A person hums into the top of a well (tube open at only one end) and finds that standing waves are established at frequencies of 40.6, 67.7, and 94.7 Hz. The frequency of 40.6 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?

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