How Do You Solve the 3D Non-Isotropic Oscillator Numerically?

[emptymaximum]

then i'll have the amplitudes left.
so how do i numerically find the amplitudes?
see, the problem says:

Find x,y,z as functions of t NUMERICALLY.

if they left out that numerically this problem would be pie.

oh well
maybe i'll still get most of the marks for it.

 

[Cyrus]

hmm, can I point out another thing... maybe it helps, you said

the general solution is x = A1 sin(╥t + φ) + A2 cos(╥t + φ), and so on for y and z with B and C replacing A respectively. φ is the phase angle.
for this problem the particle is at the origin at t=0. this forces φ = 0, and the sine term = 0. THis leaves the second term, but since x = 0 A2 = 0.

Why does that force phi to equal zero. Phi could be nonzero provided that the coefficient A1=A2 and that phi = -45 degrees, just as an example, or is there something fundamentally wrong with calling phi nonzero at t=0?

 

[Tide]

Yes, you do!

mass held by six springs and is located at the origin. Potential function is given by V = k/2 (x^2 + 4y^2 + 9z^2). at t = 0 the mass is given a push in the (1,1,1) direction imparting vo.

 

[Cyrus]

Tide, is the phase angle going to be zero, and not something like say -45? How would you determine that.

 

[emptymaximum]

ya, it's pushed in (1,1,1) doesn't mean it goes through (1,1,1).
i'm sure i asked somehwere abut that.
like (1,1,1) means that vo is non-zero and of equall magnitude in all directions. i mean as soon as it gets pushed, the restoring forces kick in and since they're not = then they pull different and it's going to be a curved path. right?φ is zero because it is like the 'original position' of the wavefunction.

 

[Tide]

No, it means that velocity is imparted of a certain magnitude and the specified direction. Yes, the restoring force "kicks in" but it is a force acting on a moving body starting at the origin with a particular velocity - I.E. your initial conditions!

 

[emptymaximum]

vo is the initial conditions?
i'm not sure i understand what you said.
imparted velocity is in all 3 directions and is of equal magnitude.
the force that acts on it is:
F = -kxi -4kyj -9kzz
i put (1,1,1) in that?
i don't get it.
i hate this numerical analysis buisiness.

 

[Tide]

Do you mind if I ask you exactly what course this is for and what your math/physics background is? It seems like we're talking past each other.

Let me cut the problem down to size for you. Your specific problem is "separable" meaning that with the potential provided and from the equations of motion the object acting under that potential undergoes motion in 3D but each dimension is independent of the other 2. In other words, you can think of it in terms of 3 separate oscillators (one for each dimension) and you can (?) solve for the motion in each of the three dimensions independently.

For example, one oscillator corresponds to the x direction with a restoring force equal to -kx which starts at x = 0 with a velocity $v_0$. You shoud be able to solve that one completely. You will have similar equations and solutions for the other 2 dimensions.

When you're all done with that then you can think about the remaining question of whether the object ever makes it back to the (3D) origin.

 

[Tide]

Why is this discussion going on in two different places?

 

[emptymaximum]

are you sure can analyse the componets seperately because:

r=f(t)i+g(t)j+h(t)k
and
F=f(x)i+g(y)j+h(z)k

the forces are position dependent, and the position varies with time. Subsequently, the forces are not time invariant.
imagine a 1-d oscillator. but now instead of pulling the mass down, pull it sideways and down.
now you have a 2-d problem for that one spring. the restoring force no longer pulls in just the x-direction. the extension in the x direction remains the same, but the force in the x-direction is less than in the 2-d case. right?
and so the spring along the x-axis will provide force in the y direction.
extended to 3-d and again the restoring force on the spring in the x-direction will have the same magnitude, but the force in the y-direction will be diminished because it must share the force in the z-direction.

as far as it retracing its path, this will occur iff the angular frequencies are commensurate:
w1/n1 = w2/n2 = w3/n3 where n1,n2,n3 are integers.
since w1 = ╥, w2 = 2╥, and w3 = 3╥ ; then they are commensurate with: n1 = 1, n2 = 2, n3 =3.
the path lies in a plane and is in the shape of a lissajous figure.
like i said in previous post, the retracing the path problem isn't the problem.
the problem is the solve numerically for x(t),y(t),z(t).

this is for mechanics 1 class.
my background according to the academic institution is 2nd year student, but I've done a lot before i ever went there.
did i post this in the wrong forums?
is it advanced?

 

[emptymaximum]

because i wasn't sure if it was advanced or not

 

[quantumdude]

Tide: Why is this discussion going on in two different places?

emptymaximum: because i wasn't sure if it was advanced or not

This appears to be above the level of "Halliday and Resnick", so we'll call it advanced. I've merged the two threads into one, and I am about to do the same with the "Death Zone" threads. One thread per topic in the future please.

Thanks,

Tom

 

[emptymaximum]

ok thanks. yes, all my stuff is beyond halliday resnick and krane.

can you help me with this mentor?

 

[quantumdude]

I haven't read the full 3 pages of this thread, so let me ask you: Have you learned Lagrangian dynamics yet?

 

[emptymaximum]

no lagrangian dynamics for this problem. that's a few chapters later.

 

[quantumdude]

OK, no problem.

Once again, I haven't read the thread, so I don't know what you did.

Given:
$V(x,y,z)=\frac{1}{2}k\left(x^2+4y^2+9z^2\right)$
$\vec{r}(0)=0\hat{i}+0\hat{j}+0\hat{k}$
$\vec{v}(0)=\frac{v_0}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Try to follow these steps.

1.) Calculate the force $\vec{F}$ from the potential $V(x,y,z)$.
2.) Set $\vec{F}$ equal to $m\vec{\ddot{x}}$ (mass times acceleration).
3.) Separate the vector equation into 3 scalar equations.
4.) Solve the scalar equations.

You should have 3 second order differential equations, which will require 6 pieces of initial data to solve completely. Fortunately, you were given that. Also fortunately, all 3 equations are basically the same.

Try to follow those steps. It shouldn't be too difficult. If you get stuck, post what you've done and where you got stuck.

 

[emptymaximum]

F = -kx$\hat{x}$ -4ky $\hat{y}$ -9kz $\hat{z}$
recall k = ╥²m, k/m = w² = ╥²
then the accelerations are:
$\ddot{x}$= -╥²Asin(╥t)
$\ddot{y}$=-4╥²Bsin(2╥t)
$\ddot{z}$=-9╥²Csin(3╥t)
my problem is finding the amplitudes of the motion. I know that A = $\dot{x_i}$ / ╥
the problem is is I'm supposed to find x(t),y(t), and z(t) NUMERICALLY.
i'm mostly confused what this solve numerically means.
this is not a computer problem. no mathematica.
how do i find the amplitudes numerically?

 

[quantumdude]

F = -kx$\hat{x}$ -4ky $\hat{y}$ -9kz $\hat{z}$
recall k = ╥²m, k/m = w² = ╥²
then the accelerations are:
$\ddot{x}$= -╥²Asin(╥t)
$\ddot{y}$=-4╥²Bsin(2╥t)
$\ddot{z}$=-9╥²Csin(3╥t)

How did you get these expressions for the acceleration? They certainly don't come out of the differential equations you were supposed to have set up.

my problem is finding the amplitudes of the motion. I know that A = $\dot{x_i}$ / ╥

Finding the amplitudes is trivial, once you have the general solution. All you have to do is apply the initial values that were given.

the problem is is I'm supposed to find x(t),y(t), and z(t) NUMERICALLY.
i'm mostly confused what this solve numerically means.
this is not a computer problem. no mathematica.
how do i find the amplitudes numerically?

You'll have to ask your teacher what is meant by that. This problem is easily solved by hand. No numerical analysis is required.

 

[emptymaximum]

question directly out of the textbook:
A small lead ball of mass m is suspended by six light springs. The stiffness constants are in the ratio 1:4:9, so that the potential energy function can be expressed as
$V(x,y,z)=\frac{1}{2}k\left(x^2+4y^2+9z^2\right)$
At time t = 0 the ball receives a push in the (1,1,1) direction that imparts to it a speed $v_0$ at the origin. If k = ╥²m, numerically find x, y, and z as functions of time t.

The expressions for acceleration come from the general solution and the IC. The general solution reads:
x = Asin(╥t+φ) + Dcos(╥t+φ)

From the IC at t=0 x=0. For the first term on the RHS to vanish φ = 0.
For the second term on the RHS to vanish D = 0. This leaves:
x = Asin(wt) so
$\dot{x}$ = wAcos(wt) and
$\ddot{x}$ = -w²sin(t)
where w = ╥

for y and z, w = 2╥ and 3╥ respectively.

is this wrong?

from what the question says does that mean numerical analysis?
or since w = ╥ i can get numbers for the amplitudes/initial speeds?

 

[quantumdude]

is this wrong?

Ach, no it's not wrong. You just worked it out a few steps more than I had expected you to.

from what the question says does that mean numerical analysis?
or since w = ╥ i can get numbers for the amplitudes/initial speeds?

You certainly can't get a number for the amplitudes, because you were given the initial velocity in terms of $v_0$. That implies that the symbol $v_0$ must appear in the amplitudes.

Again, I really have no idea of what is meant by a numerical solution in this case. You'll have to ask your teacher about that. You can however find the amplitudes easily by using the initial conditions. It looks like you've already successfully applied the condition $\vec{r}(0)=0\hat{i}+0\hat{j}+0\hat{k}$ to eliminate the coefficients of the cosines. Now use the condition $\vec{v}(0)=v_0(\hat{i}+\hat{j}+\hat{k})$ to find the amplitudes of the sines, and you're done.

 

[emptymaximum]

the amplitudes are gotten by maximizing $\dot{x}$.
this is of course when the cosine term = 1, and this happens at t=0.
at t =0 $\dot{x}$ = $\dot{x_0}$ ; then:

$\dot{x_0}$ = ╥A or A = $\dot{x_0}$/╥

right?

 

[quantumdude]

Right, but you should use the notation that was given.

Actually, there was a typo in my last post so I'll correct it now.

The initial velocity was given as:

$$\vec{v}(0)=\dot{\vec{x}}(0)=\frac{v_0}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$$

I left out the $\sqrt{3}$ last time. Anyway, the components of that vector are equal to $\dot{x}(0)$, $\dot{y}(0)$, and $\dot{z}(0)$, respectively. Use the given symbol ($v_0$), and you will have it.

 

[emptymaximum]

thanks for the help