5-6: 7 SEE ANSWER

Answer: 428 Hz, 920 Hz, 1472 Hz, and 2038 Hz

We are given the following values:

    l = 50 cm = 0.5m(l)
    cut 20 cm = 0.2 m from tip (lo)
    velocity at room temperature = 345 m/s
    l - l0 = L = 0.5m - 0.2m = 0.3m = L

Use the formula for the frequency of a conical tube to find the frequency values:

    f1 = 2.49 X (v / (4L)) = 2.49 X (345 / (4 X .3m)) = 428 Hz
    f2 = 3.20 X (v / (4L)) = 3.20 X (345 / (4 X .3m)) = 920 Hz
    f3 = 5.12 X (v / (4L)) = 5.12 X (345 / (4 X .3m)) = 1462 Hz
    f4 = 7.09 X (v / (4L)) = 7.09 X (345 / (4 X .3m)) = 2038 Hz

Therefore, the other resonant frequencies are 428 Hz, 920 Hz, 1462 Hz, and 2038 Hz.

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