Suppose the bottom support in the mass-spring chain in Example 10.40 is removed. (a) Do you pred

step 1 Good day, ladies and gentlemen. Again, we're looking at problem number 33 here. And in this prob

Suppose the bottom support in the mass-spring chain in...

Key Concepts

Mass-Spring Systems

Mass-spring systems are fundamental models in mechanics where masses are connected by springs, allowing the study of oscillatory behavior. They are governed by differential equations derived from Newton’s laws, providing insights into energy distribution, damping, and resonance phenomena which are applicable in various fields of physics and engineering.

Boundary Conditions

Boundary conditions define how the system behaves at its limits, such as fixed, free, or supported ends. Changes to these conditions, like removing a support, directly impact the system’s natural frequencies and vibration patterns by altering the constraints imposed on the masses, thereby modifying the dynamic response.

Normal Modes

Normal modes represent the intrinsic oscillatory patterns of a coupled system where all elements move synchronously at characteristic frequencies. Each mode is associated with a specific spatial pattern and eigenfrequency, simplifying the analysis of complex vibrations by enabling a modal decomposition of the system’s motion.

Eigenvalue Problems

The computation of vibrational frequencies in coupled systems is grounded in solving eigenvalue problems. Here, the system’s equations of motion are formulated in matrix form, where the eigenvalues correspond to the squared natural frequencies and the eigenvectors define the mode shapes. This method is essential for predicting how changes, such as altered boundary conditions, affect the system.

Initial Conditions and Transient Response

Imposing an initial displacement, like displacing the middle mass, sets specific initial conditions that lead to transient dynamics as the system evolves over time. The resulting motion is a superposition of normal modes, and graphing these responses provides a visual insight into energy distribution, modal interference, and the impact of boundary modifications on the system’s behavior.

Transcript

00:05 Good day, ladies and gentlemen.

00:08 Again, we're looking at problem number 33 here.

00:14 And in this problem, again, we're not really emphasizing the solving method so much.

00:20 We're more setting up a problem and then looking at what we get when we get it.

00:29 So a lot of this is actually using things from 32, which i, i didn't technically need to do here.

00:42 So i'm just sort of grabbing information for 32.

00:47 So if any of this stuff you're not aware of, it's because you haven't looked at problem number 32, or you don't know how to solve previous problems.

00:59 So i'm sort of just summarizing pretty much what needed to be done.

01:04 So in the first case, in the first part of it, when you set up the ordinary differential equation, the way put this, so the 10 here, which is the mass of the object, this is the my double prime term.

01:33 In other words, this is the force term.

01:35 This here, this is a force term.

01:36 This here, it's a see if i remember correctly.

01:44 Is this the k or the, i'm forgetting now which term this is? so one is the k and one is the b turn.

01:56 And now i can't have this written.

01:59 And i can't remember what it is.

02:04 Sorry about that.

02:05 I shouldn't have it here.

02:07 Okay.

02:08 So this is really the, i think the b term.

02:18 Let's see.

02:22 Yeah.

02:23 So the 60 here, sorry.

02:27 This is the by prime.

02:30 And it's the, i guess, as i said, it's the, it refers to the damping term, which comes from friction.

02:45 And then the stiffness term, the k, that comes from the spring so this is sort of the spring constant so you just i mean the setup of the problem is really to just input those terms into the equation and the previous one has no is undamped so this is a case where you have damping the other one is undamped and you're sort of supposed to compare between the two of them what happens.

03:19 So the first case, so now if we look at what differential equation is, yeah, okay, so what i want to do is just mention here that when you go through the solving and everything, you're going to see this is your equation here.

03:51 Then when we want to talk about setting it to zero okay we just set y equals zero and solve and what you get is this tangent 4 t to be negative three halves and then that tells you that t is equal to this but you'll um this is not four divided by pi by the way it's four pi it was supposed to be four pi um this number is negative uh which doesn't mean anything but you can add um four pi to it you can add basically by adding numbers of pi um multiples of pi to this to this initial value of t i was looking for the first time that um this would be positive so i can't remember what the number was it was something like negative 14 or something which like negative 14 seconds or something and so by adding four pi to it i get back the 1 .63 which is the first time it adds it's positive so this tells me basically that in 1 .63 seconds it gets back to zero...

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