Question

What frequencies will a 1.80 m long tube produce in the audible range at 20.0 C if:

(a) The tube is closed at one end?

(b) It is open at both ends?

Short answer

(a) The frequencies if the tube is closed at one end are47.63 n, n = 1,3,5,…

(b) The frequencies if the tube is open at both ends are 95.25 n , n = 1,2,3,…

Step-by-step solution

Given Data

The length of the tube isl = 1.80 m.

The temperature is \[T = 20.0^\circ C = 293\;K\].

Concept

The expression for the n^th overtone frequency of ear canal is given by,

\[{f_n} = n\frac{v}{{4l}}\]

Here v is the velocity of sound, l is the length of the wind instrument and f is the frequency of sound.

Calculation of the frequency for closed tube  

(a)

For a closed tube, the resonance frequencies are,

\[{f_n} = n\frac{v}{{4l}},\;n = 1,3,\;5,\;....\]

The speed of sound at t₁temperature is

\[v = 331\sqrt {\frac{T}{{273}}} \]

The fundamental frequency is,

\[\begin{aligned}{f_1} = \frac{{331\sqrt {\frac{{293}}{{273}}} }}{{4 \times 1.80}}\\ = 47.63\;Hz\end{aligned}\]

Therefore the frequencies are 47.63 n, n = 1,3,5,….

Calculation of the frequency for closed tube  

(b)

For an open tube, the resonance frequencies are,

\[{f_n} = n\frac{v}{{2l}},\;n = 1,2,\;3,\;....\]

The fundamental frequency is,

\[\begin{aligned}{f_1} = \frac{{331\sqrt {\frac{{293}}{{273}}} }}{{2 \times 1.80}}\\ = 95.25\;Hz\end{aligned}\]

Therefore the frequencies are 95.25 n , n = 1,2,3,….