For the oscillating object in Fig. E14.4, what is its maximum acceleration?

Question

Graph depicting position over time for a simple harmonic oscillator, labeled with time intervals.

Accepted Answer

Hey everyone in this problem. The figure below shows the position time graph of a particle oscillating along the horizontal plane and were asked to find the maximum acceleration of the particle. Now the graph were given has the position X and centimeters and the time t in seconds. All right, so let's recall the maximum acceleration. We're trying to find a max can be given as plus or minus the amplitude a times omega squared. So in order to find the maximum acceleration we need to find the amplitude A and the angular frequency omega while the amplitude A. Okay, this is going to be the maximum displacement from X equals zero. and our amplitude here is going to be 10cm. Okay, we see both positive and negative 10 centimeters. Okay. And so our amplitude is going to be 10 centimeters and it's important to remember when we're looking at the amplitude. It's that max displacement from X equals zero. Okay, so it's this distance here or this distance here but it's not the sum of the two. It's not the entire distance from 10 to negative 10. It's measured from that X equals zero. Okay, so our amplitude A is 10 centimeters now. We need to go ahead and find, oh my God, well one thing we know that's easy to find from a graph like this position time graph is the period T. Okay, so let's recall that we can relate omega the angular frequency to the period T through two pi over t. Now let's try to find T. Alright. So what we want to do to find the period is we want to go through one entire cycle of our wave. Okay. Of our oscillation. So if we start with this red point here okay we're positive six centimeters and we're going downwards so we follow the curve along and we want to get to the next point where we're doing the exact same thing which is right here and then we're gonna look at the time Between those two points. Now, be careful when you're doing this. Okay we're taking six cm here. The period is not until the next time we hit six cm on our graph. Okay, when we hit six cm in the same way. Okay, so on the left hand side here we've picked six cm and we can see that the position is decreasing. Okay if we were to pick six centimeters here that's when the position is increasing. So that's not the same. That's not one entire cycle. We have to wait until we're at six centimeters in the position is decreasing again. Okay. Alright and so we find that our period T. Okay, the time between these two is going to be 10 seconds. Alright, So if we get back to our calculation for angular frequency We have two pi divided by the period which we found to be 10 seconds. So this is gonna be pie Number five radiance per second and now we can go ahead and do our calculation for the maximum acceleration a max. This is going to be the amplitude 10 cm times The angular frequency high over five radiance per second. Hey, all squared. So we get an a max a maximum amplitude of 3.9 or eight centimeters per second squared. Okay. And if we look at the answer choices, these are in meters per second squared. So we're just gonna go ahead and convert that into meters per second squared. So this is going to be 3.948 centimeters per second squared times one m centimeters. Okay, The unit of centimeter will divide out. So we essentially divide by 100 and we end up with 0.39 approximately meters per second squared. And if we look at our answer traces we see that that corresponds to answer C. Okay, so the maximum acceleration here is 0.39 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

For the oscillating object in Fig. E14.4, what is its maximum acceleration?

Verified video answer for a similar problem:

Key Concepts

Simple Harmonic Motion (SHM)

Acceleration in SHM

Graph Interpretation