Question

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

Answer 1

The length of the tube is 85 cm

Step-by-step 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

`\lambda = (4L)/(3)\\\\ F= (v)/(\lambda)\\\\F_2 = (3v)/(4L)`

Third harmonic of open-closed tube is given by;

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

`\lambda = (4L)/(5)\\\\ F= (v)/(\lambda)\\\\F_3 = (5v)/(4L)`

The difference between second harmonic and first harmonic;

`F_2 -F_1 = (3v)/(4L) - (v)/(4L)\\\\F_2 -F_1 = (2v)/(4L) \\\\F_2 -F_1 =(v)/(2L)`

The difference between third harmonic and second harmonic;

`F_3 -F_2 = (5v)/(4L) - (3v)/(4L)\\\\F_3 -F_2 = (2v)/(4L) \\\\F_3 -F_2 =(v)/(2L)`

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

v / 2L.

`700H_z- 500H_z= (v)/(2L) \\\\200 = (v)/(2L)\\\\L = (v)/(2*200) \\\\L = (340)/(2*200)\\\\L = 0.85 \ m\\\\L = 85 \ cm`

Therefore, the length of the tube is 85 cm