An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period of 2.0 s and a maximum speed of 40 cm/s. b. What is the glider's position at t = 0.25 s?

Video transcript

Hey, everyone. So this problem is dealing with simple harmonic motion. Let's see what it's asking us. An object of mass 0.5 kg is attached to a spring with a spring constant of 20 newtons per meter. The object is displaced from its equilibrium position and then released from rest at time equals zero seconds. Determine the position of the object at T equals 00.45 seconds. If the object oscillates with a maximum speed of 30 centimeters per second, consider the period of oscillation is four seconds. Our multiple choice answers here are a 19. centimeters. B 14.5 centimeters c 11. centimeters or D 15. centimeters. So there are a few key equations dealing with simple harmonic motion that we need to recall to in order to solve this problem. So the first is that our maximum velocity is equal to our amplitude multiplied by our angular frequency, our position or X of T. So position as a function of time is given by that amplitude multiplied by multiplied by the cosine of omega T or omega times multiplied by time. And then in turn omega angular frequency is equal to two pi divided by T or our period. That's a capital T. And so the first thing we can do is take this first equation to solve for amplitude. And then we will use that amplitude in this second equation to solve for position because that's ultimately what the problem is asking us. So our amplitude is equal to our B max divided by omega and plugging in what we know here. This third equation we actually get V max multiplied by uppercase T divided by two pi. And so we have everything we need there to solve for our amplitude. So amplitude, our maximum speed 30 centimeters per second. I'm going to rewrite that as 300.3 m per second. So we can keep that, keep ourselves in standard units multiplied by our period, which was given as four seconds all divided by two pi. And that is 0. m. And now all we need to do is use that plug in our um time of 0.45 seconds and we have everything we need to solve for our position. So X of 0. seconds is equal to 0.191 m which we just saw for multiplied by the pros sign of hour angular frequency or two pi divided by four seconds multiplied by T 0.45 seconds. We plug that into our calculators and we get 0.145 m per second. And then we can put that back, oh Sorry, not meters for a second, just meters. And then we can put that back into centimeters. So we move that decimal place over by two and we get 14.5 centimeters. And that is the correct answer for this problem. And that aligns with answer choice B. So that's all we have for this one. We'll see you in the next video.