A particle of mass m moves under the influence of a force 𝐅=-k 𝐫, where k is a constant and 𝐫(t)

A particle of mass m moves under the influence of a force...

A particle of mass $m$ moves under the influence of a force $\mathbf{F}=-k \mathbf{r},$ where $k$ is a constant and $\mathbf{r}(t)$ is the position of the particle at time $t$ (a) Write down differential equations for the components of $\mathbf{r}(t)$ (b) Solve the equations in part (a) subject to the initial conditions $\mathbf{r}(0)=\mathbf{0}, \mathbf{r}^{\prime}(0)=2 \mathbf{j}+\mathbf{k}$

A particle of mass $m$ moves under the influence of a force $\mathbf{F}=-k \mathbf{r},$ where $k$ is a constant and $\mathbf{r}(t)$ is the position of the particle at time $t$
(a) Write down differential equations for the components of $\mathbf{r}(t)$
(b) Solve the equations in part (a) subject to the initial conditions $\mathbf{r}(0)=\mathbf{0}, \mathbf{r}^{\prime}(0)=2 \mathbf{j}+\mathbf{k}$

Key Concepts

Harmonic Oscillator

This concept refers to systems in which a particle experiences a force that is directly proportional to its displacement from equilibrium and directed towards the equilibrium position. The resulting motion is oscillatory and can be described by second order differential equations of the form x'' + (k/m)x = 0, where the angular frequency is ? = sqrt(k/m). This principle underlies the dynamics of many physical systems subject to linear restoring forces.

Decoupled Differential Equations

In problems involving forces that are functions of the position vector in each coordinate independently, the overall vector differential equation can be separated into independent scalar equations for each coordinate. This separation simplifies the solution process, as each coordinate equation (such as x'' + (k/m)x = 0, y'' + (k/m)y = 0, and z'' + (k/m)z = 0) can be solved separately using standard methods for second order linear differential equations.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that not only satisfies the differential equation itself but also meets given initial conditions such as initial position and velocity. In the context of the harmonic oscillator, the specific values provided at time t = 0 allow for the determination of the particular constants in the general solution, ensuring a unique and physically meaningful solution for the motion of the particle.

Transcript

00:01 For this problem, we are told that the force on a particle is given by f equals negative k r of t, where r of t is the path that the particle is following.

00:10 We are then asked to write and solve the differential equations for each component, where r of zero equals zero and r prime of zero equals 2j plus k.

00:19 So, writing down the differential equations, we'd have that, oh, important detail, the particle has a mass m.

00:26 So we'd have then that m times a, because force equals mass times acceleration, but acceleration is second time derivative of r.

00:37 So we'd have that m times r double prime of t.

00:41 I'll call that rx double prime of t equals negative k rx of t.

00:46 Oops, that's an f, negative k rx of t.

00:54 M, ry double prime of t equals negative k ry.

01:00 Or r y of t and m r z double prime of t equals negative k r z of t then when we want to start solving these we note that all of them have the exact same sort of form where they are second order differential equations so we make the substitution that r x y or z i'm just going to write it as r equals e to the power of m x so we'll or actually i'll have to use something else, let's call it n, eats of nx.

01:37 But actually, that should be eats of nt, because t is our independent variable here.

01:42 So we'll have m times t squared equals negative k times nothing.

01:51 So just mt squared equals negative k, which means that t squared has to equal negative k over m, which then means that we have t equals plus or minus i root k over m which when we substitute that into our equation for r and i'll write r x equals r y equals r z they will all equal well e to the power of i root k over m plus e to the power of negative i root k over m that is going to be the form where we get a combination of cosine and sine as the solution.

02:36 Specifically, we'll have that rx, ry, and rz will equal c1 times cos of root k over m, oops, root k over m, t, plus c2, sine of root k over mt.

02:59 Oops, that was a big blurry k, but anyways.

03:04 But we'll note that this is that we have, different parameters for each one of those differential equations.

03:12 So now that we have that we want to actually solve for the given components or given conditions...

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