00:02
Once again, welcome to new problem and, you know, we're still dealing with wave motion.
00:11
We're dealing with wave motion and this time we have a string, you know, think about a string and obviously it's going to have some thickness if you want to call it.
00:26
But more so, the linear density of the string, which we're calling me, is the same as 2 .0 grams per meter.
00:46
And if we have an axis, this is the y axis and this is the x axis.
00:55
And what's happening is now we're stretching this string in the positive direction of the x -axis.
01:06
We're stretching it that way.
01:11
There's a tension that we're using to stretch the string and that tension is 20 newtons.
01:18
This other side of the string is at the origin and obviously the x at the origin is going to be 0 meters.
01:27
And the string itself is tied up to a hook and this hook goes up and down like that.
01:50
It goes up and down and the frequency by which it goes up and down is 100 hertz.
01:58
You know when the hook goes up it's obviously going to come back down and it displays itself based off of some amplitude and the equivalent amplitude in this case by which it displaces itself you can see it's going to be one meter or one millimeter if you want to call it that one millimeter that's the amplitude when the time is zero seconds the this hook is right here that the lowest point right there that's at the time when it's zero seconds.
02:43
Our goal is this problem is to find the velocity and lambda of the string.
02:55
Obviously velocity is v and lambda is going to be the wavelength.
03:02
That's the first thing we want to find out.
03:05
The second thing we're going to be finding out is the amplitude, which we already call that is a.
03:13
And we also want to get phi which is the face constant.
03:22
We want to get the face constant, not just the amplitude but the face constant.
03:31
The other thing that we're looking for in this problem, other than the amplitude, the velocity, the face constant and all of that is going to be the equation of the wave.
03:47
Remember these things, a wave traveling.
03:49
There's a wave traveling.
03:52
So we want to find the equation of that way.
03:56
And finally, besides the equation of the wave, i want to find the displacement or the distance in a specified direction by which the wave bounces off of the this is the x.
04:20
So we want to find the distance up and down that x equals to meters and at 0 .5 meters and that time equals to 15 milliseconds.
04:35
So, you know, what's this height right here or that height right there? this is a wave.
04:44
You want to find that displacement if you want to call it that.
04:52
So now that we have all those pieces of information, the first thing we're doing is thinking about the wave number as 2 pi of a lambda, that's the wave number.
05:07
Lambda is a wavelength and the velocity v is radical the tension over the linear density.
05:22
The linear density, as you can recall, is 2 .0 grams per meter.
05:29
We want to change that to kilograms.
05:31
Remember one kilogram is 1 ,000 grams.
05:36
So we transform the linear density from 2 grams per meter and let it become 2 times 10 to the negative 3 kilograms for meter that's our new linear density and then the velocity is radical the tension of linear density so our velocity is going to be radical 20 newtons divided by the linear density in as i units is 2 times 10 to the negative 3 kilograms per meter and so the velocity are simplified to or the speed 100 meters per second as you can recall the second part of the problem were required to find the lambda the second part but the second aspect of of part a was supposed to find the lambda and the lambda is a wavelength we know the velocity is the frequency turns to lambda we divide both sides by the frequency so lambda is the ratio between the velocity and the frequency this is the speed we're using that interchangeably as velocity and this is the frequency of oscillation and so lambda is the same as 100 meters per second we had to find the velocity in the first step before we can resolve the lambda and this is 100 hours and so lambda ends up becoming one meter that's the wavelength that's our wavelength or lambda and part b of the problem is a little bit easier because you already have a maximum displacement defined and the problem or the wave and that's the amplitude the maximum displacement in this sense is going to be a as the same as one millimeter the next part of the problem is a requirement to determine the face difference, the face difference is still part b.
08:14
We have two issues to deal with in the second part.
08:17
Let's define the amplitude and the face constant.
08:20
Where he found the amplitude is 1 millimeter.
08:24
Now we're looking at the face constant.
08:26
The general equation of the wave are in terms of x position and time in the second.
08:35
Second is a sine sine minus omega t.
08:41
Your finite, k is the wave number.
08:50
Omega is the angular frequency and obviously finite is the initial face constant.
09:07
So our goal is to get the face constant.
09:13
When x is 0, t is going to be also 0.
09:17
So we know that the amplitude we're dealing with is 1 millimeter.
09:32
And so we want to do some substitution in the equation.
09:35
So we have negative 1 millimeter is the same as so d or d is negative 1 millimeter.
09:47
That's our displacement.
09:50
Our amplitude is 1 millimeter.
09:54
This is the minimum displacement.
09:57
And the minimum displacement happens at negative 1.
10:01
Remember this is the x -axis.
10:03
This is the maximum displacement at positive and then minimum displacement at negative.
10:10
This is negative 1.
10:12
So that's what we're going with.
10:14
That's where here we have negative 1.
10:15
The amplitude is 1 millimeter and then sign off, finite.
10:22
And the reason why it's just sign of finite because the call the equation looks like that.
10:28
A sine kx omega t is finite.
10:37
The x is 0, so that part is going to cancel.
10:42
T is 0, that part is going to cancel.
10:44
So we have this as 1 millimeter and this as negative 1 millimeter.
10:51
Then we're going to divide both sides by 1 millimeter.
10:57
And now we have final, we have to make sure this is comfortable.
11:03
We have sine of finite being the same as negative 1.
11:12
So finite which is the face constant has to be sine inverse of negative 1.
11:19
So we're looking for an angle...