In the case of a tube that is open at one end, a node is forced at the closed end (no air molecules can vibrate up and down) and an antinode occurs at the open end (here, air molecules are free to move). A different spectrum of standing waves is produced. For instance, the fundamental standing sound wave produced in a tube closed at one end is shown below. In this case, the amplitude of the standing wave is referring to the magnitude of the air pressure variations.

This standing wave is the first harmonic and one can see that the wavelength is \begin{align*}\lambda = 4L\end{align*}. Since \begin{align*}v = \lambda f\end{align*}, the frequency of oscillation is \begin{align*}f = v/4L\end{align*}. In general, the frequency of oscillation is \begin{align*}f = nv/4L\end{align*}, where \begin{align*}n\end{align*} is always odd.

**for a tube closed at one end**

\begin{align*}f =nv /4L\end{align*}, where \begin{align*}n\end{align*} is always odd

**for a tube open at both ends**

\begin{align*}f =nv /2L\end{align*}, where \begin{align*}n\end{align*} is an integer

#### Example

The objects A, B, and C below represent graduated cylinders of length 50 cm which are filled with water to the depths of 10, 20 and 30 cm, respectively as shown.

a) If you blow in each of these tubes, which (A,B,C) will produce the highest frequency sound?

The water forms the bottom of the tube and thus where the node of the wave will be. Thus the air column is where the sound wave can exist. The larger the air column, the larger the wavelength. Frequency is inversely proportional to wavelength, thus the tube with the smallest air column will have the highest frequency. So the answer is tube C.

b) What is the wavelength of the 1st harmonic (i.e. fundamental) of tube B?

The air column is 50 cm - 20 cm = 30 cm. The first harmonic has a quarter wavelength in the tube. Thus \begin{align*} \lambda = 4 \times L \end{align*}. Thus, \begin{align*} \lambda = 4 \times 30 \text{cm} = 120 \text{cm} \end{align*}

c) The speed of sound at room temperature is about 343 m/s. What is the frequency of the 1st harmonic for tube B?

Using the wave equation for the first harmonic (thus, n = 1) of a tube open at one end we get \begin{align*}f = \frac{v}{4L} = \frac{343 \text{m/s}}{1.2 \text{m}}= 286 \text{Hz}\end{align*}

### Review

- Aborigines, the native people of Australia, play an instrument called the Didgeridoo like the one shown above. The Didgeridoo produces a low pitch sound and is possibly the world’s oldest instrument. The one shown above is about 1.3 m long and open at both ends.
- Knowing that when a tube is open at both ends there must be an antinode at both ends, draw the first 3 harmonics for this instrument.
- Calculate the frequency of the first 3 harmonics assuming room temperature and thus a velocity of sound of 340 m/s. Then take a shot at deriving a generic formula for the frequency of the \begin{align*}n\end{align*}th standing wave mode for the Didgeridoo, as was done for the string tied at both ends and for the tube open at one end.

- Students are doing an experiment to determine the speed of sound in air. They hold a tuning fork above a large empty graduated cylinder and try to create resonance. The air column in the graduated cylinder can be adjusted by putting water in it. At a certain point for each tuning fork a clear resonance point is heard. The students adjust the water finely to get the peak resonance then carefully measure the air column from water to top of air column. (The assumption is that the tuning fork itself creates an anti-node and the water creates a node.) The following data were collected:

Frequency of tuning fork (Hz) |
Length of air column (cm) |
Wavelength (m) |
Speed of sound (m/s) |
---|---|---|---|

184 | 46 | ||

328 | 26 | ||

384 | 22 | ||

512 | 16 | ||

1024 | 24 |

(a) Fill out the last two columns in the data table.

(b) Explain major inconsistencies in the data or results.

(c) The graduated cylinder is 50 cm high. Were there other resonance points that could have been heard? If so what would be the length of the wavelength?

(d) What are the inherent errors in this experiment?

- Peter is playing tones by blowing across the top of a glass bottle partially filled with water. He notices that if he blows softly he hears a lower note, but if he blows harder he hears higher frequencies. (a) In the 120 cm long tubes below draw three diagrams showing the first three harmonics produced in the tube. Please draw the waves as transverse even though we know sound waves are longitudinal (reason for this, obviously, is that it is much easier to draw transverse waves rather than longitudinal). Note that the tube is CLOSED at one end and OPEN at the other. (b) Calculate the frequencies of the first three harmonics played in this tube, if the speed of sound in the tube is 340 m/s. (c) The speed of sound in carbon dioxide is lower than in air. If the bottle contained \begin{align*}CO2\end{align*} instead of air, would the frequencies found above be higher or lower? Knowing that the pitch of your voice gets higher when you inhale helium, what can we say about the speed of sound in \begin{align*}He\end{align*}.

**Review (Answers)**

- (b) 131 Hz, 262 Hz, 393 Hz; formula is same as closed at both ends
- Discuss in class
- (b) 70.8 Hz, 213 Hz, 354 Hz (c) voice gets lower pitch. Speed of sound in He must be faster by same logic.