91Ó°ÊÓ

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.\(v \propto T \propto f\).

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,

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

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 \( - 1.917\% \).