You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower

a) L = 0.75m f₁ = 113.33 Hz, f₃ = 340 Hz, b) L=1.50m f₁ = 56.67 Hz, f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1 node λ₁ = 4 L

With 2 nodes λ₂ = 4L / 3

With 3 nodes λ₃ = 4L / 5

The general term would be λ_n = 4L / n n = 1, 3, 5, ((2n + 1)

The speed of sound is

v = λ f

f = v / λ

f = v n / 4L

Let's consider each length independently

L = 0.75 m

f₁ = 340 1/4 0.75 = 113.33 n

f₁ = 113.33 Hz

f₃ = 113.33 3

f₃ = 340 Hz

L = 1.5 m

f₁ = 340 n / 4 1.5 = 56.67 n

f₁ = 56.67 Hz

f₃ = 56.67 3

f₃ = 170 Hz