The lowest temperature on the moon’s surface occurs during the two weeks that make up a lunar night. This low temperature is -261.4°F. What is this temperature in degrees Celsius and kelvins?

 L = 116.6 m

Explanation:

For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at

               λ = 4L         1st harmonic

               λ = 4L / 3    third harmonic

               λ = 4L / 5    fifth harmonic

General term

               λ = 4L / n      n = 1, 3, 5,...    odd

                                    n = (2n + 1)      n are all integers

They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency

            v = λ f

            λ = v / f

we substitute

             L =

for the first resonance n = n

             L = (2n + 1)

for the second resonance n = n + 1

             L = (2n + 3)

we have two equations with two unknowns, let's solve by equating

             (2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}

             (2n + 1) f₂ = (2n +3) f₁

              2n + 1 = (2n + 3)

              2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1

we substitute the values

              2n (1- ) = 3 -1

              2n 0.21875 = 1.34375

              n = 1.34375 / 2 0.21875

              n = 3

remember that n must be an integer.

We use one of the equations to find the length of the Tunal

                  L = (2n + 1) \frac{v}{4f_1}

                  L = (2 3 + 1)

                  L = 116.55 m