Sound Waves Legacy Problem #20 Guided Solution

This is a difficult problem, and like many difficult problems, it's not so much the mathematics that makes it difficult. What makes this problem difficult is the task of reading and deciphering the given information, trying to figure out what's going on, what's being asked for, being able to use your noodle, and knowing some good physics, physics related to standing wave patterns. What we have are two strings, string A and string B, and they're made of the same material, they all stretch the same tension, and what's important about that is that they have the same speed. However, these strings have different lengths, with A being 80 centimeters long and B being 60 centimeters long. They're vibrated at various frequencies, and as a result, there's standing wave patterns produced within them. What we're to do is to consider the first six standing wave patterns, the first six harmonics, beginning with the fundamental frequency or first harmonic on up to the sixth harmonic. What we wish to figure out is which harmonic of string A would have the same frequency as one of the harmonics of string B. Now the task of determining the answer to this question demands that you recognize that if these two strings have the same speed, then in order to have the same frequency, they must also have the same wavelength. So whatever harmonic of string A has the same wavelength as that harmonic of string B, those two harmonics will also have the same frequency. Now since the speeds aren't given, there's no possible way to calculate the frequencies, because you would need to use the wave equation to do that. So what you need to do is calculate the wavelengths for the six harmonics of string A and of string B, and from those wavelengths, wherever they are equal, one of string A is equal to one of the harmonics of string B, you know you've found your answer. So taking string A, which is the 80 centimeter length string, I can begin to sketch standing wave patterns and figure out the various wavelengths for the varying harmonics. For the first harmonic, with a half of a wavelength within the length of the 80 centimeter string, I would have a wavelength of 160 centimeters. For a wavelength of the second harmonic, it would be 80 centimeters, since a full wave stretches through that 80 centimeters length of string. And I can repeat the process, a pattern quickly develops for wavelength three, it's 53.3 repeating centimeters wavelength, that's because there's three halves of a wave within the 80 centimeters, so I divide the 80 centimeters by three halves. Wavelength four is 40 centimeters, I take the 80 centimeters and I recognize that within the 80 centimeters length of string there's two full wavelengths, I divide by two. And then finally wavelength five, 32 centimeters, divide the 80 by five halves, wavelength six is 26.6 centimeters, divide the 80 by three wavelengths, by three, since there's three wavelengths within the string. So that gives me for harmonics one through six, varying wavelengths are 160, 80, 53.3, 40, 32, and 26.6 centimeters, and I can repeat the process, it's the same logic for string B, only the length of the string is different, it's 60 centimeters. And when I repeat the process of dividing by a half, and dividing by one, dividing by three halves, and dividing by two, and dividing by five halves, and dividing by three, I get these wavelengths, going from the first to the sixth, I get 120 centimeters, 60 centimeters, 40 centimeters, 30 centimeters, 24 centimeters, and 20 centimeters. Now that you've listed all the information, you can now answer the question, and that is for string A, the fourth harmonic has the same wavelength as the third harmonic of string B, and therefore, those two harmonics must also have the same frequency.