A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. The student reports hearing two successive resonances at 53.95 Hz and 59.00 Hz. (Assume that the speed of sound in air is 343 m/s.)

(a) How deep is the well?

(b) How many antinodes are in the standing wave at 53.95 Hz?

a) L = 33.369 m, b) 21

Explanation:

The analysis of the ocean depth can be performed assuming that at the bottom of the ocean there is a node and the surface must have a belly, so the expression for resonance is

λ = 4 L / n

n = 1, 3, 5, ...

The speed of the wave is

v = λ f

v = 4L / n f

L = n v / 4f

Let's write the expression for the two frequencies

L = n₁ 343/4 53.95

L = n₁ 1,589

L = n₂ 343/4 59

L = n₂ 1.4539

Let's solve the two equations

n₁ 1,589 = n₂ 1,459

n₁ / n₂ = 1.4539 / 1.589

n₁ / n2 = 0.91498

Since the two frequencies are very close the whole numbers must be of consecutive resonances, let's test what values give this value

n₁ n₂ n₁ / n₂

1 3 0.3

3 5 0.6

5 7 0.7

7 9 0.77

9 11 0.8

17 19 0.89

19 21 0.905

21 23 0.913

23 25 0.92

Therefore the relation of the nodes is n₁ = 21 and n₂ = 23

Let's calculate

L = n₁ 1,589

L = 21 1,589

L = 33.369 m

b) the number of node and nodes is equal therefore there are 21 antinode