Two masses m1 and m2 rest on a horizontal frictionless...

Two masses $m_{1}$ and $m_{2}$ rest on a horizontal frictionless plane. The masses are attached to fixed walls by springs whose spring constants are $k_{1}$ and $k_{3}$. (See Figure $9.7 .5 .$ The masses are connected by a spring whose spring constant is $k_{2}$. Determine a first-order system of differential equations that governs the motion of the system. (FIGURE CAN'T COPY)

Key Concepts

State Space Representation

State space representation involves defining state variables (like displacement and velocity) for each component of the system, allowing the use of a structured framework to analyze and solve the dynamic behavior of the system through first-order differential equations.

Newton's Second Law

This law forms the foundation for deriving the motion equations by stating that the net force acting on a mass is equal to the mass multiplied by its acceleration. It is used to set up the dynamic equations for mechanical systems such as those involving multiple masses and springs.

Hooke's Law

Hooke's Law describes how the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This principle is crucial in establishing the force balance on each mass in a spring-mass system.

Coupled Oscillators

Coupled oscillators refer to systems where individual oscillators (masses attached to springs) are interconnected in a way that the motion of one affects the others. The resulting equations are interdependent, leading to a system of differential equations for the entire system.

First-Order Differential Equations

Many physical systems are modeled by second-order differential equations, but these can be reformulated as a set of first-order equations by introducing auxiliary variables (such as velocities). This transformation is useful for both analytical and numerical analysis of the system.

Transcript

00:01 So for this problem, we've been given this two mass free spring coupled equation.

00:09 And i've drawn it on the screen here.

00:12 And we've been asked to try and determine what the differential equations governing the system are.

00:20 So here we've just written out.

00:22 So when this is an equilibrium position, so we have that the extensions x and y are equal to zero.

00:29 But let's suppose that we move this spring this way, so x is non -zero.

00:39 The change in x is simply delta x.

00:43 So it's simply the extension of this left -hand spring.

00:49 So that's delta x.

00:51 But now the length of the middle spring has automatically got smaller.

00:56 So the change, if the spring, if the middle spring was to go to the right -hand side, it's not just dependent on the position of y, it's also dependent on position of x.

01:08 So the change in y is delta x is delta y minus delta x because if the position of the left hand spring moves more than y, the springs cross over.

01:26 And so the m2 has actually moved to the left.

01:30 So here we see that that's determined the change of the first two springs.

01:40 Now, simply for the last spring, the spring k3, this is only dependent on the last mass, and this changes with length minus y.

01:52 And that's important.

01:53 So now we know that for the first spring, it's dependent only on the position of x.

01:58 Middle spring, it's dependent only on the position, it's dependent on both positions, and for the last spring, it's again dependent only on the y position.

02:08 So now we can use hux law and newton's second law to get some differential equations.

02:14 So we know that from newton's second law, let's look at the first mass, that m1 times the acceleration, so b squared x t squared, is equal to the sum of the forces.

02:32 F1 is going in the opposite direction to what we're travelling...

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