A tube, open at only one end, is cut into two shorter (non-equal) lengths. The piece open at both ends has a fundamental frequency of 425 Hz, while the piece open only at one end has a fundamental frequency of 675 Hz, both in room temperature air. What is the fundamental frequency of the original tube?

f = 158.6Hz

Explanation

Given Data

f₀=425 Hz ... Open at both ends has a fundamental frequency

f₁=675 Hz ... open only at one end has a fundamental frequency

To find

Fundamental frequency of the original tube

Solution

Fundamental frequency with open ends given as

f = v / (2·L)

where v speed of sound,

and L is length

And

Fundamental frequency of a tube closed at one end is given as

f = v / (4·L)

For the piece with two open ends:

f₀ = v / (2·L₀)

L₀ = v / (2·f₀)

For the other piece

f₁ = v / (4·L₁)

L₁ = v / (4·f₁)

The fundamental frequency of the original tube is that of a closed end tube of length L₀ + L₁.

f = v / (4· (L₀ + L₁)

f = v / (4· (v / (2·f₀) + v / (4·f₁))

f = 1 / (2/f₀ + 1/f₁)

f = 1 / (2 / 425Hz + 1 / 625Hz)

f = 158.6Hz