rm{^\circ C}}}} - {f_{{\rm{22^\circ C}}}}\)

Substitute the values in the above equation,

\(\begin{array}{c}\Delta f = \frac{{{v_{{\rm{11^\circ C}}}}}}{\lambda } - \frac{{{v_{{\rm{22^\circ C}}}}}}{\lambda }\\ = \frac{{{v_{{\rm{11^\circ C}}}} - {v_{{\rm{22^\circ C}}}}}}{\lambda }\end{array}\)

Calculation of the percentage change in the frequency.

The ratio of change in frequency to the original frequency is given as,

\(\begin{array}{c}\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \frac{{\frac{{{v_{{\rm{11^\circ C}}}} - {v_{{\rm{22^\circ C}}}}}}{\lambda }}}{{\frac{{{v_{{\rm{22^\circ C}}}}}}{\lambda }}}\\ = \frac{{{v_{{\rm{11^\circ C}}}}}}{{{v_{{\rm{22^\circ C}}}}}} - 1\end{array}\)…… (i)

The speed of sound at\(22{\rm{^\circ C}}\)can be written as,

\({v_{22{\rm{^\circ C}}}} = 331 + 0.6{T_{22{\rm{^\circ C}}}}\)

The speed of sound at\(11{\rm{^\circ C}}\)can be written as,

\({v_{{\rm{11^\circ C}}}} = 331 + 0.6{T_{{\rm{11^\circ C}}}}\)

Substitute the values in equation (i),

\(\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \frac{{331 + 0.6{T_{{\rm{11^\circ C}}}}}}{{331 + 0.6{T_{{\rm{22^\circ C}}}}}} - 1\)

The percentage difference is,

\(\begin{array}{c}\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \left( {\frac{{331 + 0.6\left( {11{\rm{^\circ C}}} \right)}}{{331 + 0.6\left( {{\rm{22^\circ C}}} \right)}} - 1} \right) \times 100\% \\ = - 0.01917 \times 100\% \\ = - 1.917\% \end{array}\)

Thus, the percentage frequency be off at \(11{\rm{^\circ C}}\) is \( - 1.917\% \).