A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found by using the energy approach used in Section 14.3, and which cannot? Explain.

Video transcript

Hey everyone in this problem. A spring mass system is used to simulate the motion of the human hamstring. The hamstring executes a simple harmonic with an amplitude of 15 cm and an angular frequency of 6.28 radiance per second. Now we're asked to calculate a bunch of things so let's go through them one at a time. So we're asked to calculate the maximum magnitude of the acceleration in the speed at the equilibrium position. Were asked to calculate the acceleration and speed at the position x equals negative 7.5 cm. Has to calculate the time needed to go directly from X equals zero centimeters two X equals positive 7.5 centimeters. And finally were asked which of the quantities requested previously can be determined using the conservation of mechanical energy. Alright so let's start with this first one here and the first part of that question ask us, ask us sorry the maximum magnitude of the acceleration. Now recall that we have a max. The maximum acceleration can be written as plus or minus a the amplitude times. Oh my guesswork. Alright well we're given an amplitude of 15 centimeters. Were given an angular frequency omega of 6.28 radiance per second. So we can go ahead plug these values into our equation for our maximum acceleration and solve So our amplitude is 15 cm. We're gonna multiply this by one m per centimeters. Okay so this is like dividing by 100 to convert from centimeters to meters we want in our standard unit. Okay and then we're gonna multiply by the angular frequency of 6. radiance for a second. And this is quantity that Omega is squared. So 6.28 radiance per second squared. And when we work this out We're gonna get a value of 5. 157, m/s squared. Alright, so we found the first thing we were looking for and we found that the maximum acceleration is approximately 5.9 m per second squared. Okay, so in our answer traces we can already eliminate C because that one found 3.9 m per second squared, which is different. Alright, so we've done part one Of part one. Now let's do part two. It's asking for the speed at the equilibrium position. All right. So we want to find the speed at the equilibrium position. Well, let's go back to our conservation of mechanical energy. When we have conservation of mechanical energy we have 1/ mv squared. That's one half. Okay. X squared. Okay, kinetic energy potential energy and this is equal to one half. Okay, A squared. Now when we're talking about being at the equilibrium position, Okay, That means that X is equal to zero. So the term in the middle goes away, we're left with one half mv squared. And what about the speed? What do we know about the speed at equilibrium? Well, when we're at equilibrium this is actually our maximum speed. So the speed we're finding is going to be V max And that's equal to 1/2 K. A. Sward. And you may have recalled already the formula or the equation that you can use for B max. This is just showing how we get there. Okay? Um And I would just check with your professor on whether they want you to derive this or whether they want you to go ahead and use that equation if you remember. Okay. Alright. So, we can divide out, the one has divide by mass. We get V max squared is equal to K over M times a squared. And if we take the square root, we're gonna find that V max is equal to a times the square root of K over M. We know that the square square root of K over M is equal to the angular frequency omega. So we end up with a omega. All right, So, in our problem, okay, again, we know and we know Michael, we use them. When we calculated a max, we're gonna do the same thing here. So, we're gonna have 15 centimeters times one m per 100 centimeters, dividing by 100 to convert from centimeters to meters. And then we're gonna multiply by the angular frequency 6. radiance per second. And this is going to give us A maximum speed or speed at equilibrium of 0.94, two meters per second. Alright. So, we've done part two of part one. So back to let's look back at our question. So we've done this part. We've done this part. What's next? Now? We're asked to find the acceleration and the speed at a particular position. Alright, let's do that. So part two and then part one of this were asked to find the acceleration first at a particular position. Now recall that the acceleration is related to the position through the following. The acceleration is equal to negative K over M times a position X. So if we're at a particular position we can figure out our X value. Okay, what about this? Negative K over M. We don't have information about K or M. But recall, okay. And we just use it above that K. The square root of K over M is equal to the angular frequency omega. So we can write this as negative omega squared times X. Those are values that we know. Okay, so we get negative the angular frequency 6. radiance per second. All squared times the position. And were asked to determine this at 7. centimeters at negative 7.5 centimeters. Sorry? Okay, so this is gonna be negative 7.5 centimeters. And again we want to convert to meters. So we're gonna divide by 100 to convert to meters. And when we work this out we're gonna get an acceleration of 2. meters per second squared. So we found our acceleration at that particular point. What about our speed? Mm Well, let's go back to our conservation of mechanical energy again. So we had one half and v squared plus one half K X squared is equal to one half K A squared. And what we want to do is we want to solve for V. Now we can divide out the one half that's everywhere first then we have to divide by M. Okay, so let's do that. So we get V squared is equal to we divided at the one half. Okay, we have K A squared minus K X squared. And we're dividing by now again recall K over M is going to be omega squared. So when we take the square root, okay, we can factor out the K over M. We're gonna end up with omega Actually let's leave the square. Omega squared times a squared minus x squared. Okay, and when we take the square root we get omega times the square root of a squared minus X squared. Okay, so again, you may have recalled this formula right off the bat for speed. I'm just showing where it came from through the conservation of mechanical energy. And again, just double check should you be using this formula directly or should you be deriving it when you're doing a problem? Alright, so let's use that. We're gonna use it on the right hand side with our values. So V is equal to omega times the square root of a squared minus X squared. Now we know these values. We know that omega. The angular frequency is 6.28 radiance per second. Our amplitude a We know is cm. So we're gonna have zero point whoops. 15 cm. We're going to divide by 100 to turn it into meters. Okay, I'm not going to write out the entire thing like this just for the sake of space. But when we divide by 100 it's gonna turn to a meter square that and then same thing with our XK were given our exposition of negative 7. centimeters. So we have negative 7.5 divided by 100. That's going to be in meters and we square. All right now, this is just a bunch of things. Kind of multiplied, add together, take the square root, we can do this on our calculator and we will get a value for the speed of 0.816 m/s. Okay, so that's the speed at that particular position of negative 7. cm. Alright, so we found our speed here. We found our acceleration right above in part one. We found our maximum speed and our maximum acceleration And it doesn't hurt to to check when you're finding speed and acceleration at particular points. You can check that it's lower than the maximum speed and acceleration you found. Okay, so we found an acceleration of 2.9. That is less than the maximum. We found a 5.9, that's a good check. Um just that nothing went wrong. If you found a speed or acceleration bigger than the maximum then something may have gone wrong with either of those calculations. Alright, so we've done part two, we found the acceleration and the speed at that particular position. Now we're on to part three. Part three asks us the time needed to go from X equals zero centimeters two X equals positive 7.5 centimeters. All right, so part three Now we're looking at comparing time and position. Let's recall that we can write the position with respect to time. Okay. When we're looking at Spring mass systems as a sign, omega T. Okay. In this case we have at time T equals zero. Our position is zero. Okay, so we're using sign. Um and sometimes it will be a co sign there instead depending on the type of wave you have. Alright, so we want to find the time T when we're at 7. cm. So let's just fill in what we know and see if it's enough. So we have 7.5 cm is our position. We know our amplitude is 15 cm and then we have sine of omega but we know omega is 6. radiance per second times T. Alright, so now we see that T. Is the only thing in this equation we don't know. Okay so we can solve for this. So if we divide by 15 centimeters on the left hand side, we're gonna have one half on the right hand side, we get signed A 6.28 radiance per second times T. This tells us that T. Is going to be equal to while we take the inverse sine inverse Of one half. Okay then we're going to be left with 6.28 times t. So then we need to divide by 6.28 radiance per second. And when we do this we're gonna get a time of 0.08 seconds. Alright, so that's our time. The time that it takes to go from zero centimeters to 7.5 centimeters. Alright, we're getting close, we only have one question left to answer. Okay, we've done part three. The only thing we have left to do is part four which asks which quantities requested can be determined using conservation of mechanical energy. Alright, so we kind of found three things. We found speed. We found acceleration and we found time. Now we know that for speed we can get there using conservation of mechanical energy because we did exactly that. Okay, so for speed we can definitely definitely use conservation of mechanical energy. We can also find acceleration through conservation of mechanical energy. Okay, Because the conservation of mechanical energy involves forces. Okay, And we know that we can relate forces to mass times acceleration through Newton's laws. Okay. So we know that we can relate the acceleration to that conservation of mechanical energy as well. However, in our conservation of mechanical energy equation which you can see here, it does not involve t doesn't involve time at all. So we can't find time using that. Alright, so let's go back up here and pick one of these answers. So we found a maximum acceleration of 5.9 m per second squared approximately. So we can eliminate C. Then we found a maximum speed or speed at equilibrium of 0.94 m per second. Okay? So we can eliminate B. Because it has a different answer. So let me just write this out. We can eliminate this one can eliminate this one. So we're looking at A. Or D. We found that the acceleration at negative 7.5 centimeters was 2.96 m per second squared and the speed at that position was 0. m per second. So those are our next two values. 2.96 And 0.82, which is the same for answer A&D. We found That our time to go from zero cm to 7.5 cm with 0.08 seconds. Okay? So if we go back up, we want the next value to be 0.08 seconds and we see that a. Has that indeed desert. Okay. And then we look at the last answer. It also has speed and acceleration, which was our answer for which values can be determined through the conservation of mechanical energy. And so we have answer choice a here. Alright. Thanks everyone for watching. I hope this video helped see you in the next one.