Chapter 12, Problem 33P

(a)

To determine

To Find: The length that must open organ pipe.

Answer to Problem 33P

The length of the open organ pipe is $0.583\text{m}$ .

Explanation of Solution

Given:

Frequency, $f\text{=294 Hz}$

Temperature, $T=20°\text{C}$

Formula used:

The frequency is calculated as,

  $f=\frac{v}{2L}$

Where $f$ is the frequency, $v$ is the speed and $L$ is the organ pipe length

Calculation:

The fundamental frequency for the open pipe is given as,

  $\begin{array}{l}L=\frac{v}{2f}\\ =\frac{\text{343}\text{m/s}}{\text{2×294}\text{Hz}}\text{=0}\text{.583m}\end{array}$

Conclusion:

The length of open organ pipe is $0.583\text{m}$ .

(b)

To determine

To Find: The fundamental frequency, if the pipe is filled with helium.

Answer to Problem 33P

  $f_{He}=862\text{ Hz}$

Explanation of Solution

Given:

Velocity, when filled with helium gas at $20°\text{C}$ , $v=1005\text{ m/s}$

Length of pipe, $L=0.42\text{ m}$

The speed of the sound in helium is, $v_{\text{He}}=1005\text{m/s}$

Formula used:

The frequency is given as,

  $f_h=\frac{V_{helium}}{2L}\text{ Hz}$ .

Calculation:

Velocity of the helium gas at $20°\text{C}$ , $v=1005\text{ m/s}$

The fundamental frequency for open organ pipe is calculated as,

  $\begin{array}{l}{f}_h=\frac{V_{helium}}{2L}\text{ Hz}\\ f_h=\frac{1005\text{ m/s}}{2\times 0.583\text{ m}}\text{ Hz}\\ f_h=862\text{ Hz}\end{array}$

The fundamental frequency is $862\text{ Hz}$

Conclusion:

The fundamental frequency of the organ pipe is $862\text{ Hz}$