00:01
We're using an experiment here to determine the diameter of the sphere.
00:10
So there is a string tied to a frictionless pulley.
00:20
And then the sphere at the end is either dunked in water or not dunked in water.
00:33
All right.
00:36
So let's think about what's going on here.
00:38
First of all, i'm going to draw a force diagram.
00:41
So this is unsubmerged.
00:53
I'm going to write t sub u is going to directly overcome mass of the sphere times gravity.
01:10
So t sub u equals mg.
01:17
Submerged.
01:23
We've got the tension, t sub s.
01:30
We have m sub g also, but we have a third force, also acting upward, which is the buoyant force, equals the volume of water that is displaced.
01:54
Well, the volume of water is the volume of the sphere, i mean the weight of water that is displaced.
02:06
So the volume of the water in the sphere is four -thirds pi r cubed.
02:19
So that volume times the density of water, volume times the acceleration due to gravity.
02:34
So that would be mass times acceleration due to gravity.
02:38
Gravity.
02:39
We know that r is d over 2, so i'm just going to substitute that in.
02:47
So r cubed would be d cubed over 2 cubed.
02:53
2 cubed is 2 times 2 is 4 times 2 is 8.
02:56
So 4 8ths is 1ā2.
03:06
G density water.
03:13
All right.
03:14
So our tension plus the buoyant force equals mass times gravity.
03:37
So the tension is going to be mg.
03:43
I was using a capital m.
03:46
I haven't used a lowercase m yet.
03:52
Minus one -sixth pi, d -cued, g, density water.
04:08
And that's t -sub -s, t -subs -s.
04:11
And so, when it is unsubmerged, we are seeing the third harmonic.
04:27
So unsubmerged, i'm going to use, f sub m equals m times v over 2l.
04:45
All right.
04:47
So f sub unsubmerged is the third harmonic.
05:00
Three times v.
05:05
V is the square root of t over mu.
05:12
Is t submerged over to l.
05:20
F fifth harmonic is five square root t sub s.
05:33
The other one should have been unsubmerged, u unsubmerged, right, over mu over to l.
06:08
So the frequency of the third harmonic over three would be the frequency of the first harmonic, but no, that's not, i don't think that's heading in the right direction.
06:28
Because the first harmonic is going to be different depending on the tension.
06:38
All right.
06:38
So what is telling us is that these are the same.
06:41
They exactly match.
06:43
So three square root of t sub u over mu over to l equals five square root of t sub s over mu over to l.
07:12
Well, 2 to l, l, l, mew.
07:23
So we had some things cancel out there.
07:25
3 square root of t sub mu.
07:34
You know what? i'm just going to keep going with this.
07:35
T sub u equals 5 square root of t sub s.
07:43
And so t sub s is the one that has the diameter of the sphere in it so i want to do square root of t sub s over t sub u equals three -fifths and now i can square both sides t -s over t -sub u equals nine -fifteenths just squaring both sides or t sub s equals 9 .15th t sub u.
08:23
T sub s is m g minus one -sixth pi, d cubed, g, density of water...