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22 November, 22:26

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?

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  1. 22 November, 22:45
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    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
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