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?
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
A) A buoy rises in the water as a boat speeds past.
Explanation:
The passing boat transfers energy in the form of a wave. Other options illustrate other physics concepts like gravity (falling egg) or Newton's law (for every action, there is an equal and opposite reaction).
