In FIGURE EX10.27, what is the maximum speed of a 2.0 g particle that oscillates between x = 2.0mm and x = 8.0 mm

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Hello everyone. Let's go through this practice problem. The figure below shows the potential energy with respect to the position of an object undergoing oscillatory motion between X equals one millimeter and X equals five millimeters with a mass of g. What is the maximum speed that this object can achieve? We're showing a little graph of the potential energy with respect to the position of the mass. And we're given four multiple choice options. Option, a 54.8 m per second. Option B 33.4 m per second. Option C 64.7 m per second and option D 44.3 m per second. So in this problem, we're looking for the maximum speed that the object can have. And we're supposed to find that using a graph that tells us about energy. So we'll want to have some relation between the speed of the object and its energy. And recall that speed is related to kinetic energy specifically. So let's pull up our trusty old formula for that recall that kinetic energy of a particle is equal to one half multiplied by the mass of the particle multiplied by the square of its speed. Since the ultimate goal of this problem is to find the speed, let's algebraically rewrite this equation to solve for V. So dividing both sides of the equation by one half M, we find that V squared is equal to two multiplied by the kinetic energy divided by the mass. And then taking the square root of both sides, we find that the speed is equal to the square root of two K divided by M all underneath a square root. And since we're looking for the maximum speed, specifically, this equation will only work if we have the maximum kinetic energy. So we want to first use this graph to determine the maximum amount of kinetic energy of the object has this graph only tells us about potential energy. So recall the law of conservation of energy, which states that the energy should be conserved as it can't be created or destroyed. And if we're dealing with mechanical energy, then the energy of the system is pretty much just going to be a sum of the kinetic energy and potential energy at any given point in time. This is where it's useful to consider what's happening in the problem. We're told that the particle is undergoing an oscillation. And we're even specifically told the two points that the object is oscillating between one millimeter and five millimeters. And if you consider what happens when a particle is oscillating back and forth between two positions, then this becomes a little bit more clear because when something is moving in one direction and then changes direction, it first has to slow down to a stop and before it is able to start traveling in the opposite direction. So that means that when this particle is at those two positions, one millimeter and five millimeters, that is at a point where it has no kinetic energy. Because those are the points where its speed has slowed down to a stop such that all the energy in the system at that point is just stored in the potential energy. So if we look at the graphs, we can see that at those two points, one millimeter and five millimeters, the potential energy is the same, it is at six jewels. And because the object must be at rest at those two points and therefore have no kinetic energy, that means that the total energy of the system is gonna be equal to zero jewels of kinetic energy plus six jewels of potential energy, which means the total energy of the system is just six jewels. Another thing we can learn from looking at the graph is that there are two points where the potential energy reaches zero. That means that those points, all of that potential energy, all six jewels must have been converted into kinetic energy. And due to the law of conservation of energy, that also must be the maximum kinetic energy because it it can't there can't be more than six Jews of kinetic energy unless there's some additional energy being added to the system. And we're not told anything about that. We're assuming this is an isolated system. So what that means is that the maximum kinetic energy of the system must be equal to six jewels, which is something that we can put into the formula for the maximum speed that we discussed earlier. So to recap the maximum speed of the object is equal to the square root of two multiplied the maximum kinetic energy which we just discussed is going to be six jewels. OK? And then this is being divided by the mass of the particle, which is given to us in the problem as 4 g. Now, in order for our units to be consistent, we're going to want to convert this into kilograms. So we divide this by 1000. So instead of 4 g, we can write this as 0. kg. And if we put this into a calculator, then we find a maximum speed of approximately 54.8 m per second. So that is our answer to the problem. And if you look at our multiple choice options, we can see that this agrees with option A which means that option A is the correct answer to the problem. And really that is it for this problem. So that is all for this video. I hope this video helps you out. Um if you want more experience, please consider checking out some of our other videos which will give you more practice with these types of problems. That's all for now. I hope you all have a lovely day.

In FIGURE EX10.27, what is the maximum speed of a 2.0 g particle that oscillates between x = 2.0mm and x = 8.0 mm

Verified Solution

Key Concepts

Simple Harmonic Motion (SHM)

Maximum Speed in SHM

Amplitude