An object in simple harmonic motion has an amplitude of 8.0 cm, n angular frequency of 0.25 rad/s, and a phase constant of π rad. Draw a velocity graph showing two cycles of the motion.

Question

Graph showing velocity over time for two cycles of simple harmonic motion.

Accepted Answer

Hey, everyone in this problem, we have a floating cylinder that's undergoing simple harmonic motion with an amplitude of 0.6 centimeters. We have a is equal to 0.6 centimeters. OK. The angular frequency of the motion is 4. radiance per second. OK. The angular frequency we write as omega that is 4.45 radiance per second. OK. And the phase constant we're told is pi divided by three and the phase constant is given by five. So we have P is equal to pi divided by three. Now, we're asked to draw about two cycles K of the cylinders velocity time graph starting at T equals zero seconds. We're told to assume that the motion follows X of T is equal to a cosine of omega T plus five. So looking at X of T and we have a, we have omega, we have five. So we have all of the constants in this equation. So that's great. The question is asking us to draw a velocity time girl. OK. So how can we relate the velocity to the position? OK. Because remember that X of T represents the position. Now, we're called the velocity V F T is related to the position through the der so V F T is equal to D X divided by D T. Yes, we're gonna take the derivative of our position function X of T with respect to the type the amplitude out in front is a constant. So it's gonna stay, then we take the derivative of cosa. Now the derivative of cosine is negative sign. We're multiplying by negative sign and we keep everything inside just like it was inside the cosine. So we have negative sign of Omega T plus five. And then by the chain rule, we need to multiply by the derivative of what's inside of the function. OK. So inside of our cosine function, originally, we had Omega T plus five taking the derivative of that with respect to T OK. Omega T, the derivative is gonna be omega and then plus five, that's just a constant. The derivative is going to be zero. And so we're gonna multiply by just oh my God. And that's the derivative that's our velocity. Now, we can simplify this, forget that the velocity V T is equal to negative A omega multiplied by sine omega T plus five. OK. So just rearranging there to have those constants together at the beginning. OK. So we've written V in terms of a omega and five. And again, we know those values. So let's go ahead and substitute them in and our equation becomes V F T is equal to negative 0. m. Sorry, that should be centimeters. We were given our amplitude in centimeters. So we have negative 0.6 centimeters multiplied by 4.45 radiance per second multiplied by sign of 4.45 radiance per second plus pi divided by three. All right. So this is our velocity equation. We'll do one more step. We'll just simplify that constant open front V F T is going to be equal to negative 2.67 centimeters per second, multiplied by sine of 4. radiance per second T plus pi divided by three. All right. And this is the equation that we want to graph. And we were asked for the velocity time graph. So this is the equation we want to graph. Now the most straightforward way to do this, we're just gonna write out a table of values K with T and V T in order to figure out how we should graph this. And what we need to do though is figure out what T values, how long do we have to go in time to get the two full cycles that we're looking for? Well, let's think about the period. OK. Recall that the period T is related to the angular frequency through the following T is equal to two hi, divided by Omega. Hm. In this case, Omega is 4.45. So we have two pi divided by 4. a radiance per second, which gives us a period of about 1. seconds. Ok? So if our period is about 1.4 seconds, that's the time it takes to complete one cycle. And so the time to take to complete two cycles, it's gonna be about 2.8 seconds and we're just gonna round up 23 seconds. Ok? So we're gonna draw a three seconds here and that's gonna cover those two full cycles. So let's get to our table of values. We're gonna leave up our equation so we can see it. We have the time and then we have the velocity of V T and the time T is gonna be given in seconds and the velocity is gonna be in centimeters per second. OK? Because that's that unit we have in that constant out in front. All right. Yeah, we're told to start at time equals zero seconds. We wanna go up to three seconds. Let's go on 0.25 increments. Um Just so that we're getting a lot of detail, we can kind of see how this curve is changing, but we don't have too many calculations either. So 0.25, 0.5, 0.751, 1.251 point five, 1. two, 2.25, 2.52 point 75 and three. Those are all of the key values that we're going to evaluate. Now if we substitute T is equal to zero into our equation for the velocity, what do we get? Well, we get negative 2.67. Hey, and I'm dropping the units because we know it's gonna be centimeters per second multiplied by sign. If T is equal to zero, then we just get sign of pi divided by three. And this is gonna be equal to approximately negative to 0.31 meters per second. Moving to the next value T is equal to 0.25. We have negative 2.67 multiplied by sine 4.45, multiplied by 0.25 plus pi divided by three. And if we work this out, this is gonna give us negative 2.22, whoops, not meters per second centimeters per second on both of these centimeters per second and always double check your unit so that you aren't making mistakes like this. All right now, we can keep going and I'm gonna stop writing out the entire expression just to save a little bit of space here. When we plug in T equals 0.5, we do it the exact same way we'll get negative 2.67, multiplied by sine of 4.45, multiplied by 0.5 plus pi divided by three. We're gonna get that. This is equal to zero 0.3477 A. All our units are in centimeters per second. And we're just gonna keep on going substituting in these values 0.75, we get 2.58 and we have a time of one. We get a velocity of 1. for 1.25. Our velocity is gonna be zero, negative sorry, 0.856 a time of 1.5, we get negative 2. moving down in our table. And we are just continuing this same pattern. We're substituting in the T value into our equation for the velocity. And we're recording what that velocity is for 1.75 seconds. The velocity is negative 1. for two seconds. We have 1.33, 2.25. We have a velocity of 2.66 for 2.5. We have a velocity of 1.2 54 for 2.75 seconds, negative 1.7572. And finally, for three seconds, our velocity is gonna be negative 2.58. And you can see that we're going almost through two cycles or just over two cycles. OK. Remember that our amplitude is 0. m. OK. When we look at the velocity equation, let me go back to it. OK. This constant out in front of our sign is gonna represent the amplitude on our graph. Our velocity time graph. OK. So we have negative 2.67. So you can see that we're starting almost at that maximum negative value. OK. Our velocity is increasing, it becomes non negative, it increases up to that really close to that maximum. And then it comes back down to negative, back down to the neg the maximum negative value before it increases again, back up to that maximum and comes back down. OK. So you can see that we're going through those cycles. Now, let's plot this out on our graph and we're going to have on the X axis the time in seconds and on the Y axis, we're gonna have the velocity V T in centimeters per second. Now, our time goes from 0 to 3 seconds. So we're gonna go ahead and say that every three ticks is gonna be one second. So we have one second, two seconds, three seconds, our maximum Y volumes about just under three. We're gonna do the same spacing on the Y axis. I'm in the negative direction as well. All right. Now, I'm gonna take blue and draw our function in blue just so we can distinguish it from these um black axis labels. And we're just gonna plot all of these points when we have a time equals zero. We found that the velocity was negative 2.31. OK. So that's gonna be right around here when we had a velo or a time of 0.25, we had a velocity of negative 2.22. OK. So we're increasing just a little bit when we have a time of 0.5, we have a velocity that's 0.3 or seven. So that's increased a lot. It's gonna be right above that first line in the positive quadrant at 0.75 seconds. We have a velocity of 2.58, sorry 2.528. So 0.75 and 2.528, you can see that we're just going through and plotting all of these points that we wrote down one second and you can follow along plotting all of these points. OK? 1.25 seconds, we're back down 1.5 seconds. We're at negative 2.647, 1.75 seconds, negative 1. at two seconds. Velocity is 1. at 2.25 seconds. The velocity is 2. at 2.5 seconds of velocity is just above one at 2.75 seconds. It's back down about negative 1. and finally at three seconds, it's at negative 2.58. We've just approximately put all of these points on our curve and we are now gonna connect them and it is gonna look something like this and you're gonna start to see that sine wave in those cycles that we drew. OK. Function goes up it back down and repeats that same process. And you can see that we have two full cycles drawn here. OK. So that is an approximate graph of what the velocity time curve looks like in this situation. Thanks everyone for watching. I hope this video helped see you in the next one.

An object in simple harmonic motion has an amplitude of 8.0 cm, n angular frequency of 0.25 rad/s, and a phase constant of π rad. Draw a velocity graph showing two cycles of the motion.

Verified Solution

Key Concepts

Simple Harmonic Motion (SHM)

Velocity in SHM

Graphing SHM