An open-pipe resonator has a fundamental frequency of 120 Hz. What is the frequency of its third overtone?

1. For an open-pipe resonator, the resonant frequencies are f_n = n x v / (2L), where f1 is the fundamental, f2 is the first overtone, f3 is the second overtone, and f4 is the third overtone. 2. Since f1 = v / (2L) = 120 Hz, the third overtone corresponds to n = 4, so f4 = 4 x f1. 3. Substitute: f4 = 4 x 120 Hz = 480 Hz, matching option A. 4. Option B, 360 Hz, would instead result from using n = 3, which is the second overtone, not the third. _Source: OpenStax Physics (CC BY 4.0), Ch 14 "Sound", section Open-Pipe and Closed-Pipe Resonators_