3-d non- isotropic oscillator

[emptymaximum]

ok. 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. find x(t) y(t) z(t) numerically if k = m(pi^2). part b: will it every get back to origin, if so what t does it return with v = vo.
what does it mean solve numerically? t is in a cosine.
i don't get it.

what i did is n my post #2.

 

[Tide]

Your vector expression for the acceleration is not correct.

 

[emptymaximum]

how?
if F=-grad(V) then F=-k(xx+4yy+9zz)
ma=-k(xx+4yy+9zz)

so if k = m╥² then:

a = -╥²(xx+4yy+9zz) [eqs 1.(1,2,3)]]

you can also go like:

a+(k/m)(xx+4yy+9zz) = 0 [eqs 2]

x = Acos(╥t) ; y = Bcos(2╥t) ; z = Ccos(3╥t) ;where ╥ = √(k/m) [eqs 3.(1,2,3)]

let xDoubleDot be d²x/dt² then

xDD = -╥²Acos(╥t) ; yDD = -4╥²Bcos(2╥t) ; zDD = -9╥²Ccos(3╥t) [eqs 4.(1,2,3)]

combining [eqs 1.1] and [eqs 4.1] gives:

-╥²x = -╥²Acos(╥t) ; which reduces to:

x = Acos(╥t) ; which is just [eqs 3.1], the x solution to the homogenous DE described by [eqs 2].

so clearly my expression for the acceleration isn't wrong.

 

[Cyrus]

Just to point out, I would not recomdend using x,y,z in bold, but rather i,j,k instead.

 

[emptymaximum]

oh. in the first one it should be (x+4y+9z)r
guess i did write it down wrong.HOW DO I FIND THE AMPLITUDES?
i know they come from IC. it isn't going to go through (1,1,1) right? it just gets pushed that way? it says that to show that vo has no cos sin buisiness and is nonzero and of equall magnitude in all directions?

 

[emptymaximum]

i don't much care for i j k. never used it yet, not about to start now.
if it's so i don't confuse myself, I'm not worried. i actually get less confused like this.
if there's some other kind of reason, please tell me.

 

[emptymaximum]

3-d non-isotropic oscillator

ok. 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. find x(t) y(t) z(t) numerically if k = m╥². part b: will it every get back to origin, if so what t does it return with v = vo.
what does it mean solve numerically? t is in a cosine.
i don't get it.
F = -kxx -4kyy -9kzz = ma
a + (k/m)(x+4y+9z)r = 0
ok so then is it like:
a+╥²(x+4y+9z)r = 0
take this and separate and integrate? so there's no trig functions?
how do i find the amplitudes of the trig functions? I know they come from IC. it isn't going to go through (1,1,1) right? it just gets pushed that way? it says that to show that vo has no cos sin buisiness and is nonzero and of equal magnitude in all directions?
this thing is driving bonkers and it shouldn't be.

 

[Cyrus]

well, its so that when taking derivatives you don't mistakenly take the derivative of a unit orthognal pointing vector. As for your acceleration term, I get:

$$[ai+aj+ak] + k/m [xi+4yj+9zk] =0$$

your sign for k/m is wrong, its a + not a -

 

[Tide]

First, $(x - 4y - 9z)\vec r$ is NOT the same as $x \hat i - 4y \hat j - 9z \hat k$.

Second, the vector x is understood to mean $x \hat i$ so when you write x x people will interpret that as $x^2 \hat i$. As an alternative you could write the unit vector as $\hat x$ where the circumflex or hat tells the reader that you're talking about a unit vector.

With regard to your problem, you determine the amplitudes by applying the initial conditions (you know both the initial position and initial velocity).

So, can you determine whether the mass will return to its starting postion? :)

 

[emptymaximum]

it's right in my second post.
and that's what I've been working wiht.

 

[Tide]

It was wrong in the first and I explained why.

 

[emptymaximum]

r is the unit vector i guess how you guys write it is (i+j+k).
see that's why i use x y z. i know that it isn't the same as times i just wrote it short like. then the x gets timesd on the x and the like...
how do you make it write math style?
i can do circonflex but only on like ê â and î. i can do dot notation on those, maybe i should use hthose. does this thing recognise mathematica fonts?

 

[Cyrus]

r is not (i+j+k)

r is <i,j,k>/ |i,j,k|

(for the unit pointing vector)

 

[Tide]

No, r is not the unit vector. It's the position vector given as $x \hat i + y \hat j + z \hat k$. If you want to specify a unit vector you can use $\hat r$. I recommend adopting the conventions used by most people since otherwise both you and they will be confused.

To use the math symbols you'll need to learn a little LaTex which you can find out about by clicking on any equation someone has posted. A popup will appear with a link leading you to some reference material.

 

[emptymaximum]

so:

.5mv²+.5kr² = .5mvo²

v²+╥²r² = vo²

(xD)²+╥²x² = (xDo)²

how do i solve that for x(t) numerically?
the re-tracing part isn't bothering me.
it's the solving for x(t) numerically. how?

 

[emptymaximum]

if i would have known how to put a hat on it i would have.
sorry.
and
r hat = (i hat + j hat + k hat)
is what i suppose i should have wrote.

 

[Tide]

The potential energy is not $\frac {1}{2}k r^2$.

You were on the right track in #3.

 

[emptymaximum]

sorry. i wrote it wrong.
potential energy = $\frac {1}{2}k (x^2+4y^2+9z^2)$

that still doesn't really tell me anything about:
what does solve numerically mean?

 

[Tide]

WTG! You figured out the LaTeX!

I am not sure what the authors of your problem mean by "solve numerically" but it usually means that you use approximation methods to find the solutions. However, in this case you don't need to resort to numerical methods since you can solve the equations directly and exactly. Does your textbook offer any guidance?

 

[emptymaximum]

from the post #5:
xDo = yDo = zDo

so at t = 0

xDo = -╥A yDo = -2╥B zDo = -3╥C

all = each other then A =2B =3C ?

and i get the numerical value of the amplitude by...

 

[emptymaximum]

no guidance in the textbook.
word on the street is it's a pretty popular book. Fowles and Cassiday, Analytical Mechanics.

as for the latex, i couldn't find where it says how to do dot notation. i'll look for that in other posts some other time.

 

[abszero]

It means you're going to need to bust ou MATLAB to bust out "solving numerically". Also, what you have written down for the second line of your equations is incorrect. You need to break Newton's laws down into its components. If you know Lagrange's equations I recommend you go through that. It might make things a little clearer.

 

[Physics Monkey]

Note that the motion is completely separable so you can solve each direction independently. Are you sure that "solve numerically" means you have to use a computer? This problem is fairly easy and no computer is actually required.

 

[emptymaximum]

still need help please

 

[emptymaximum]

no compy required. there are computer problems also but they are listed seperately. no mathematica here.
i know what i wrote up there is wrong, it should also be + not -.

and if this problem is so fairly easy why is it driving me bonkers. I'm over 10 hours now.
that isn't right.
if someone could offfer up a strategy or something that would be super.

x = Asin╥t y = Bsin(2╥t) z = sin(3╥t)

get the speeds and accelerations from that is pie.

how i number solve A,B,C is the problem. i know original speed/ ╥ = A
how do i number solve original speed then?
this thing doesn't actually go through (1,1,1) does it? it just gets pushed that way?
/s

 

[Tide]

I think you were on the right track with post #3.

 

[emptymaximum]

ok. where was that track going? some kind of hint.
please?
i'm into hour 12

 

[Cyrus]

Im trying to follow you along here without knowing the material; and it seems that you stated:

"x = Acos(╥t) ; y = Bcos(2╥t) ; z = Ccos(3╥t) "

However, this makes no sense to me, as at t=0, (x,y,z)=(0,0,0) and your coefficients A,B and C will all subseqently be zero. Are you sure your equation is correct?

 

[emptymaximum]

:grumpy:

the coefficients are the amplitudes of the motion.

when x=A,y=B, or z=C, the particle has reached a turning point in the motion.

xdot = 0 ydot = 0 zdot = 0

oh. i know what i did. i think i wrote the cosine terms for position. the position should be sine, the velocity is described by cosine, and the acceleration is -sine. reason:

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.
dx/dt and then you'll see that at t=0 xDotInitial = ╥A1
re-arranging the terms gives A1 = xDotInitial/╥
now if i knew xDotInitial, i could numerically solve this, but i don't so i don't know what the hell I'm supposed to do.

my whole problem with this isn't so much conceptual.

my problem is HOW DO I NUMERICALLY SOLVE.
i need vInitial or the amplitudes.
HOW?
i have multiple expressions for each, but they all involve the other.

 

[Cyrus]

Was I wrong in pointing that out?

 

[Cyrus]

oh, and by the way,

the velocity is given as:

[ (v_0 / sqrt(3))i + (v_0 / sqrt(3)) j + (v_0 / sqrt(3)) k ]

if that amounts to anything...

 

[emptymaximum]

you weren't wrong pointing that out.

how do you get that for velocity?

 

[Cyrus]

make a unit pointing vector in the direction of (1,1,1), so its (1,1,1)/ |1,1,1| = <sqrt(3),sqrt(3),sqrt(3)> now you just multiply that by the magnitude of v.

I may not know the material, but I can still TRY to help...

 

[emptymaximum]

oh. right. the old rHAT = xHAT + yHAT + zHAT
that's useless here because:
i don't know what vInitial is.
sucks.

and that's the problem I'm having.

 

[Cyrus]

you can't leave it as v_0 and have it factor itself out after doing some later computations?