You have two air columns that are each 2.390 m long. One column is open at both

Question

You have two air columns that are each 2.390 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 14.00°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

lowest frequency    
Hz second lowest frequency     Hz highest frequency    
Hz

Explanation / Answer

speed of sound at 20 C,

v = vo + 0.6*T

= 331 + 0.6*14

= 339.4 m/s

a) fundamental frequency, fo = v/(2*L)

= 339.4/(2*2.39)

= 71.00 Hz

so,

lowest frequency   = 71.00 Hz

second lowest frequency = 2*fo

= 2*71

= 142.0 Hz

f_highest = n*fo

==> n = f_gighest/fo

= 20000/71

= 208.6

so, n = 281.7

highest frequency = 281*71

= 19951 hz

b)
fundamental frequency, fo = v/(4*L)

= 339.4/(4*2.39)

= 35.50 Hz

so,

lowest frequency   = 35.50 Hz

second lowest frequency = 3*fo

= 3*35.50

= 106.5 Hz

f_highest = n*fo

==> n = f_gighest/fo

= 20000/35.5

= 417.2

so, n = 563

highest frequency = 563*35.5

= 19986.5 hz

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