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Simple Harmonic Motion Example 2 (Take-Home Practice)A 3.00-kg diving board above a swimming pool is set up at a equilibrium position: The board is tilted by a dive...

Question

Simple Harmonic Motion Example 2 (Take-Home Practice)A 3.00-kg diving board above a swimming pool is set up at a equilibrium position: The board is tilted by a diver by 0.150 m below the equilibrium point: The diver jumps off and then the board is released. Ignore friction. F: (a) the spring stiffness constant k; (b) the amplitude of the horizontal oscillation A; (c) the period T and frequency f; (d) the magnitude of the maximum velocity Vmax; (e) the magnitude of the maximum acceleration amax o

Simple Harmonic Motion Example 2 (Take-Home Practice) A 3.00-kg diving board above a swimming pool is set up at a equilibrium position: The board is tilted by a diver by 0.150 m below the equilibrium point: The diver jumps off and then the board is released. Ignore friction. F: (a) the spring stiffness constant k; (b) the amplitude of the horizontal oscillation A; (c) the period T and frequency f; (d) the magnitude of the maximum velocity Vmax; (e) the magnitude of the maximum acceleration amax of the mass.



Answers

A mass on a spring oscillates with simple harmonic motion of amplitude $A$ about the equilibrium position $x=0 .$ Its maximum speed is $v_{\max }$ and its maximum acceleration is $a_{\mathrm{max}}$ (a) What is the speed of the mass at $x=0 ?$ (b) What is the acceleration of the mass at $x=0 ?$ (c) What is the speed of the mass at $x=A ?$ (d) What is the acceleration of the mass at $x=A ?$

And this problem we're dealing with an object you passed the spring. This means that its motion is simple Armani motion meaning that it oscillates or it's going to move back and forth. So if you think of a wave you can picture like this. So this object is actually going like this, like a way up and down, up and down. So this right here with the amplitude a of this way for the height of the motion and this problem, we're told that a the amplitude is 12 centimeters or we divide by 100 to get 1000.12 meters, and we're also given the maximum acceleration, which is a Mac. And that's 100 and eight centimeters per second squared but divided by 100. So we can't 1.8 meters per second squared. And we're also told to keep em as a variable, meaning don't change it. And we don't know a specific numerical value for it apart. A. You want to find the period and that's found by T is equal to two pi the out of my w, the angular frequency. And that's what we don't know Here. We do know that a max, The maximum acceleration. It's found by a the amplitude times. Jeff, you squared from here. Let's solve for W So you have dub you is equal to a max divided by a in the it's it's a square. You take the square root of both sides to get rid of that too. So this is the square root. So we know a max and we know a the attitude. So just looking our numbers. So it's 1.8 the better by appointment to And that, of course, is square root for the angular frequency W we have three rads per second, and that's gonna be crucial for our other calculations. No, you just put that in our period. So the now t two pi about a rectory. So we get T is equal to 2.9 seconds. Next. For part be, we want to find just the frequency of this motion. But we know that the angular frequency W is equal to two pi climes f the frequency in which is what we're looking for. So then just saw for so we can't we divide both sides by two pi. So then this just after frequency is equal to W the angular frequency that better by two pi. And of course, we know W which is just three. That's three divided by two points. So apart me frequency, his 0.4 eight hurts. Next report. See, you want to find the velocity, Max The max that's found by a damp it two times w the angle of frequency. And of course, we know both those numbers. So both of them in So it's 0.12 times three. So were the max with a maximum velocity. And we have 0.36 community per second. And for D Do you want to find energy of this motion 100 singing with simple harmonic motion? The energy is one, huh? Okay. The spring, constant times a squared, a being the amplitude. But we don't know. OK, we do know that another form of the angular frequency is k over and square. Root the square both sides and you have w squared is equal to king over and so multiple able sign by. And so you have a K is equal to w squares times m Well, we do know w squared and remember em. Keep is a variable. Just leave it alone. So you have three square times in, Okay, We get nine. And so we can just plug this in. So we have our energy is now one half, 9 a.m. Times a squared looking a rally for a You have one half my ember that a is 0.12 squared. So for the energy in E, we get 0.0 six by and Jules remember em and leave alone now for he I wanted to find K It's green constant, but we just found it about here, right here. Okay. You know, like, you know that all the K spring constant is nine times and and finally for f looking at these variables, we know that the period frequency and maximum velocity don't depend on knowing a numerical value for em. But for the energy and the spring Constant K, it does depend on yeah,

So we're given an object that undergo simple harmonic motion with an amplitude of 12 centimeters, which we might want to use meters, sometimes to on the acceleration is given by 108 centimetres per second squared restaurant A whole bunch of things that first goes to find the period trying to find the period need to relate acceleration and angular accelerant, acceleration and the imp itude. And we know that Omega's a equals a omega squared. So omega is the square root of X l maximum acceleration over that. If you plug that in you actually with three parts, right? So now we're just gonna kind of go down the list. We know that the period is two pi over Omega, right? We're just two. Pi over three is about 2.1 seconds. It doesn't depend on the mass. The linear frequency is one over that. So it's about 0.48 hurts doesn't depend on the mass. The maximum speed that this object will go is a Omega A is 12 centimeters and omega is three. So that's 36 centimetres per second. Still doesn't mean of the mass, uh, next rest to find the energy that's given by 1/2 a k A squared, um, or you could do 1/2 him V squared. Either one watch. You get it. In this case, we haven't found K yet, so I'm gonna use the 2nd 1 We have v here that we can just plug in. Um, and if you do it appropriately in s I where I'll use meters per second, I actually find that this is 0.65 times the mass right, which we leave is a variable here. And then finally be spring constant, which you could get from equating the energy here or just rearranging our angular frequency one. And so you end up with nine times in, right. And so you will notice The only dependence is on em are right here. Right? The energy and spring constant. Both are linearly proportional, but otherwise, there's no other, uh, ems anywhere. Right? Nothing else depends on it.

Our guys. So the question was asking us the amplitude, the time, period and the frequency of emotion. Well, but amplitude since we know the initial species row at the position, that is Gero Poin want to draw me away from the equilibrium position So they MP is just simply equal to 0.12 meter. Okay, so the empty was just a good your point. Want to? Here he goes at a position There's no speed by the acceleration is at a maximum. So that's the amplitude of emotion. Okay? And to find the time period, we know that you know what a complete one time period you have to pass through the equilibrium point twice and in the question you were saying that after a 0.800 2nd that this placements found to be a 0.12 meters away from equilibrium position. So this is once angle to make sure you pass it equally being twice you have, you have a non 0.800 2nd on it which this is two off the 0.800 seconds and just two times the tea which will give us 1.6 seconds. So 1.6 seconds is the time period. And to find the frequency of the motion, we know that the frequency off the motion is equal to while, over time period, which is long over 1.6 seconds. And this would be best 0.6 to 5 hers. And that's a frequency off the motion. And as my answers with a question, Thank you.

So for part A, we can say at X equals zero, this would be the equilibrium position. There is no potential energy stored. All of the energy is in kinetic energy. And so we can say the speed of the mess is at its maximum value where the velocities equaling the maximum velocity now four part to be it, we can say that at X equal zero the acceleration of the mass is equaling zero meters per second squared four parts. See, we know that at ex equaling the amplitude here there it is all in the energy of the system is all in the form of potential energy rather than kinetic energy. So here we can say that the velocity of the masses going to simply be equal to zero meters per second. However, we can say that then for part D at X equaling a, the acceleration would be equal to the negative maximum acceleration because at the at the at the point where X equal a at where X equals a, the mass is at its maximum amplitude. It is changing direction from a positive to a negative velocity and therefore the acceleration is negative. However, the magnitude of the negative acceleration is the maximum acceleration. That is the end of the solution. Thank you for watching.


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