Chapter SH, Problem 10.4APP

To determine

To calculate: Number of times the wavelength of a sound wave in aluminum is longer than wavelength of a soundwave in air.

Answer to Problem 10.4APP

The wavelength of a sound wave in aluminum is longer than wavelength of a soundwave in air by $\underset{\_}{14.7\text{ times}}$ .

Explanation of Solution

Given:

Speed of sound wave in air at $0°\text{C}$ = $331\text{ m/s}$

Speed of sound wave in aluminum at $0°\text{C}$ = $4877\text{ m/s}$

Both waves have same frequency.

Formula used:

Speed is given by,

  $v=f\lambda$

where f is frequency and $\lambda$ is wavelength.

Calculation:

Wavelength of sound wave in air is calculated as:

  $\begin{array}{l}{v}_{\text{air}}=f{\lambda }_{\text{air}}\\ {\lambda }_{\text{air}}=\frac{v_{\text{air}}}{f}\\ {\lambda }_{\text{air}}=\frac{331\text{ m/s}}{f}\end{array}$

Wavelength of sound wave in aluminum is calculated as:

  $\begin{array}{l}{v}_{\text{aluminum}}=f{\lambda }_{\text{aluminum}}\\ {\lambda }_{\text{aluminum}}=\frac{v_{\text{aluminum}}}{f}\\ {\lambda }_{\text{aluminum}}=\frac{\text{4877 m/s}}{f}\end{array}$

Number of times the wavelength of a sound wave in aluminum is longer than wavelength of a soundwave in air is calculated as:

  $\begin{array}{l}\frac{{\lambda }_{\text{aluminum}}}{{\lambda }_{\text{air}}}\\ =\frac{\frac{\text{4877 m/s}}{f}}{\frac{331\text{ m/s}}{f}}\\ =\frac{\text{4877 m/s}}{f}\times \frac{f}{331\text{ m/s}}\\ =14.7\end{array}$

Conclusion:

The wavelength of a sound wave in aluminum is longer than wavelength of a soundwave in air by $\underset{\_}{14.7\text{ times}}$ .