g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

The difference between second harmonic and first harmonic;

The difference between third harmonic and second harmonic;

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

Therefore, the length of the tube is 85 cm

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