Question

An open organ pipe has a fundamental frequency of 262 Hz at room temperature (20oC). What is the length of the pipe?

Answer 1

Given:

Fundamental frequency = 262 Hz.

Temperature = 20 degrees celcius

Let's find the length of the pipe.

Apply the resonant frequency:

$f_o=\frac{v}{2L}$

Where:

L is the length.

To find the length, rewrite the equation for L:

$L=\frac{v}{2f_o}$

Where v is te speed, to find the speed, we have:

$\begin{gathered} v=331\sqrt[]{1+\frac{T}{273}} \\ \\ \text{Where:} \\ T=20^0c \\ \\ v=331\sqrt[]{1+\frac{20}{273}} \\ \\ v=331\sqrt[]{1+0.07326} \\ \\ v=331(1.03598) \\ \\ v=342.9\text{ m/s} \end{gathered}$

Thus, to find the length, we have:

$\begin{gathered} L=\frac{342.9}{2\times262} \\ \\ L=\frac{349.2}{524} \\ \\ L=0.67\text{ m} \end{gathered}$

Therefore, the length of the pipe is 0.67 m.

ANSWER:

0.67 m