Question

# You have two air columns that are each 2.470 m long. One column is open at...

You have two air columns that are each 2.470 m long. One column is open at both ends and the other is closed at one end. You wish to determine the frequencies you can produce in the audible range (20 Hz–20,000 Hz) on a day when the temperature of the air is at 24.00°C. (Give your answers to at least four significant figures. Assume that the speed of sound at 0° C is exactly 331 m/s.)

(a) in the column that is open at both ends

 lowest frequency second lowest frequency highest frequency How can you determine the largest value for n for the audible range? Hz

(b) in the column that is closed at one end

 lowest frequency second lowest frequency How is the length of the air column closed at one end and the speed of sound related to the resonant frequencies? Hz highest frequency How can you determine the largest value for n for the audible range? Hz

speed of sound at ttemperature Tis

v = 331 + 0.6 T

= 331 +0.6(24)

=345.4 m/s

(a)

if column open at both ends the lowest frequency is

f1= v/ 2L = 345.4 m/s/ 2( 2.470) = 69.919 Hz

second lowest

f2 = 2f1 = 2( 69.919) = 139.83 Hz

since the greatest audiable range is

n= 20000 Hz/69.919 = 286

But it's not all since the fundamental frequency is in the audible range. Therefore, the total overtone is: 286 - 1 =

highest frequncy is

f= 286 f1

= 286( 69.91Hz)

=19994.26 Hz

(b)

if the column is closed at one end

f1 = v/4L = 345.4 m/s/4 ( 2.470)

=34.95 Hz

second lowest frequency

f3 = 3f1 = 3( 34.95 Hz) = 104.87 Hz

n = 20000 Hz/34.95 Hz = 572

However, in the closed pipe, only odd harmonics are present. Otherwise, the fundamental frequency is in the audible range; thus, the total overtones is:
(572 - 1) = 571

f= nf1 = 571 ( 34.95 Hz) = 19956.45 Hz

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