For the oscillating object in Fig. E14.4

, what is (a) its maximum speed?

Video transcript

Hey everyone in this problem we have a position time graph of a particle attached to a horizontal spring shown in the image. Okay but we're asked to find the objects. Maximum speed. Alright so let's recall the maximum speed V max is given by plus or minus the amplitude times. Oh my God. Alright so what we need to do is we need to find the amplitude A and we need to find omega in order to calculate our maximum speed. Okay let's start with the amplitude. Okay now the amplitude A is going to be the maximum displacement From x equals zero. So if we look at our graph, okay the maximum value on our graph is at 4cm. Okay the minimum is at negative four cm. Okay and so the maximum displacement from x equals zero. It's going to be this distance of four cm. Equivalently this distance of four centimeters or amplitude A is going to be equal to four centimeters and just be careful. It's not that entire distance from the maximum to the minimum. It's the distance, maximum displacement from X equals zero to that. Either maximum or minimum. Okay so it's four centimeters not eight. Alright so we found our max displacement. Now we need to find omega. Okay now one thing when we get graphs of position time like this, we want to think of things that are simple to find off our graph and one thing that is relatively simple to find a bar graph is the period T. Okay so we want to find omega we know that we'll be able to find our period T. Okay so we're called that we can relate omega to the period T. Through omega is equal to two pi over tea. Alright so let's go to our diagram and try to figure out this period. Okay so we want to go through and pick points that are in the same spot of the cycle. Okay? We want to go through an entire cycle and choose the next point. That's at the same position. Alright so let's pick the point here on the y axis. So at T equals zero. We have positioned four centimeters. Now we want to go through one complete oscillation, one complete cycle. So we're gonna go all the way down to the minimum, back up to the maximum and choose this corresponding point here. Okay, that's one complete cycle. And our period is going to be the time between those two points. Hm. Alright so we'll see that our period t. Okay we're starting at zero seconds and we're going all the way up to approximately 12.5 seconds. So our period T. Is going to be 12.5 seconds. Just that time it takes to complete one oscillation or one cycle. Alright so now we can find our angular frequency omega. It's going to be two pi over the period 12. seconds. Which gives an omega value of 0.5 0- x five radiance per second. So we have a we have omega. Now we can calculate that maximum speed we're looking for using a omega. So R V max Again is a omega which is going to be equal to four cm times the angular frequency. 0.50- radiance per second. Which gives a maximum speed of 2.1 centimeters per second. Okay. Now if we look at our answer traces we see that they're all in meters per second. We have centimeters per second. And this is why it's important to carry your units throughout the problem. Okay. The graph we were given gave us the position in centimeters. Okay? Which is what we wrote out when we had our amplitude a Okay, Which is how we get to the units of centimeters per second for speed. V. Okay. So it's important to write out those units so that when you work out the problem, you know what unit you have for your final answer? Okay. Now we want to convert this to meters per second to compare to these answer choices. Okay this is gonna be 2.1 centimeters per second. Okay. And we can multiply by one m per 100 centimeters. The centimeters will cancel. Okay? And we're left with 0. approximately meters per second. Okay? So if we look at our answer choices we see that we have be the objects. Maximum speed is going to be 0.02 m/s. Thanks everyone for watching. I hope this video helped see you in the next one.