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## Direct and inverse variation

Current time:0:00Total duration:4:42

# 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.