Chapter 15, Problem 62A

(a)

To determine

To calculate: The length of the shortest organ pipe that will resonate at the given frequency.

Answer to Problem 62A

The length of the organ pipe is $\underset{\_}{\text{10}\text{.45 m}}$

Explanation of Solution

Given:

Frequency of the lowest note = $\text{16}\text{.4 Hz}$ .

Formula used:

Wavelength is given by the formula:

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

Where $v$ is velocity and $f$ is frequency.

Length of an open pipe is given by the formula:

$L=\frac{\lambda }{2}$

Where $\lambda$ is wavelength.

Calculation:

Wavelength is calculated as:

$\begin{array}{l}\lambda =\frac{v}{f}\\ =\frac{343\text{ m/s}}{16.4\text{ Hz}}\\ =20.9\text{ m}\end{array}$

Length of the organ pipe is calculated as:

$\begin{array}{l}L=\frac{\lambda }{2}\\ \text{ =}\frac{20.9\text{ m}}{2}\\ \text{ =10}\text{.45 m}\end{array}$

Conclusion:

The length of the organ pipe is $\underset{\_}{\text{10}\text{.45 m}}$

(b)

To determine

To calculate: The pitch if the organ pipe is closed.

Answer to Problem 62A

The pitch if the organ pipe is closed is $\underset{\_}{8.20\text{ Hz}}$ .

Explanation of Solution

Given:

Frequency of the lowest note of open pipe = $\text{16}\text{.4 Hz}$ .

Formula used:

The lowest frequency of an open pipe is given by the formula:

$f=\frac{v}{2L}$

Where $v$ is velocity and $L$ is length.

The lowest frequency of a closed pipe is given by the formula:

$f=\frac{v}{4L}$

Where $v$ is velocity and $L$ is length.

Calculation:

Let frequency of the closed pipe be $f_a$ and frequency of the open pipe be $f_b$ .

The lowest frequency of the closed pipe can be calculated as:

$\begin{array}{l}{f}_b=\frac{v}{4L}\\ \text{ =}\frac{1}{2}\cdot \frac{v}{2L}\\ \text{ =}\frac{1}{2}\cdot f_a\\ \text{ =}\frac{1}{2}\cdot (16.4\text{ Hz})\\ =8.20\text{ Hz}\end{array}$

Conclusion:

The pitch if the organ pipe is closed is $\underset{\_}{8.20\text{ Hz}}$ .