 Let's solve a question on finding average velocity and instantaneous velocity from a displacement time graph for a simple harmonic motion. So here we have a student who extends, then releases a mass attached to a spring and a graph of the masses displacement over time is shown below. So we can see a displacement time graph and we need to figure out average velocity between zero and three seconds. So zero and three seconds and the instantaneous velocity v at time two seconds, at time t two seconds. As always pause the video first, try this one on your own. Alright, let's first look at the average velocity. Now from what we know, from what we know average velocity, we can write average velocity. Average velocity is this is equal to delta x divided by delta t. This is a change in position divided by the time interval. We know the time interval here is between zero and three. So this is this is three minus zero and delta x, this would be the position of position of the mass at time three seconds. And that we can see is plus one centimeters. So let me write one minus the position of the mass at time zero and that is four. We can see it is with it is at four centimeters. So this is four. So when you work this out, this becomes minus three divided by three and this comes out to be equal to minus one centimeters per second. All right, now let's move on to the second part. We need to figure out the instantaneous velocity at time two seconds. So if you want to figure out the instantaneous velocity at time two seconds, we need to figure out the slope of the tangent at this particular point at t equals to two second at this point and the tangent tangent is a line which just touches the curve at the point of interest. So let's let's try and draw a tangent tangent, which is touching the curve at at at this at this point, which is two comma four, it looks somewhat somewhat like this touching the just touching the curve at t equals to four. And now now we have drawn the tangent line slightly extended because we need to figure out, figure out its slope. We can plug in values, we can try and figure out its slope. So when we do that, when we do that, this would be slope. This would be, we know that slope is y2 minus y1 divided by x2 minus x1. So for this line, for the red line, y2, we can take two points, we can take, we can take this point right here and this point right here. So for y2, it's starting from five, ending at three. So this is three minus five, y2 is three and y1 is, y1 is five starting from five. Now x2, we don't really know what x2 is, but we can still try and estimate this can be somewhat around. This can be, let's say this can be 2.25, could be 2.25. And x1, again, we don't know what that is. So we can still again try and estimate. This can be, this can be 1.75, approximately 1.75. So now we can write 1.75 in place of, in place of x1. And this comes out to be equal to minus two, minus two divided by 0.5. So when you work this out, this is, this is minus four, minus four centimeters per second. We have a minus sign because we can see that the slope, the slope of the tangent, it will be negative. The red line will not have a positive slope. It's a, it's a negative slope. So the instantaneous velocity at two seconds will be, will be negative. And it is negative four centimeters per second.