A 0.500-kg mass on a spring has velocity as a function of time given by vx(t) = -(3.60 cm/s) sin[(4.71 rad/s)t - (pi/2)]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

Video transcript

Hey everyone in this problem. An object of mass 1.2 kg is attached to an ideal spring. The objects velocity as a function of time follows a relationship V sub X of T is equal to negative 6.2 centimeters per second times sine of 7.53 radiance per second, T minus pi over four were asked to calculate the period of the motion and the springs force constant. Alright, so we're given this equation. The velocity is a function of time. Let's just write out that equation in general. Okay, so in general V sub X of T is going to be equal to negative a omega. Okay, the amplitude times the angular frequency times sine of omega T plus the phase angle fly. Alright, so we're giving this equation and we want to calculate the period. So how can we relate the period to this equation? Well, let's recall that the angular frequency omega can be written as two pi over the period T. Okay, so we're looking for the period T. And we can see that from our equation. We're going to be able to find this value of omega. All right, so in our equation omega appears twice. Okay, The first time it appears it's with the amplitude a multiplied together. So we can't really determine which portion of that is amplitude and which portion is omega. Okay, but in the second part. Okay, inside the sign we have omega in front of the T. Okay, it's the coefficient with the variable T. And in our equation we see that that's 7.53 radiance per second. Okay so we know that the angular frequency Omega is 7. radiance per second. This is gonna be equal to two pi divided by the period T. And if we isolate T we get T. It's going to be equal to two pi Over 7.53 radiance per second. Okay. two pi divided by 7.53. The unit of radiant cancels. We're left with 0.834 seconds. Okay so we have a period of 0.834 seconds. Okay so we're done part one of the problems. If we look at our answer traces we see that we have either answer A. B. Or F. That has the period of 10.834 seconds. Okay so we've eliminated the other options. And now let's move to this um springs force constant to do the second part of this problem. So the springs force constant is given by K. How can we relate K. To some of these other values in our equation. Well let's recall that the angular frequency omega. Okay. Above we used that it was equal to two pi over tea to recall that it is equal to the square root of K. Over M. As well. Okay so we know omega. We found it in this first part of the problem from our equation. And we're looking for. Kay the springs force constant and we know the mass were given the mass. and the problem so we can solve for K. So we got 7.53 radiance per second is equal to the square root of K. Divided by the mass. Now if we go back up to the problem, You see that the mass is 1.2 kg. So we get mass 1.2 kilograms. Now we're solving for K. So if we square everything, we get 7.53 radiance per second squared is equal to K over 1. kg. Okay. And multiplying by 1.2 kg, we get that K is equal to .04108. And our unit here is going to be Newton Her meter. Alright, so now we have our force constant K. We found our period. So let's go back to our answer choices. And we see that we have answer choice A and we found that the period is 0.834 seconds. And this spring's force force constant K. Is approximately 68 Newtons per meter. Thanks everyone for watching. I hope this video helped see you in the next one