In a pipe closed at one end and filled with air, a 384-Hz tuning fork resonates when the pipe is 22 cm long; this tuning fork does not resonate for any smaller pipes. a. State three other lengths at which this pipe will resonate with the 384-Hz tuning fork. b. The end of the pipe that was closed is now opened, so that the pope is open at both ends. Describe any changes in the lengths of pipe that will resonate with the 384-Hz tuning fork. c. The air in the closed pipe is replaced with helium. Describe an experiment that would use the pipe to determine the speed of sound in helium.

a) 384, 1153 and 1819.9 Hz, b) L = 0.44 m

Explanation:

In a one-tube system open at one end and closed at the other =, at the open end you have an antinode (belly) and at the closed end a node, so the fundamental resonance is

fundamental λ = 4L

3rd harmonic λ = 4L / 3

5th harmonic λ = 4L / 5

general term λ = 4L / n n = 1, 3, 5, ...

a) the frequencies for which the pipe resonates are

fundamental λ = 4 0.22 = 0.88m

with the resonance frequency we can find the speed of sound in the air

v = λ f

v = 0.88 384

v = 337.9 m / s

this is the speed that we will use in all the rest of the problem

for the 3rd harmonic

λ = 4 0.22 / 3 = 0.293 m

f = 337.9 / 0.293

f = 1153.3 Hz

for the 5th harmonic

λ = 4 0.22 / 5 = 0.176 m

f = 337.9 / 0.176

f = 1919.9 Hz

b) In the case of the two open ends there are bellies in both, for which

fundamental λ = 2L

2 harmonic λ = 2L / 2 = L

3 harmonic λ = 2L / 3

General term λ = 2L / n n = 1, 2, 3, ...

let's find the wavelength for the resonance

v = λ f

λ = v / f

λ = 337.9 / 384

λ = 0.8799 m

let's find the tube length for this resonance

λ = 2L

L = λ / 2

L = 0.8799 / 2

L = 0.44 m

therefore for the same resonance the tube must be twice as long

c) If the air is exchanged for helium we look for the length of the tube to find how long the resonance of 384 Hz is, so if the system is open at one end and closed at the other

v = (4L) f

v = 4 L 384

where L is measured