 Creating musical sounds

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# 5.10 Vibrating air column: end effects

In the previous two sections on standing waves in cylindrical tubes, we assumed that at an open end there must be a pressure node. In fact, the pressure node (and the corresponding displacement antinode) actually lies a small distance outside the tube. The effect is that the air column behaves as though it were a little longer than it really is by an amount called the end correction. Because of this end correction, the resonance frequencies will be a little lower than originally expected.

For a cylindrical tube of radius r, the end correction is approximately 0.6 × r. This amount should be added to the geometrical length of the tube for each open end to find the effective length of the air column. In other words, for a tube closed at one end and open at the other, an end correction of 0.6r must be added to find the effective air-column length. For a tube open at both ends, an end correction of 1.2r must be added to find the effective air-column length. The effective air-column length should then be used in the expressions for the resonance frequencies developed in the previous two sections.

## Activity 20

(a) Calculate the effective length of a 1.5 m long cylindrical tube of radius 0.02 m that is open at one end and closed at the other.

(b) Using your result from part (a), calculate the first three resonance frequencies of the tube. Assume that the speed of sound is 340 m/s.

(c) Compare your answers with those from Activity 19.

### Answer

(a) The total end correction for a 1.5 m long cylindrical tube of 0.02 m radius that is open at one end and closed at the other is 0.6 × r = 0.6 × 0.02 = 0.012 m. Therefore, the effective length of the tube is 1.5 + 0.012 = 1.512 m.

(b) Using the above value for the effective length, the first three resonance frequencies are determined as follows:

First resonance frequency f1 = v/4L = 340/(4 × 1.512) = 56.2 Hz.

Second resonance frequency f2 = 3v/4L = (3 × 340)/(4 × 1.512) = 168.7 Hz.

Third resonance frequency f3 = 5v/4L = (5 × 340)/(4 × 1.512) = 281.1 Hz.

(c) Comparing these results with those of Activity 19 shows that when the end correction is taken into account, the calculated resonance frequencies are slightly lower.

TA212_2

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