00:01
All right, now we're being asked about a violin.
00:05
Without putting your finger on it, it plays at a frequency of 440 hertz, which happens to be an a note, 30 centimeters, which is 0 .30 meters.
00:30
So what if you want to change the frequency? i'm going to call it f subc.
00:38
To 523 hertz.
00:48
Okay.
00:51
Well, we know that f sub a is v over to l.
00:59
And we know that f sub c is v over to, we're asked for how far from the end of the string.
01:21
So this would be l minus how far from the end of the string, where x is how far it is from the end of the string.
01:33
Okay, solving the first one for v gives me 2lf sub a, substituting that into the second equation gives me 2l f sub a over 2 l minus x.
02:01
The twos cancel out.
02:04
We know what l is.
02:06
We know what f sub a and f sub c are.
02:10
So we need to figure out what x is.
02:13
So let's just keep solving this some more.
02:19
Minus x equals l f sub a over f sub c.
02:28
Just rearranging a little bit...