Problem

Engineering Application Near the top of the Citig…

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Problem 35 Hard Difficulty

16.5 Energy and the Simple Harmonic Oscillator
The length of nylon rope from which a mountain climber is suspended has a force constant of $1.40 \times 10^{4} \mathrm{N} / \mathrm{m}$ . (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 $\mathrm{kg} ?$ (b) How much would this rope stretch to break the climber's fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy. (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

Answer

a) $f=1.99 \mathrm{Hz}$
b) $x=0.16 \mathrm{m}$
c) $f_{2}=1.4 \mathrm{Hz}, x_{2}=0.23 \mathrm{m}$

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Chapter 16

Oscillatory Motion and Waves

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Video Transcript

All right. So we have a climber who is dangling on a rope, Um, in the rope has a force constant associated with it. That is 14,000 Newtons per meter. Um and then they also tell us that our climber and his gear together, um weighed 90 kilograms home, and then it's asking us to find the oscillation of climber bobbing up and down. So the fact we're looking for frequency of oscillation and that we have a force constant like that cruises into the fact we're gonna be treating the rope like a spring. So the equation we're going to use that ties in spring constants and mass dangling to the frequency is the square root of K just the spring constant. We're for our cases, the force constant of the rope que over em equals two pi times the frequency, and we're gonna divide both sides by two pi, and then we're gonna plug in 14,000 for K. We're gonna plug in 90 for the mass. So we get Warren over two pi times the square root of 14,000 over 90. That's gonna equal our frequency. And that is gonna equal 1.985 hertz. So the answer is the first question is the frequency of the bobbing or the isolation 1.985 hertz? The second question, it's asking us, um, so the climber falls off the wall and it falls. He falls two meters, and then the rope starts catching him. And then it's asking us, how much does the rope stretch? Um, and then it tells us to use conservation of energy as a So we're going to say that the potential elastic energy is going to equal the potential gravitational energy. Eso the potential elastic is 1/2 que x squared, and then our gravitational is gonna be in G h. Um, but what's tricky here is, if our climbers falling two meters and then falls how much ever the rope stretches on top of that, that means our HR height isn't gonna be two meters. It's gonna be two meters, plus the amount that the ropes stretched. Um, so we can say that the amount that are climber falls for the height. It's gonna be two plus x. So we get 1/2 kick square equals MGI times two plus X and then if we multiply that out or distributed, and then we move it over to the left side. We get 1/2 CAC squared minus M g X minus two and she and that's gonna equal zero. And we have a quadratic, and we also need to plug in what we know. So you know that K is 14,000 times 1/2 It's give me 7000 x squared. Um, Then we're gonna do end times G g is the gravitational constant, 9.8 meters per second squared times are mass, which is 90. I'm so if we multiply those together we get 808 to x minus two mg. So it's just that times two just gonna be 17 40 or 64. Sorry equals zero. At this point, we can use the quadratic formula to figure out what X is. If we do that, we get to X values. Our 1st 1 is zero point 569 Our 2nd 1 is a negative x value. I'm so we can just ignore that. We're only looking for the positive one. Um, so the amount of the rope stretches is going to be 0.569 meters. Um, parte si asked us to go back and do part A and Part B again, but now we have twice as much rope out. Um, this is a little tricky, but what happens is because we have twice as much rope out. It takes half assed much force to stretch the rope one meter. And so what happens is our que value, which was 14,000. We now have half of that for a new K value. I'm sorry, Kate Prime We could call it is gonna be 7000 now, 7000 Newtons per meter. So we're going to go back and do the 1st 2 problems with 7000 is or cable. Um, so for the first part, we're doing the exact same thing. 1/2 pi square root of Now we're plugging in 7000 instead. So 7000 over 90. And that's gonna equal frequency. In this case, that equals 1.4 04 hurts. Um, And then for the second part, uh, we're gonna go back in and change our K value to 7000. So we get 1/2. Okay, So 7001 half I'm 7000 times x squared. Um, and then everything else stays the same. So we get minus 882 minus 17. 64 and that's all gonna equal zero. Um, and then this is the same thing again where we can just plug this in the quadratic Formula one have 10,500. And if we plug this into the quadratic formula, we get two X values again, one of which is negative. We don't care about that one. And then the one we do care about 0.847 meters. And so that is the amount that the rope would stretch if we

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