.111 AT&T; 11:33 PM loncapa.tcc.fl.edu 21-7,8 Due in2 Standing Wave hours, 25 mi

Question

.111 AT&T; 11:33 PM loncapa.tcc.fl.edu 21-7,8 Due in2 Standing Wave hours, 25 minutes in a Pipe The amplitude of a standing sound wave in a long pipe closed at the left end and open at the right end is sketched below. The vertical axis is the maximum displacement of the air, and the horizontal axis is along the length of the pipe. What is the harmonic number for the mode of oscillation illustrated? Submit Answer Tries 0/15 The length of the pipe is 0.440 m. What is the pitch (frequency) of the sound? Assume that the speed of sound is 330 m/s. Submit Answer Tries 0/15

Explanation / Answer

It is the 5th harmonic.

The closed end must have a node.
The open end must have an antinode.

If the pipe has length L the fundamental (lowest resonance, 1st harmonic), has wavelength where L = (1/4)
= 4L

The next possible harmonic has L = (3/4)
= 4L/3
Note the wavelength is 3 times smaller thnan the fundamental, so the frequency has increased 3 times. So this is the 3rd hamronic.
(There is no 2nd harmonic for a pipe closed at one end and open at the other. It wold require = 2L which is not possible.)

The next possible harmonic has L = (5/4)
= 4L/5
Note the wavelength is 5 times smaller thnan the fundamental, so the frequency has increased 5 times. So this is the 5th hamronic.
(There is no 4th harmonic.)

B) L = 0.440 m

v = 330 m/s

= 4L/5 = 0.352 m

f = v/ = 330/0.350 = 942.86 Hz

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