Astronauts in space cannot weigh themselves by standing on a b...

Question

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in FIGURE P15.46. What is her speed when the spring's length is 1.2 m?

Accepted Answer

Hey, everyone in this problem, an engineer is testing the suspension system of a new car model. He attaches one end of a large spring to the car's bumper and the other to a fixed point. The car is pulled away from the fixed point and then released. OK. We're given a figure that shows the graph of spring length versus time. And we are asked to determine the speed of the car when the length of the spring is 1.5 m. Now, this diagram we're given, we have time t in seconds. We have the length L in meters and this is a periodic function. OK? It looks like a sign or cosine wave. We're given four answer choices. So we have a 0.76 m per second. B 0.28 m per second. C 0.67 m per second and D 0.82 m per second. Now, we want to find the speed at a particular length. So let's recall it the position for this periodic motion X of T can be written as a cosine of omega T plus five plus X E Q. OK. Where A is the amplitude omega is the angular frequency five is the phase angle. And this X E Q is our equilibrium position. Now you'll notice from the graph that this is not centered around the X axis. OK. So the equilibrium position is not zero. And so we have this extra addition at the end of our question. So this is the position, but we wanna find the velocity. Now recall that the velocity is related to the position to the derivative. So V T is equal to D X T divided by D T taking the derivative. OK. A is a constant. So we're gonna have a, we leave it alone multiplied by the derivative of cosine which is gonna be negative sign. OK. With all of the same stuff inside of it omega G plus five. Using the chain rule. We multiply that by the derivative of what's inside of the cosine, we have omega T plus five, the derivative of Omega T is going to be a omega, the derivative of five which is a constant is just going to be zero. So we multiply by Omega and then the derivative of this X equilibrium just the constant that's being added. And so that's gonna be zero. So we get that our velocity is gonna be equal to negative A omega multiplied by sine omega T plus five. So in order to find the velocity V T, we need to know the amplitude A, we need to know the angular frequency omega, we need to know this phase of and we need to know the time point that we're looking at. OK. We were given the length but we weren't given the time point. So we have a lot of things to calculate before we can get back to this velocity equation. And we're just gonna work through them one by one. So our amplitude A this is just gonna be one half of the height of our curve. OK? Or one half of that total displacement. So in this case, one half multiplied by that maximum length is 1.8 m. The minimum length is 0.6 m. So it's gonna be one half of that distance between the maximum and the minimum. Recall that the amplitude is measured from the equilibrium position. OK? It's not measured from maximum to minimum, it's measured from the equilibrium. And so it's gonna be half of that total from the maximum to minimum. And in this case, that is gonna be equal to 0.6 m. So we have our amplitude A let's move to omega our angular frequency recall that Omega is equal to two pi divided by the period T and the period T is something we can figure out from the graph. OK. So recall that the period is the time it takes to complete one cycle. So at T equals zero, we are at the bottom of our graph over our curve and it is increasing, the length is increasing, OK? It increases up to a maximum. It decreases down back to that minimum where it is then going to start increasing again. Ok. So this is a point where we've completed one full cycle and that is at four seconds. So we've gone from zero seconds to four seconds. And so the period is four seconds. That means that Omega is gonna be equal to two pi divided by four seconds, which gives us pi over two. And our unit here is radiance per second. So let me put, put boxes around all of this information we're finding out. So we don't lose track of it. We found our amplitude is 0.6 m. Our angular frequency is pi over two radiant per second. Now we need to do the phase angle fire and this equilibrium position. Now let's start with that equilibrium physician and that's gonna be halfway between that maximum and minimum. So we can bind this just looking at our graph and that's gonna be 1.2 m. OK. So X equilibrium is equal to 1.2 m. Again, that halfway point between the maximum and minimum length, all right on to the phase angle F five. Now we have our equation X of T and at this point, we have the amplitude of 0.6 m multiplied by cosine of our angular frequency pi over two multiplied by the time T pull us our phase angle phi pull us 1.2 m. Now when T is equal to zero, we know that the position is 0.6 m according to our graph and that is a Y intercept. So X at zero, it is gonna be 0.6 m multiplied by cosine. But why plus 1.2 m? And again, we know that this is equal to 0.6 m according to our graph. Now, we can move our 1.2 m to the right hand side by subtracting that's gonna give us negative 0.6 m. Then we divide by 0.6 m to isolate this cosine of five. And we have that cosine of five that is equal to negative one cos api equals negative one which tells us that P P is equal to pi OK. So we're getting there, we found all of these constants. We need, we have the amplitude, the angular frequency, this equilibrium position and our phase angle five. What's left to find in order to calculate our velocity is the time point that we're interested in. And I mentioned this earlier, but we were given the position or the length but not the time point. So we can use our X of T equation that position equation to figure out the time point when we have the length we're interested in, we have of T is equal to 0.6 m multiplied by cosine of pi divided by two T plus. Oops Nothi anymore. We know what is plus pi plus 1.2 m. Now, the position we're interested in is at 1.5 m. So we're gonna set that as our X of T, we have 1.5 m is equal to 0.6 m multiplied by cosine of pi over two T plus pi plus 1.2 m. We subtract 1.2 m from both sides. We get 0.3 m is equal to 0.6 m multiplied by cosine of pi over two T plus pi dividing by 0.6. We have that cosine of pi over two T pi is equal to one half. Now we have cosine is equal to positive one half. OK? This can happen if we're in quadrant four or wearing quadrant one. Hm. So four in quadrant one, we get a positive angle. We have pi over two T class pie is equal to pi over three. This tells us that the time T is gonna be equal to pi over three. We're gonna subtract pie. OK, which is equivalent to three pi over three. And we're gonna divide all of that by pie over two to isolate for tea. And this gives us a T value of negative four thirds seconds. Now we have pi over three minus three pi over three gives us negative two pi over three. The P is divide out negative P two pi over three divided by half gives us negative four thirds. OK. So that's the first possibility we get this negative time, which is not what we want. OK. We don't have those negative times on our graph doesn't really make sense in the context of an experiment. So let's look at the other possible angle that we get. OK. And the other possible angle is if we're in quadrant four and we have the pie over two T plus Pi is equal to negative. What's sorry? Two pie minus pi over three. So it's negative pi over three. But we've gone all the way around that entire two pi minus that pi over three. Hi. What this tells us is that Pi over two multiplied by T is equal to two pi minus 5/ minus pi. OK. We have two pi minus pi which gives us pie minus pi over three. So we have two pi over three. T is equal to two pi over three divided by pi over two. Again, the pies cancel two, divided by three, divided by half, gives us four thirds seconds. Now, this is a better time. OK? This is a positive time. And that makes more sense in the context of this problem. And so we're gonna use this time now, what we want to calculate is that velocity? OK. That's what the question is asking. Let's go back up. We have our velocity equation from differentiating our position equation. So the velocity V T is going to be equal to negative A Omega multiplied by sine of Omega T plus five. So at 4.3 or sorry, four third seconds, four divided by three seconds, forget that this is equal to negative 0. multiplied by Omega which is pi over two for the sake of space. I'm leaving out units here, but we have meters multiplied by radiance per second multiplied by sign of pie over two multiplied by four thirds pie. So we have negative 0.6, multiplied by pi over two, multiplied by sign. We have four pi divided by six, which is two pie over three plus pie is gonna give us five pi divided by three. And what we end up with is 0.8162. The unit on our amplitude was meters, then we had a per second in our angular frequency. And so our final unit is gonna be meters per second. And that is that velocity or the speed that we were looking at when the length of the spring is 1.5 m. Yeah. So we had to do a lot of work there. But what it came down to was finding all of those constants to use in our equation, the amplitude, the angular frequency or equilibrium position in the phase angle. And then finding the time when we had the length of spring that we were interested in. OK. We needed all of those to substitute in to our velocity equation to find that speed. And we found that to two significant digits, that speed is 0.82 m per second, which corresponds with answer choice D thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Simple Harmonic Motion (SHM)

Spring Constant (k)

Velocity in Oscillatory Motion

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