Weighing Astronauts. This procedure has been used to 'weigh' astr...

Question

Weighing Astronauts. This procedure has been used to 'weigh' astronauts in space: A 42.5-kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

Accepted Answer

Hey everyone in this problem. Simple Harmonic motion is used to determine the massive objects. One procedure involves attaching a 6.8 kg container at the end of a spring. The empty container completes an oscillation in 0.8 seconds. When a metal block is placed in the container, the period of oscillation becomes 1.9 seconds and were asked to determine the mass of that block. All right, so let's think about this. Let's just write out what we're given. Okay? So we're given the mass of just the container. Okay? We're going to call this M. C. Which is 6.8 kilograms. And the period of just the container Is equal to 0.8 seconds and were given the time to complete an oscillation. Okay? And recall that the time to complete an oscillation is the period. Okay, so the period is 0.8 seconds. Then we have the mass. Okay. Afterwards, we're just gonna leave it as m for an hour to come back to that. We're gonna talk about the period. Hey, we're gonna call this T. Two. So the period, the second period, Which is 1.9 seconds. This is gonna be the period of the block in the container together. Okay, so we're talking about the mass related to this period. This is going to be the mass of the container plus the mass of the block. So 6.8 kg plus the mass of the block that we're looking for? MB. Alright, so how can we relate the mass to the period T Well, let's recall that the period t be given by two pi over the angular frequency omega. And also recalled that the angular frequency omega can be written as a square root of K over M. So we can write the period T. As two pi over over the square root of K. Over M. K. Where K is the spring force constant and M. Is the mass. All right. So, we know the period T. We know the mass M. Or the mass M. Is what we're trying to find if we're thinking about the second case. Okay, so K is the only thing in here that we don't necessarily know. So if we isolate for K, we square both sides. We get T squared. K over M. Is equal to four high squared. And we can write this in terms of K. As K is equal to four pi squared um Over T squared. Alright, now, we know that K. The spring force content is the same in both of these scenarios. Okay, we have the same spring. So when we have just the container on the spring, the spring force constant. K. Is going to be equal to the spring forest constant. When we add the block inside of the container. Okay, it's the same spring. It's the same force constant. So what that tells us is that four pi squared M. C. The mass of the container over the period of the container squared is going to be equal to that same quantity for that second mass and period four pi squared um Two divided by T two squared. So the spring constant K. Is equal for both of these situations, which tells us that the right hand side of this equation has to be equal for the two situations as well. Now, what we could have done is we could have solved for K. K. We know M and T squared for the first situation, we could have solved for K and then substituted that into the other equation to find em to. Um but this works as well and this one um I find it's a little bit simpler because you can Divide out this four pi squared now. Okay. And you don't have to worry about doing that on your calculator. It's just a little bit simpler of a calculation. So M C over T C squared is equal to M two over T two squared. And now we just need to fill in what we know again, M2 is what we're looking for right now because MB that mass of the block is part of that second mass. Okay. And we know all of the other values. So we get 6.8 kg divided by 0.8 seconds squared Is equal to M to over 1. seconds. All squared. So we get that M2 is equal to the unit of second squared will cancel. We're gonna be left with 38. 2 5 kg. Okay. Alright, So you might be tempted to go ahead and answer right away. Look at the answer choices and you say, okay, answer D Because we have 38.4 approximately kilograms. But remember that what we just found is the mass M two. That's the mass of the container, plus the mass of the block that we added. Okay, it's not the mass of the block that we're looking for. Okay, remember that M to 6.8 kg plus the mass of the block. So we have to do a little bit more work here. Okay, so M two is equal to 6.8 kg plus the massive a block. And be Okay. So, the mass of the block M. B is going to be M238. kg -6.8 kg. The mass of that original container, which gives us a mass for the block of 31. kilograms. The mass of the block. So, if we go back up to our answer choices, We see that we are going to have answer choice. E we have approximately 31.6 kg for the massive a block. Thanks everyone for watching. I hope this video helped see you in the next one

Key Concepts

Simple Harmonic Motion

Period of Oscillation

Mass-Spring System

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