A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length?
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This is Giancoli Answers with Mr. Dychko. For a guitar string, the nth harmonic is n times the speed of the wave on the string divided by 2 l, where n is any natural number, 1, 2, 3, 4 and so on. Now, we're told that the third harmonic is something that we know, it's 54 hertz, and this is when the string is not fingered. And we'll use this to create an expression for l that we can then plug in to the formula for the fundamental when it is fingered because we know that the length then is 70% of what it is before. So, the third harmonic is 3 times v over 2 l and multiply both sides by l over f3 and that gives us l is 3 v over 2 times the third harmonic. And then the fundamental when the string is fingered is v over 2 times this new length, l fingered, and that l fingered is 0.7 times l. I guess I like to do my substitutions in red. So, let's go like that. Where l is 2 f3 over 3 v after you take its reciprocal. So, if we're gonna divide by l, let's instead multiply by its reciprocal because that's a lot easier to look at. So, we said v over 2 times 0.7, here copied. And then we're multiplying by the reciprocal of l. So, 2 times f3 over 3 v and, I don't know if this is being silly but let's just make that in red because that's the way I'd do it, there. And so that's a substitution in red there, 2 f3 over 3 v, the v's cancel, the 2's cancel, we're left with f3 over 0.7 times 3 is going to be the fundamental when fingered. So, this is f3 is the third harmonic when not fingered and that's 540 hertz divided by 0.7 times 3 gives us about 260 hertz will be the fundamental when it is fingered, leaving 70% of the length.
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