A 0.400-kg object undergoing SHM has ax = -1.80 m/s^2 when x = 0.300 m. What is the time for one oscillation?

Video transcript

Hey everyone in this problem we have a block of mass 0.65 kg attached at the end of an ideal horizontal spring undergo simple harmonic motion. When x is equal to negative 0.13 m. Its acceleration is 2.8 m per second squared. And were asked to determine the time taken by the block to complete one oscillation. Okay. All right. So let's think about what this means. Okay, what is it that we're actually trying to figure out? They were asked to find the time taken. Okay, by the block to complete one oscillation. So the time taken. Okay, to complete one oscillation, what does that mean? Well, that's gonna be our period. It's the time taken to complete one oscillation. This is just the period. Okay. Which is given by capital T. Okay, so what we're actually looking for is the period T. All right. How can we find the period T. Well, we're given some information about acceleration and position. And let's recall that we have the following equation that the acceleration is related to the position through the following A is equal to negative K. X. Over M. Now we have an acceleration and we have a position X. Okay, we also have a mass M. So we could use this to solve for K. How can we relate that to the period? We're trying to find well, let's recall that the period T. Is equal to two pi over the angular frequency omega, but the angular frequency omega can be written as a square root of K over M. Hm. So we can write the period T. Then this two pi divided by the square root of the springs. Force constant K. Divided by the mass. Hm. Alright so we can go about this two ways we know em so we can substitute A. M. And X into the equation on the left hand side and solve for K. And then we can move to the right hand side. We'll have a value for K. And M. That we can substitute here to find T. Or. Okay, you can notice it on the right hand side. We have this value of K. Over M. So instead of substituting M and solving for K, we can just solve for K over M. And use that in that equation. Okay so let's do that. So our acceleration 2.8 m per second squared. Our force constant K. We don't know. Okay so we have negative K over M times the position X. Negative 0.13 m. So we have a negative times a negative. It's gonna give us a positive and we have that K over M is going to be equal to 2.8 m per second squared Divided by 0.13 m. The unit of meter will cancel, we're gonna be left with 21.53846. And our unit again meters cancels and we're left with one over second squared. Alright so we have our value of K over M. Now we can use it to find the period T. So we have the period T. Is equal to two pi over the square root of K. zero g. So this is gonna be the square root of 21. seconds to the negative two. And if we work this out on our calculator We're gonna get to the period T. is equal to 1.35 seconds. Alright? So if you look at our answer choices, we found that the time taken by the block to complete one oscillation that's equivalent to the period is going to be C1. seconds. Thanks everyone for watching. I hope this video helped see you in the next one.