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Question Video: Estimating the Magnitude of the Displacement Associated with Brownian Motion Physics

If an object exhibits Brownian motion that has an average speed of 1.2 cm/s, which of the following is most likely to be the magnitude of its displacement after 60 s? [A] 0 cm [B] 1.2 cm [C] 34 cm [D] 72 cm


Video Transcript

If an object exhibits Brownian motion that has an average speed of 1.2 centimeters per second, which of the following is most likely to be the magnitude of its displacement after 60 seconds? (A) Zero centimeters, (B) 1.2 centimeters, (C) 34 centimeters, (D) 72 centimeters.

All right, so in this exercise, we’re starting with some object. And let’s say this is our object. And we’re told that it exhibits Brownian motion. This has a very specific meaning. It refers to the apparently random motion of particles in fluids. So, for example, if our object were a particle of a gas among many, many gas particles, then through collisions with other particles, this object would have its direction of travel changed over and over and over again. If we were to track the path followed by our particle as it goes through all these motions, that path would seem to be random. This is the Brownian motion that our problem statement is talking about.

We can see that over the course of all its motion, our object will have some average speed. And that average speed is given as 1.2 centimeters per second. This means that if we were to measure the entire distance traveled by our particle as it moves and then divide that distance by the time it takes our object to move that far, the result, the average speed of the object, would be 1.2 centimeters per second.

Our question then goes on to say, if this were to continue for 60 seconds, then which of the options we’re given is most likely to be the magnitude of our object’s displacement? Now, let’s recall that the displacement of an object is the straight-line distance between its starting and ending locations. So if our object started here, and then we’ll say that it ends here, then the displacement of our object is equal to this distance in that direction.

Recall that displacement is a vector having magnitude and direction. But then in our case, we’re just focusing on the absolute value of our displacement. We just want to know its magnitude. So if we let an object move apparently randomly at an average speed of 1.2 centimeters per second for 60 seconds, the question is, what will the magnitude of its displacement most likely be?

Now, often, when we think of object motion taking place, we tend to think of that motion possibly happening in three different dimensions. The object could move up and down. That’s one dimension. Or it could move right and left. That’s another. Or it could move what we could call into or out of the screen. A particle of gas, for example, in a three-dimensional container could move in all these ways as well as in any combination of these directions.

But let’s say the object we’re working with is constrained to move only to the left or to the right. This is only for the sake of example, by the way. And our analysis will work just as well for an object moving in three dimensions. So if we say that our object starts here, then the only way it can move from that point is either to the left or to the right. And let’s say further that the length of each one of these arrows we’ve drawn in represents the distance our object would move in one second. In other words, each of these distances is 1.2 centimeters, since our object moves at 1.2 centimeters every second.

Okay, so here’s where our object starts. And let’s say in the first second it moves to the right, so this way. Now, it was equally likely that the object move to the left because it’s exhibiting Brownian motion. And that means that when our object is here, after one second has elapsed, once again, it can only either move right or left. And the probability of it moving in either of those directions is the same.

Let’s say that, for this next interval of one second, the object again moves to the right. We can see that at this instant in time, its total displacement is 2.4 centimeters to the right. But only two of our 60 seconds has elapsed, so we’ll let time continue to unfold. When our object is at this point, it once again can move either to the left or to the right. And let’s say that now it moves to the left, 1.2 centimeters back towards where it started. So our object is now back here, and its total displacement has decreased.

Now, we could do this for 57 more seconds. But before we get that far along, we can start to see a general principle for the motion of this object. At any given point in time, it’s equally likely to be moving to the left as it is to the right. That has a very important implication. It means that if we were to total up all the distance to the right that our object travels over this time interval of 60 seconds. Say that that was equal to this distance. Then we would expect over the same time interval to have the object move the same distance in the opposite direction to the left.

The reason for this, as we said, is that the particle is equally likely to move in either direction. So over this relatively longer time interval, our particle would move some total distance to the right. And then it would likely move that same total distance back to the left. And therefore, it’s most likely that this object ends up in the same location where it began. We see that that agrees with option (A), which says that the magnitude of this object’s displacement after 60 seconds is most likely to be zero centimeters.

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