Hey, guys. Let's take a look at this example here. We've got a mass on a spring, and it's pulled 1 meter away from its equilibrium position. So we've got 1 meter, and we're just gonna draw a little box here. And then you release it from rest. This is a mass spring system. Right? So it's just gonna go and reach the other side where now the distance here, the displacement is gonna be negative 1 meter. And then it's just gonna go back and forth forever. So here in this first part, we are asked to calculate the amplitude, but we know that the amplitude is just the maximum displacements on either side. So that means that the amplitude of this is just 1 meter.
So the second question now, we're asked to find the period. So that is going to be that letter T. So what does the period look like? Well, we're told that the mass takes 2 seconds to reach the maximum displacement on the other side. So if you release it over here, then this time that it takes for you to go all the way over to the other side was equal to 2 seconds. Now the question is, is that the whole period? No. It's not. Because we said that the period is equal to the time that it takes you to complete one whole entire cycle here. So this 2 seconds really only represents a half period. This thing has to go all the way back to the other side for another 2 seconds, and that's going to be another half period, which means that these points in between here are actually quarter periods. Right? So this is a quarter. This is a quarter. These are all quarter periods. And this is the smallest sort of division that you can make. So you've got half periods and quarter periods. So we're told that it's 2 seconds to get to the other side, which means that the full period of this motion is going to be 4 seconds.
Now we've got, in this third part here, the angular frequency of the motion. So how do we relate angular frequency to the period? Well, we've got an equation up here that can do that. So if we're looking for the angular omega, we can use either 2 pi times the frequency, or we can use 2 pi divided by the period, which we actually know. So I'm just going to go ahead and use that. I've got omega equals 2 pi divided by the full period of the cycle, which is 4. And so what you should get is 1.57, and that's going to be radians per second. So that's it for this one. Let's keep going with some more examples.
ω = 2 π T
