Answer to Question #132466 in Physics for Tex

Question

The frequency of a fundamental note from a closed pipe is 200Hz. What is the frequency of the next possible note from the same pipe?

Answer 1

The fundamental frequency (with $n = 1$ ) in a closed pipe is given by the following expression:

$$f_1 = \dfrac{nv}{2L} = \dfrac{v}{2L}$$

where $v = 340 m/s$ is the speed of sound in air. Then the length of the pipe will be:

$$L = \dfrac{v}{2f_1}$$

The next possible note will appear at $n = 2$ .Then:

$$f_2 = \dfrac{2v}{2L} = 2f_1 = 2\cdot 200 Hz = 400Hz$$

Answer. 400 Hz.