You have two air columns that are each 2.430 m long. One column is open at both

Question

You have two air columns that are each 2.430 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

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

Explanation / Answer

speed of sound = 331 * sqrt(temperature / 273)

24 degree C = 297.15 K

speed of sound = 331 * sqrt(297.15 / 273)

speed of sound = 345.33 m/s

frequency will be lowest when wavelength will be highest and vice versa

frequency = speed / wavelength

frequency = 345.33 / wavelength

for open end pipe

wavelength for nth harmonic =  (2 / n) * L

20 = 345.33 / ((2 / n) * 2.43)

n = 0.28

so for n = 1 the frequency will be lowest

lowest frequency = 345.33 / ((2 / 1) * 2.43)

lowest frequency = 71.055 Hz

second lowest frequency = 345.33 / ((2 / 2) * 2.43)

second lowest frequency = 142.11 Hz

20000 = 345.33 / ((2 / n) * 2.43)

n = 281.47

so for 281 harmonic the frequency will be highest

highest frequency = 345.33 / ((2 / 281) * 2.43)

highest frequency = 19966.6111 Hz

for closed end pipe

wavelength for nth harmonic =  (4 / n) * L

20 = 345.33 / ((4 / n) * 2.43)

n = 0.56

for n = 1 the frequency will be lowest so,

lowest frequency = 345.33 / ((4 / 1) * 2.43)

lowest frequency = 35.527 Hz

second lowest frequency = 345.33 / ((4 / 3) * 2.43)

second lowest frequency = 106.583 Hz

20000 = 345.33 / ((4 / n) * 2.43)

n = 562.94

so for n = 561 the frequency will be highest

highest frequency = 345.33 / ((4 / 561) * 2.43)

highest frequency = 19931.083 Hz

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