From this question, we're gonna need to break it into a couple of parts, and it's gonna include a few different concepts. So we have a ball that's attached to a spring, and the unstructured spring length, which I call l, is 0.2 meters. We didn't start to whirl it in a circle where the circle is parallel to the ground. So just think of a thinking about holding it and then whirling it above your head in a circle. Um, and the velocity of it moving around 10 gentle in the circle V is three meters per second. When we're doing this, the supplies a centripetal force to the ball and that causes it to stretch the spring and the string stretches and amount x one of 0.1 meters. So the radius of orbit is going to be the length. The strength are the spring unstrap etched plus the length of the stretching of the springs. That's gonna be the total radius of orbit of this ball moving in a circle. That's gonna be X one plus out. So as you're doing this, if the spring isn't stretching anymore than that x one amount, the force from the centripetal acceleration f sub c. It's going to be an equilibrium with the force from the spring. And since it's a situation one, which is why I called the X one for displacement will call this s as someone for the force of the spring and situation. One quill centripetal force is equal to the mass times the velocity squared, divided by our and the force of the spring and situation one is the spring constant K times displacement X one from hooks law So we can rearrange this to solve for Kay. We find that kay here is equal to ah in the squared divided by X one times are where are is equal to X one plus l Now we don't know him, but that's okay because we can use this in the second part of the situation. Since it's the same spring, the spring constant K is gonna be the same, and our mass values are gonna end up canceling because really, what we want to find it. This is the displacement in the next situation. So now we're gonna take the same ball and the same spring, and we're going to hang it in gravitational equilibrium. So we're just gonna hang it horizontally. And so now the force of gravity s a G is an equilibrium with the force from the spring of suggest. But this this situation to something called f sub s too. So what we have here force of gravity is mass times acceleration due to gravity G 9.8 meters per second squared Mrs Equal to K. It's the same K as previously because it's the same spring times the displacement next to so If we saw for the displacement x two, we find that it's equal to M g over K. But if we go back, we just found an expression for Kay. We thought it was n v squared over x one, plus her ex one times x one plus l Sophie plugged that in. We find that this is equal to m times g times X one times x one close l divided by n v squared. So now conveniently the masses cancel. So it's okay that we didn't know that mass term because it's the same ball. So that same mass So this cancel their And so we're left with, uh, all values that we know since G is just the acceleration due to gravity and that's a constant. So we plugged the values into this expression and we find the displacement here is equal to 2.29 times 10 to the minus three meters. We can go ahead and box this and as our solution to the question.