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# Inverse variation word problem: string vibration

## Video transcript

We're told in this question that on a string instrument, the length of a string-- so let's call that l-- the length of a string l varies inversely as the frequency, so varies inversely as the frequency. So l is going to be equal to some constant times the inverse of the frequency. And I'll use f for frequency-- the frequency of its vibrations. And then they tell us the vibrations are what give the string instruments their sound. That's nice. An 11-inch-- and it's actually the vibrations of the string affecting the air, and then the air compressions eventually get to our eardrum. And that's actually what gives us the perception of the sound. But we don't want to delve too much into the physics of it. An 11-inch inches string has a frequency. So 11-inch, so this is its length. So the 11-inch string has a frequency of 400 cycles per second. So this right here is the frequency. A cycle per second is also called a Hertz. Find the constant of proportionality, and then find the frequency of a 10-inch string. So they say an 11-inch string. So 11-inch string is equal to some constant of proportionality times 1/400 cycles per second. So one over 400 cycles per second. I'll write second as sec. So to solve for the constant of proportionality, we need to multiply both sides by 400 cycles per second. Multiply the left hand side by 400 cycles per second, and the left hand side becomes 400 times 11. Well, 4 times 11 is 44. So 400 times 11 is 4,400. And then we have in our units, just for out of interest, our units are cycles times inches per second. So it's cycles times inches in the numerator of our units divided by seconds. And that is equal to our constant of proportionality. So we can say that the length is equal to 4,400 times-- or 4,400 cycles times inches per second-- I want to get the units right-- per second, times 1 over the frequency. So we solved for our constant of proportionality. And then we can use this to find the frequency of a 10-inch string. So now we're talking about a situation where our length is 10 inches, so 10 inches. So we get 10 inches are equal to 4,400-- as you can imagine, these units are a little cumbersome-- but 4,400 cycles times inches per second times 1 over the frequency. And so we could do a couple of things. We could just multiply both sides of this equation by the frequency so that it gets out of the denominator. So let's do that. Let's multiply both sides by the frequency. And then we could also divide both sides by 10 inches to get rid of this. And then we'll just have frequency on the left hand side. So we divide both sides by 10 inches. And then we divide it by 10 inches. The left hand side, we're just left with the frequency. On the right hand side, we have 4,400 divided by 10. 4,400 divided by 10 is 440. And then you have cycles, inches over second divided by inches. Well, the inches cancel out with inches. And you're just left with cycles per second. And then obviously, 1/f times f cancels out to just being 1. So we get our frequency when our string is when our string is 10 inches long, our frequency has increased to 440 cycles per second. From when it was 11 inches, when it was slightly longer, our frequency was 400 cycles per second. When our string got little bit shorter, one inch shorter, now our frequency increased by 40 cycles per second.