FIGURE EX15.7 is the position-versus-time graph of a particle in simple harmonic motion. c. What is vₘₐₓ?

Video transcript

Welcome back everyone. We are given this position time representation of simple harmonic motion. And we are tasked with finding what is the peak speed for our graph? We know that the velocity at any given point is going to be equal to the negative of amplitude times our our omega times the sign of omega T plus our phase constant. This comes from the derivative of position for simple harmonic motion. Now, when can we maximize this value? Well, we can maximize this value when our amplitude times our omega is at its peak. Therefore, our maximum velocity will be equal to our amplitude times our omega. So we just have to figure out those terms. Well, what is our amplitude here? Our amplitude is just going to be the physical peak of position. So looking at our graph here, we have peaks at four and negative four. The absolute value of those uh A K or displacement from equilibrium is going to be equal to four centimeters or point oh four m. Now, we know for omega, we know that Omega is equal to two pi times the frequency. And we know that frequency is equal to one over the period, combining these two, we get that Omega is equal to two pi over the period. But what is the period? The period is the time it takes for simple harmonic motion to complete one full cycle. So one full cycle will be from 0 to 4 as you can see by the motion along the graph. So our period therefore is four. So if we plug in four, we get that Omega is equal to two pi divided by four. So now we can say that V max is equal to 0. times two pi over four, which gives us a final answer of 0.63 m per second, which corresponds to our answer choice. A thank you all so much for watching. I hope this video helped. We will see you all in the next one.