00:01
So i wrote down the pertinent information.
00:04
We've got m.
00:06
We've got m, which is the mass attached to the end.
00:12
We don't know the length of this wire.
00:20
And i'm going to say that l is the total length.
00:28
Okay.
00:30
So let's work with the pulley situation first.
00:36
We know that v is the square root of t over m, but t equals m times g.
00:53
So v is the square root of mg over mu.
01:04
Now we also know that the second frequency is 100 hertz.
01:12
So that would be 2v over 2l.
01:23
This is for the second mode.
01:26
2v over 2l.
01:29
So f sub 2 is v over l actually.
01:36
So that's going to be 1 over l square root of mg over mu.
01:45
Where we know m, we know mu.
01:49
We don't know l.
01:51
We don't know g, but we do know f2.
01:54
So there's two unknowns here.
01:58
So let's switch to the pendulum.
02:05
Now, the period of a pendulum is 2 pi square root of l over g.
02:22
And the period for this is 3 .5.
02:28
314 seconds per 100 oscillations.
02:34
So the period is going to be 314 divided by 100 seconds.
02:44
314 divided by 100, which is just going to be 3 .14 seconds, which is basically pi.
02:55
So now what we're saying is that pi, equals 2 pi square root of l over g cancel out pi um and i want to solve this for l so if i square both sides let me just write this again one half equals the square root of l over g and so one four fourth is going to be l over g.
03:40
And so l is equal to g over 4.
03:45
L is g over 4.
03:47
All right, so we can substitute that back into the other equation.
03:52
F2 equals 1 over l.
03:58
That would just be 4 over g, because l is g over 4, square root of m g over b.
04:08
M.
04:13
Going farther, square root of g, square root of m over mu.
04:26
So the square root of g is 4 over f2 square root of m over mu.
04:37
And so g is squaring both sides, 16 over f2 squared, m over for a mu.
04:51
So let's put that into a calculator.
04:57
16 over f2 squared, f2 was 100 hertz.
05:14
And then that is times square root of m over mu.
05:28
M is 1 .25 and mu is 0 .001.
05:43
Okay.
05:44
This is giving me an answer 0 .057 meters per second squared.
05:57
That doesn't seem reasonable to me, just because g on earth is 9 .81.
06:09
So let's just think about what i did again here.
06:19
So going through, i wrote this correctly and i wrote that correctly.
06:41
Wait a minute f2 is 100 which is a square root of if v is a square root of t over mu um all right v is mg over mu that makes sense only i'm questioning this l right here when this is vibrating it's only going to be vibrating over half of the string so so perhaps what the original question was was only for half of the string, since it's hanging over a pulley.
07:41
So this l over here is really half of the length.
07:55
And so now f is going to be v over one half l, so there'd have to be a two there.
08:12
But then on the right side with the pendulum, we found out that l is g over 4.
08:20
So there'd be a 2 down here.
08:25
And now this, trying to think of, okay, yeah, 4 divided by 2, it's going to be 2, 2, 8.
08:51
Now that gives us a g that's half, which seems even farther away...