Answer:
a) 384, 1153 and 1819.9 Hz, b) L = 0.44 m
Step-by-step 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