00:01
So in this problem we have a diagram, and i'm going to reproduce the diagram in pretty good detail, because i think this problem is going to be kind of confusing.
00:21
This is the x -axis.
00:28
Now, on the y -axis, this is where it's attached up at the top and the bottom, but it starts here, the particle, and it's unstretched, and then it comes out to here.
00:43
And we're saying that this distance that it is stretched is x.
00:54
Now, it's unstretched distance for the spring was l.
01:00
L.
01:02
So when i draw the spring this way and the second spring this way, then the first thing that i want to do is i want to find this distance, i'm going to call this l2.
01:34
So using the pythagorean theorem, l2 is the square root of l squared plus x squared.
01:51
We also know that force equals kx, but this force is going to be in this direction, or this direction.
02:13
And i actually want to write delta x, so we don't get confused between x's and whatever in here.
02:21
This is just the generic formula.
02:24
So the change in the length.
02:27
So in this problem, the force is k times the change in the length, which would be l2 minus l.
02:42
But l2 is the square root of l squared plus x squared minus l.
02:57
Now, there is a component of each of these forces that is in the negative x direction.
03:11
The force in the x direction equals that that force times the cosine of theta.
03:32
But we know that the cosine of theta is just the adjacent which is x over the hypotenuse which is l2 adjacent over hypotenus.
03:54
And we also know that l2 is the square root of l squared plus x squared.
04:08
Now in the x direction this force is negative.
04:27
So the force is in the negative direction and there are two of them equal to k.
04:47
So i'm just putting this into here and multiplying by negative two.
04:56
So that's going to be i still have the x.
05:01
Then i have the square root of l squared plus x squared minus l over square root of l squared minus x squared.
05:25
All right, negative 2kx.
05:37
I'm just going to distribute this to the first term and the second term, one minus l over the square root of l squared minus x squared.
05:57
And that is in the x direction.
06:05
So that is our answer.
06:16
And now there's a part b.
06:17
You got to go to the next page to see part b.
06:20
We need to find the potential energy.
06:41
All right.
06:45
Well, the potential energy is the, opposite of the integral of the force d x.
07:02
So the potential energy is.
07:07
I'm going to distribute the 2kx again.
07:15
The integral of 2kx minus 2klx over the square root of l squared minus x squared d x all right so um let's see if we can do this u of x is well the integral of x dx is is x squared over two so that first term becomes k x squared squared squared but the second the second term is minus 2kl times the integral of x over the square root of l squared minus x squared dx.
08:49
I just want to make sure that i didn't make any mistakes already.
08:57
Negative 2kx and it looks like i did no i didn't k x 2k x and minus no i didn't make any mistakes yet minus 2kl the integral of x over l squared minus x squared dx all right so now the difficulty is what is the integral of x over, whoops, the square root of l squared minus x squared dx? well, i'm going to start by letting x equal l tangent of theta.
10:05
Therefore, dx would be l, driven the tangent is the secant squared of theta d theta.
10:23
So if i substitute that in, i get l tangent of theta over the square root of l squared minus l squared, squared, tangent squared of theta.
10:55
And i replace dx with l secant squared of theta d theta.
11:07
All right, so that's gonna be the integral.
11:15
Notice that the numerator, well, numerator's still just l tangent of theta, but i have two l's there, so i'm gonna write l squared.
11:28
And then i've got secant squared of theta, d theta.
11:37
But the denominator is now, i'm going to factor out l squared.
11:43
So it's going to be the square root of l squared times one minus tangent squared of theta.
11:58
All right.
12:00
The square root of l squared is just l.
12:02
So that's going to leave me with an l, only one l, tangent of theta, secant squared of theta, over the square root of 1 minus the tangent squared of theta.
12:23
Well, that would be the square root, 1 minus the tangent squared is the sequence squared, de theta, which is going to give me l times the integral.
12:39
Secant squared, no, the square root of the secant squared is just the secant.
12:44
So this is just going to be the tangent of theta times the secant of theta, d theta.
12:58
But now, l tangent is sine over cosine.
13:10
Seacant is 1 over cosine.
13:16
So this is the same as l integral of the sine.
13:23
Of theta over the cosine squared of theta d theta.
13:34
So now i'm going to let u equal the cosine of theta.
13:44
So du, derivative of the cosine, is the opposite of the sign, d theta.
13:55
So now it becomes l integral.
13:59
In the numerator, i have sine of theta d theta.
14:06
You can see it right here, sine of theta d theta.
14:09
That's the opposite of du.
14:13
So i can write, oops, opposite of du.
14:24
And then in the denominator, cosine squared of theta is u squared.
14:30
So that's going to be negative l times the integral of u.
14:37
U to the negative 2 power du, which is negative l times u to the negative 1 power over negative 1.
14:58
Let me think about this again.
15:07
This is negative l to you negative 1 power over negative 1.
15:14
Okay, good.
15:16
The negatives cancel out, and it's just l over u, which is the same as l over u was the cosine of theta.
15:48
Way up above we said x equals l tangent theta.
16:03
If i draw a right triangle, with this being tangent, i mean with this being theta, tangent is opposite over adjacent.
16:27
Using the pythagorean theorem, this is the square root of l squared plus.
16:34
X squared.
16:39
So the cosine of theta would be the adjacent over the hypotenous.
16:56
Cosine is adjacent over hypotenous.
17:06
So this now simplifies to l over l over square root of l squared plus x squared.
17:22
Because the cosine of theta is the adjacent over hypotenuse l over the square root of l squared plus x squared, which just leaves me with square root of l squared plus x squared.
17:59
So i went through all of that, and i got the square root of l squared plus x squared.
18:16
So let's keep going here.
18:22
Equals k x squared minus 2k l, and then the integral of x over the square root of l squared minus x squared d x turned out to be the square root of l squared plus x squared.
19:02
Okay, i am somehow missing a term.
19:14
All right, i think that what i'm missing is that when i take this integral of way down here, that i have to add a constant, because this is not a definite integral.
19:45
So that's going to be l over u plus a constant, which gives me the square root of l squared plus x squared plus a constant, constant.
20:12
And so up here, when i took this integral, regardless, there's going to be a constant...
