Question

A vibrating tuning fork is held above a column of air, as shown in the diagrams above. The reservoir is raised and lowered to change the water level, and thus the length of the column of air. 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 L₂ = 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, given that the frequency in the water is the same as the frequency in air.

(d) The water level is lowered again until a third resonance is heard. Calculate the length L₃ of the air column that produces 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?

_____ Too high _____ Too low _____ Still correct Justify your answer.

Answer 1

(a) The shortest length makes the fundamental which looks like this and is ¼ of the wavelength. This length is known to be 0.25m. So L₁ = ¼ λ … λ = 4L₁ = 1m.

Note: This is a real experiment, and in the reality of the experiment it is known that the antinode of the wave actually forms slightly above the top of the air column (you would not know this unless you actually performed this experiment). For this reason, the above answer is technically not correct as the tube length is slightly less than 1/4 of the wavelength. The better way to answer this question is to use the two values they give you for each consecutive harmonic. You are given the length of the first frequency (fundamental), and the length of the second frequency (third harmonic). Based on the known shapes of these harmonics, the difference in lengths between these two harmonics is equal to ½ the wavelength of the wave. Applying this → 

ΔL = ½ λ … 0.8–0.25 = ½ λ λ_actual = 1.1 m.

Unfortunately the AP exam scored this question assuming you knew about the correction; though you received 3 out of 4 points for using the solution initially presented. We teachers, the authors of this solution guide, feel this is a bit much to ask for.

(b) Using v = f λ with the actual λ … (340) = f (1.1) … f = 312 Hz.

(c) v = f λ … (1490) = (312) λ_water … λ_water = 4.8 m

(d) Referring to the shapes of these harmonics is useful. The second length L₃ was the 3^rd harmonic. The next harmonic (5^th) will occur by adding another ½λ to the wave (based on how it looks you can see this). This will give a total length of L₂ + ½ λ = (0.8) + ½ (1.1) = 1.35 m

(e) As temperature increases, the speed of sound in air increases, so the speed used in part (b) was too low. Since f = v_air / λ, that lower speed of sound yielded a frequency that was too low.