00:02
Okay, so for this problem, we have an object with a mass m and a charge q suspended vertically on a spring with a spring constant k.
00:12
In our before situation, the string stretches a distance x not, and in our after situation, our spring stretches double that length, so 2x0.
00:26
So what happens after that causes it to stretch this double is that we have an electric field here pointing downwards that's uniform and exerting a force on the charge.
00:41
So the first question is based on this picture, what sign is the charge of this object? and to do, to answer this question, we have to remember one crucial fact.
00:56
And that is that electric fields point in direction that a positive charge goes when the charge, when the field is applied to that charge.
01:29
So here we see there's an electric field pointing down, and then what direction does this charge go? it goes down as well.
01:37
So we know that positive charges move in the same direction as the electric field points, and negative charges would move in the opposite direction.
01:48
So here we have a downward electric field and downward motion of this charge, so we can conclude that this object has a positive charge.
02:04
All right.
02:06
The next question is to find the strength of the electric field, that e, in terms of the mass of the object, the acceleration due to gravity, or little g, and the charge of the object.
02:24
So what we're going to do here is draw free body force diagrams for each situation here and see if we can find something out.
02:36
We know that in the before situation and the after situation these objects are stationary or there's no forces acting on it they're just sitting there or no external or unbalanced forces so all the forces acting on the object before and after are all going to cancel out so let's write out these forces we know that gravity is always going to act on the object and in the before situation that's getting balanced out by this upward force caused by the spring.
03:16
And we know the force due to a spring is always the spring constant times the amount it's stretched out.
03:26
So in this situation, the spring is going to exert an upward force again of double what it was before, because now it's stretched out, double the amount.
03:38
And then for this after situation, there's also an electric force acting downward on this object, and we know the force due to an electric force, it's going to be the charge of the object, times the strength of the electric field.
03:57
So in this before situation, we know that, so all of these forces need to cancel out and balance each other, right? so we know that k, x ,0 is going to be equal to mg here, because the force is going up equal to the force is going down.
04:17
And then a similar situation here, 2k, x -not, is going to equal m -g, plus this additional downward force, qe, to balance it out.
04:31
So now they're asking the strength of the electric field, so let's try to solve for that.
04:36
If we manipulate this equation a little bit, we see that qe is equal to sorry, 2kx0 minus m .g.
05:00
And then we can make a substitution here now because of this equation.
05:06
We know that kx not is equal to mg.
05:13
So this becomes qe equals 2mg minus m.
05:23
Which is just equal to 1mg.
05:29
So then our electric field, once we solve it by itself, is going to be mg over q.
05:40
Awesome.
05:42
So now the next question is, if we have this after situation and we displace in a certain amount, like if we stretch the spring a little bit and then release it, why is the motion going to be what's considered simple harmonic motion? and to answer this, we have to know what simple harmonic motion is, right? well, this is what it looks like.
06:05
You get this oscillatory motion, kind of like a sine wave, where the objects can go up, then down, then up, then down again.
06:15
So that's simple harmonic motion, and what we need to ask ourselves is what forces what forces are necessary to produce this motion? like what force is acting on the object to make it go up and down like that? so when we have a distance or a position, a y position here, versus time graph, to find the force, to get the force versus time graph or the force acting on this object, we know we have to look at the curvature of this graph.
06:50
Like how much is this graph curving.
06:53
Up here, it's curving quite a lot.
06:57
It's going downward.
06:58
But then here, we see that this line is kind of straight.
07:02
So there's going to be very little curvature here.
07:05
So if we graph that curvature, in this region, the, in this region, the curvature is concave down, so that's negative curvature...