Question

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?

Answer 1

Answer:

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