00:02
All right, so let's draw a little picture, first of all.
00:10
We have a steel wire connected to a copper wire.
00:19
And let's say we like maybe just jostle this just a little bit, and we send a pulse right through.
00:33
We want to know how long will it take for this pulse to go from one end, like from here, all the way to the other end right here and and what we know is the length of the steel wire and we also know the length of the copper wire the length of the steel wire is 30 meters and the length of the copper wire is 20 meters um we also have the diameter of both wires the diameter of the steel wire is equal to the diameter of the copper wire so i'll just call d.
01:36
That is 10 to the minus third meters.
01:40
It's, it says, it's indicated that it is a one millimeter.
01:46
We also know the tension.
01:50
The tension is 150 new, newtons.
01:56
It's the same in both wires.
02:01
All right.
02:01
So each, the pulse will have a different speed for each wire, right? so there, there will be a v steel and of v and a v copper.
02:21
They'll equal, you know, square root of t over mu, okay, each one will be like have its own mu.
02:38
Now, mu steel, right, will be, it'll be the mass of the steel wire over the length of the steel wire.
02:49
We don't know the mass of the steel wire, but they gave us the diameter of the wire.
02:57
So why would they give us the diameter? well, if you think about it, they gave it to us because the density of steel is a known quantity.
03:07
It's something we can look up.
03:10
Okay.
03:11
So actually, the density of steel is equal to the mass of steel over the volume of the steel.
03:22
Wire.
03:25
Now, that would make the mass of steel equal to the density of steel times the volume of the steel wire, which is equal to the density of steel.
03:45
Now, the volume of the steel wire is going to be the length of the wire times the area of the circle that is at the the top and bottom of the wire, right? so the area of that circle is pi and then it's pi r squared, right? but instead of r, they've given us the diameter.
04:16
So that is just d over two.
04:19
R is d over two squared.
04:22
And then times the length of the steel wire.
04:27
This is the mass of the copper wire is row copper, pi.
04:40
Let me, okay, you know, let me rewrite it.
04:49
Let me just write it a little different.
04:52
How about we'll do pi over four? rho s, d squared, l, s.
05:04
Okay.
05:08
That will make mu s equal to pi over four, by the way, this is row c, row c, and lc.
05:30
That would make mu s, pi over four, row s, d squared, ls, mass over length.
05:44
That's mass over the length.
05:47
So the length is just gone.
05:52
Likewise, mu copper is pi over four, row copper, is pi over four, row copper d squared.
06:05
Now, we go back to our velocities here.
06:09
The velocity of steel is the square root of tension over mu -s.
06:19
So that would put four here and everything else at the bottom...