https://youtu.be/-XlwZFI-lB4
PHYS 1101: Lecture Twenty-Three, Part Five
Okay, let’s put these ideas together in a last example. The problem reads, “I’ve got a computer that has to be used in a satellite and it has to be able to withstand accelerations up to 25 times acceleration due to gravity.” That’s a lot. I need to be able to test to see whether it meets the specification. If I just dropped my computer or threw it off a cliff, it would only accelerate at 1 g. I need to test 25 gs and see if it holds up.
So here’s what they do. They bolt this computer to a frame and they’re going to vibrate it back and forth. They’re going to subject it to simple harmonic motion, oscillating it back and forth about some equilibrium point. This is going to be my positive x direction, and they can oscillate it back and forth with a frequency of 9.5 Hz. Given that information, what’s the minimum amplitude of vibration that I need in this test in order to simulate this 25 times g as being my maximum acceleration?
They’re asking for an amplitude of vibration. That’s the variable A. I can picture that I’m looking at this computer as though it’s attached to a spring, and I’ve plucked this spring or this system so it’s oscillating back and forth. I need to know if the frequency is 9.5 Hz, what amplitude do I need in order that I achieve this acceleration. Well, the acceleration is the largest out at the turning points, out at the two extremes where this computer has to whip around, turn around to then move in the opposite direction. Let me draw that on here for you. Right here and right here, I have the maximum acceleration values.
How big is that? Well, the analysis I went through just above says my maximum acceleration for simple harmonic motion is set by the amplitude times Omega2. I want to solve for A. I know what this is. This has to be 25 times the acceleration due to gravity. 25 times g. 25 times 9.8 meters per second2. What’s Omega? That’s the last thing that I need. If I get a number for this, I’ll be set. A will just be this number divided by Omega2.
If you go back to our relationships, I defined for you that the frequency is cycles per second. That’s equivalent to 1 over the period. Well, what’s the period? That was related to the angular frequency, remember? If you look back at that previous equation, Omega is set by 2π divided by the period. So I can rewrite this. Maybe then it will be clearer if I just write this as 2π times 1 over the period.
Maybe from that you recognize 1 over the period. 1 over the period, that’s equivalent to f. I’m going to make that substitution because that’s a quantity I know. So this is equivalent to 2π times the frequency. Omega is 2π times the frequency. So let me plug in those numbers and let’s get a number for Omega. Omega then is 2π times the frequency. I need to multiply 2π then times my 9.5 Hz. 2π times 9.5 Hz gives me 59.8, and then the units that I have here now are radians per second.
Okay? So now I have a value for Omega. Let’s go back to our final equation, the equation that defines our maximum acceleration in terms of the amplitude and Omega2. Amax is equal to the amplitude, the variable I want times Omega2. Just do a little algebra and rearrange that. My amplitude then has to be Amax, which is 25 times g divided by Omega2. Plug in numbers for that and A is 25 times my 9.8, and then I have to divide by 59.8 radians per second2. When you multiply that out I end up with 0.069 meters. That’s a very small, small amplitude. That’s about 7 centimeters. If you know how big a centimeter is, there’s about 2.5 centimeters in an inch. So that’s about less than 3 inches is the distance, the amplitude of this oscillation motion. If I just go several inches on either side at about 10 cycles per second frequency I end up with a huge acceleration at these turning points of almost 25 g. So that’s the answer to this problem. And that brings us to the end of lecture 20.