# DE for motion of a particle

#### bcjochim07

1. Homework Statement
A particle of mass m is confined to a horizontal plane. It is elastically bound to its equilibrium position by an isotropic elastic force,

F=mω^2r
Where r is the displacement of the particle from equilibrium and ω is a real constant parameter.

From Newton's law, we obtain the equation of motion:

r'' + ω^2r = 0.

In rect. coordinates r(t) = ix(t) + jy(t).

Obtain expressions for x(t) and y(t).

2. Homework Equations

3. The Attempt at a Solution

The x-component of force :

Fx = -x(t)mω^2 = mx''(t)
x''(t) + x(t)ω^2 = 0
The solution to this DE is x(t) = c1cos(ωt) + c2sin(ωt)
Then I'll apply the initial conditions given in the problem to obtain the constants.

Here's the part I'm unsure about
When I solve the differential equation for the y component I get
y(t) = a1cos(ωt) + a2sin(ωt). I'm thinking I have to multiply this by t so that the solutions are linearly independent:

y(t) = a1*tcos(ωt) + a2*tsin(ωt). Is this correct?

Related Advanced Physics Homework Help News on Phys.org

#### George Jones

Staff Emeritus
Gold Member
Here's the part I'm unsure about
When I solve the differential equation for the y component I get
y(t) = a1cos(ωt) + a2sin(ωt). I'm thinking I have to multiply this by t so that the solutions are linearly independent:

y(t) = a1*tcos(ωt) + a2*tsin(ωt). Is this correct?
No. Don't multiply by t.

#### bcjochim07

I don't understand why. Could you please explain?

Thanks.

#### HallsofIvy

Homework Helper
You don't understand why? What makes you think you should? It certainly can't be because x depends on those functions- the y-coordinate has nothing to do with the x-coordinate. They are completely indepedent.

#### VKint

There's no need to multiply by $$t$$. Yes, $$x(t)$$ and $$y(t)$$ are solutions to the same ODE, but there's no reason for them to be linearly independent; you're not going to use them as a basis in which to express other solutions of the ODE. In fact, you already used such a basis to get $$x$$ and $$y$$, i.e., $$\cos(\omega t)$$ and $$\sin(\omega t)$$.

You're probably confusing this with a similar-looking situation encountered with some second-order linear ODEs, which can be converted to a system of coupled linear first-order equations; if the associated matrix of this system is degenerate (that is, has repeated eigenvalues), then a basis for the solution space may include such elements as $$t \sin(\omega t)$$. (The reason for this is best understood as a consequence of the fact that the "degenerate" real canonical form of a matrix is the sum of the identity and a nilpotent matrix. Thus, when exponentiated, such a matrix yields a prefactor of $$e^t$$ times the identity and a sum of higher-order terms in $$t$$ that eventually terminates.)

As a final note, it's usually preferable to express solutions like $$c_1 \cos(\omega t) + c_2 \sin(\omega t)$$ in the form $$A \cos(\omega t + \phi)$$ for some phase $$\phi$$.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving