Chapter 12, Problem 31P

To determine

To Find: The percentage that the frequency will be off.

Answer to Problem 31P

The percentage that the frequency off is $-2.6\%$ .

Explanation of Solution

Given:

Organ in tune $\text{= 20°C}$

Frequency off at $\text{= 5°C}$

Formula used:

The frequency fraction is determined as,

  $\frac{\Delta f}{f}=\frac{V_2}{V_1}$

Where, $\Delta f$ is the change in frequency, $V_1$ and $V_2$ are the initial and final volume

Calculation:

As the temperature is changes, the sound speed is also changed. So, the frequency of the organ pipe is also changed. Let us consider the length of the pipe is unchanged. So, the frequency for each case is,

  $\begin{array}{l}{f}_{20}=\frac{v_{20}}{\lambda }\\ and,\\ f_{5.0}=\frac{v_{5.0}}{\lambda }\end{array}$

The change in frequency is,

  $\begin{array}{c}\Delta f=f_{5.0}-f_{20}\\ =\frac{v_{5.0}-v_{20}}{\lambda }\end{array}$

Now use this formula and other values in the above formula and the frequency percentage is given as,

  $\begin{array}{c}\frac{\Delta f}{f}=\frac{\frac{v_{5.0}-v_{20}}{\lambda }}{\frac{v_{20}}{\lambda }}\\ =\frac{v_{5.0}}{v_{20}}-1\\ =\frac{331+0.60\left(5.0\right)}{331+0.60\left(20\right)}-1\\ =-2.6\times {10}^{-2}\\ =-2.6\%\end{array}$

Conclusion:

The frequency percentage is $-2.6\%$