Example E: A vibrating tuning fork is held above a column of air. The reservoir
ID: 1874180 • Letter: E
Question
Example E: A vibrating tuning fork is held above a column of air. The reservoir is raised and lowered to change the water level. The shortest length of air column that produces a resonance is L1-0.25 m, and the next resonance is heard when the air column is L 0.80 m long. The speed of sound in air at 20°C is 343 m/s and the speed of sound in water is 1490 m/s a) Calculate the wavelength of the standing sound wave produced by this tuning fork. b) Calculate the frequency of the tuning fork that produces the standing wave, assuming the air is at 20°C c Calculate the wavelength of the sound waves produced by this tuning fork in the water d) The water level is lowered again until a third resonance is heard. Calculate the length Lj of the air column uces this third resonance. e) The student performing this experiment determines that the temperature of the room is actually slightly higher than 20°C. Is the calculation of the frequency in part (b) too high, too low, or still correct?Explanation / Answer
Solution:
Let us go to the basics first.
The usual assumption is that the standing wave in the pipe has an antinode at the open end and a node at the water. Thus, the shortest air column corresponds to 1/4 wave, and the next column to 3/4 waves.
The wave equation is v = f where v is the velocity of sound in air, f the frequency, and the wavelength.
(a) 1/4 = .25 m so = 1.0 m (Answer)
3/4 = .80 m so = 1.1 m (Answer)
So the wavelength is just over one meter. (Answer)
We'll just use 1 m for the rest of the problem.
(b) Using the wave equation, 343 m/s = f ( 1.0 m ) so f = about 343 Hz (Answer)
(c) = speed / frequency = 1490 / 343 = 4.34 m (Answer)
(d) The next resonance occurs when the length of the air column is 5/4
5/4 (1.0 m ) = about 1.25 m. (Answer)
Thanks!!!
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