Question
(a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?
Final Answer
Solution video
OpenStax College Physics for AP® Courses, Chapter 17, Problem 42 (Problems & Exercises)
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This is College Physics Answers with Shaun Dychko. A tube is open on both ends and has a length of 0.672 meters and the speed of sound we are told is 344 meters per second. The frequency of the harmonics is n which is the harmonic we are interested in harmonic 1, harmonic 2, harmonic 3 and so on. multiplied by the speed of sound divided by 2 times the length of the tube and the fundamental is another name for harmonic number 1. So we replace the letter n with the number 1 for part (a) when we are finding the frequency of the fundamental. So we have 1 over 2 times 344 meters per second divided by 0.672 meters which is 256 hertz. The second harmonic means n is the number 2 so we have 2 over 2 times 344 meters per second divided by 0.672 meters and that is 512 hertz; This you could also say is the first overtone.