Chapter 15, Problem 58A

To determine

To calculate: The bar’s lowest frequency of oscillation.

Answer to Problem 58A

The bar’s lowest frequency of oscillationis $2675\text{ Hz}$ .

Explanation of Solution

Given:

Length of aluminum rod = $1.2\text{ m}$ .

Speed of sound in aluminum = $6420\text{ m/s}$

The aluminum rod oscillates like an open pipe.

There is a pressure antinode in the center of the bar.

Formula used:

Wavelength of an open pipe is given by,

$\lambda =2L$

Where $L$ is length of the pipe.

Frequency is given by,

$f=\frac{v}{\lambda }$

Where $v$ is velocity and $\lambda$ is wavelength.

Calculation:

Wavelength of the air column is calculated as:

$\begin{array}{l}\lambda =2L\\ =2\times 1.2\text{ m}\\ =2.4\text{ m}\end{array}$

The frequency of oscillation is calculated as:

$\begin{array}{l}f=\frac{v}{\lambda }\\ =\frac{6420\text{ m/s}}{2.4\text{ m}}\\ =2675\text{ Hz}\end{array}$

Conclusion:

The bar’s lowest frequency of oscillationis $2675\text{ Hz}$ .