An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 5.0 Hz and 6.4 Hz , and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 333 m/s .

 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

we need answer choices