A closed-pipe resonator has a length of 0.672 m, and the speed of sound is 344 m/s. What is its fundamental frequency?

1. For a closed-pipe resonator, the fundamental frequency satisfies f1 = v / (4L). 2. Identify the knowns: v = 344 m/s and L = 0.672 m. 3. Substitute the SI values: f1 = 344 m/s / (4 x 0.672 m) = 344 / 2.688. 4. Compute: f1 = 128.0 Hz, matching option A. 5. Option B, 64 Hz, is half the correct value, consistent with using 8L instead of 4L in the denominator. 6. Options C and D, 256 Hz and 512 Hz, are double and four times the correct value respectively, consistent with dropping a factor from the denominator. _Source: OpenStax Physics (CC BY 4.0), Ch 14 "Sound", section Open-Pipe and Closed-Pipe Resonators_