Finding the Third Overtone in an Open-Pipe Resonator

If a tube that’s open at both ends has a fundamental frequency of 120 Hz, what is the frequency of its third overtone?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Since we already know the value of the fundamental frequency (n = 1), we can solve for the third overtone (n = 4) using the equation $f_{n} = n \frac{v}{2 L}$ .
Solution
Since fundamental frequency (n = 1) is

$f_{1} = \frac{v}{2 L},$
14.17

and

$f_{4} = 4 \frac{v}{2 L}, f_{4} = 4 f_{1} = 4 (120 Hz)=480 Hz .$
14.18

0 of 2 steps open

How did it go?