Answer:
Explanation:
The resonance in a tube that is open at one end and closed at the other, we can find it because in the closed part you have a node and in the open part you have a belly, so for the fundamental frequency you have ¼ Lam, the different resonances are:
Fundamental λ = 4L
3 harmonica λ = 4L / 3
5 Harmonica λ = 4L / 5
General harmonica λ = 4L / (2n-1) n = 1, 2, 3
Let's apply this equation to our case
The speed of sound is given by
v = λ f
λ = v / f
Let's look for wavelengths
λ₁ = 40.6 / 343 = 0.1184 m
λ₂ = 67.7 / 343 = 0.1974 m
λ₃ = 94.7 / 343 = 0.2761 m
Since the wavelengths are close we can assume that it corresponds to consecutive integers, where for the first one it corresponds to the integer n
λ ₁ = 4L / (2n-1)
λ₂ = 4L / (2 (n + 1) -1) = 4L / (2n +1)
λ₃ = 4L / (2 (n + 2) -1) = 4L / (2n + 3)
Let's clear in the first and second equations
2n-1 = 4L / λ₁
2n +1 = 4L / λ₂
Let's solve the system of equations
4L / λ₁ + 1 = 4L / λ₂ -1
4L / λ₂ – 4L / λ₁ = 2
2L (1 / λ₂ - 1 / λ₁) = 1
1 / L = 2 (1 / λ₂ -1 / λ₁)
1 / L = 2 (1 / 0.1974 - 1 / 0.1184)
1 / L = 2 (5,066 - 8,446) = -6.76
L = 0.1479 m
