OpenStax College Physics for AP® Courses, Chapter 17, Problem 45 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We model the flute as a tube open at both ends and has some length L and we can see that half of the wavelength will fit in this length L when the wavelength is added longest. So, that’s the fundamental wavelength or the wavelength of the first harmonic and so that means lambda one is going to be two times L and so the fundamental frequency is v over lambda one and because we have this wave speed formula, v equals frequency times lambda which we can then solve for f and say its v divided by lambda and so that’s v over two L and the… cause we are told what the fundamental frequency of this flute is and we are told that the temperature as well so we substitute this in red in place of v so that’s the speed of sound in terms of the air temperature and then we are gonna solve this all for L and so we will multiply both sides by L and divide both sides by fundamental frequency and so L is 331 meter per second divided by two times the fundamental frequency times square root of temperature in absolute kelvin divided by 273, so that’s 331 meter per second divided by two times the frequency of the metal which is 262 hertz times square root of 20 degrees Celsius converted into kelvin by adding 273.15 divided by 273 and we end up with 65.4 centimetre, must be the length of flute.