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Find motion of a particle subject to a given potential

  • Thread starter Lizwi
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  • #1
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Homework Statement



Fx=-dV/dt where V=-(Voa2)/(a2+x2)

Homework Equations



How to solve this equation?

The Attempt at a Solution


Fx=ma =md2x/dt = -dV/dt =-d-(Voa2)/(a2+x2)/dt. If I differentiate -(Voa2)/(a2+x2) I get more complicated results which Im unable to solve. PLEASE HELP ME cause I studied how to solve second order differential equations but this look dufferent.
 

Answers and Replies

  • #2
Simon Bridge
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So, a is acceleration, F is a force Vo is V-subscript-0 (V0) and you want to solve for something unspecified, is this correct?

Your notation is confusing - do you mean bold-face to be a vector? In which case, how can you have a scalar equal a vector? Shouldn't acceleration also be a vector?

If all these can be taken in the x direction ... ditching the bold-face:[tex]ma = V_0\frac{a^2}{a^2+x^2}[/tex]... perhaps you want to find x as a function of time?

Then why not rearrange so all the powers of "a" are on the left and all the others are on the right, substitute a=d²x/dt² and solve the differential equation?
Don't you get a quadratic in "a"?
 
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  • #3
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Oh no, please ignore bold face. Vo is V0 that's when I get lost after I have substitute a = d2x/dt2 I get md2x/dt2 = d(V0a2/(a2+x2) I don't know how to contnue.
 
  • #4
Simon Bridge
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You are still getting your notation messed up. When you sub in a=d²x/dt² you are supposed to do it for all the "a"'s and you left out the ones on the RHS... your problem is that some are in the denominator.

Start with the version of the equation I showed you (last post).
Try to simplify it before substituting the d/dt's. Hint: eliminate the denominator on the RHS.

[also - if you quote post #2 you'll see how I wrote the equation so nicely: try to use that method when you reply ok?]
 
  • #5
vela
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I think you mean
$$F_x = -\frac{d}{dx} V(x).$$ The derivative is not with respect to time.
 
  • #6
Simon Bridge
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darn ... should have spotted that.
 
  • #7
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Ok, I'll do it this way, $$F_x = -\frac{d}{dx}V(x) = -\frac{d}{dx} \left(-\frac{V_0 a^2}{a^2+x^2}\right)$$ because ##V=-V_0a^2/(a^2+x^2)##.

Because ##F_x=ma## and ##a=d^2x/dt^2##,
$$ m\frac{d^2x}{dt^2} = -\frac{d}{dx}\left(-\frac{V_0 a^2}{a^2+x^2}\right)$$ and then after that I don't know what to do. Please.

Mod note: I fixed your equations for you. Use LaTeX for all of the formatting. Don't mix it with the BBcode tags.
 
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  • #8
Simon Bridge
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What are you trying to solve for?
(You will end up with one of the variables on the RHS to be equal to some function of everything else ... which is it? Previously I just assumed you needed x(t).)

Check your original question and see if that is really intended to be a time derivative or not. Don't just go along with what we say.

What is "a" a function of?

You know how to differentiate a quotient right?
 
  • #9
turin
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Lizwi, please define all of your variables/symbols. It isn't clear to us which are constants, independent variables, dependent variables, etc.. For example, I have no idea why vela suggests the x derivative instead of the time derivative, except that the problem would then look much more standard to us physicists. But then, you aren't even solving a differential equation, it just becomes a simple Calculus 1 problem (and I wouldn't exactly qualify this as mathematical methods for physicists in the advanced physics forum).
 
  • #10
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Hi, the problem is axactly as follows: A point mass m moves in one dimension under the influence of a force F[itex]_{x}[/itex] that has a potential energy V(x). Recall that the relation between these is F[itex]_{x}[/itex]= -[itex]\frac{dv}{dx}[/itex]. Take the specific potential energy V(x) = -[itex]\frac{V_{0}a^{2}}{a^{2}+x^{2}}[/itex] where V[itex]_{0}[/itex] is positive. Sketch V. Write the equation F[itex]_{x}[/itex] ma[itex]_{x}[/itex]. There is an equilibrium point at x=0, and if the motion is over only small distances you can do a power series expansion of F[itex]_{x}[/itex] about x=0. What is the differential equation now? Keep just the lowest order non-vanishing term in the expansion for the force and solve that equation subjet to the initial conditions that at time t=0, x(0)=x[itex]_{0}[/itex] and V[itex]_{x}[/itex](0)=0.


I don't know why should I use power series expansion, I have no Idea. Please!
 
  • #11
Simon Bridge
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There is a lot to get confused about here:
Looks like the "a" in the equation for V(x) has units of distance ... i.e. it is not an acceleration ... is that correct? You have been asked about this several time but for some reason have not replied. Isn't V(x) potential rather than potential energy? Also - I think you have mixed the upper and lower case "v"'s ... I think [itex]v_x(0)=0[/itex] is supposed to be the particle speed at [itex]t=0[/itex].

If "a" is just a constant then:[tex]F_x=V_0 \frac{d}{dx}\frac{a^2}{a^2+x^2}[/tex]... you can do that - it's just the quotient rule.

You are supposed to use a power-series expansion to simplify the equation so it is easier to handle.

You have three stages ...
1. plot V(x)
2. find the relation for [itex]F_x[/itex]
3. put [itex]F_x=ma_x[/itex] and solve for x(t).
... you are expected to evaluate step 3 using a power-series approximation.
 
  • #12
vela
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I don't know why should I use power series expansion, I have no Idea. Please!
Because the problem says to. It's a common method used, so you need to learn how to do it.

Isn't V(x) potential rather than potential energy?
It's potential energy, which is commonly denoted by V.
 
  • #13
Simon Bridge
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It's potential energy, which is commonly denoted by V.
http://en.wikipedia.org/wiki/Potential_energy#Relation_between_potential_energy.2C_potential_and_force
Actually a "V" (as well as [itex]\phi[/itex]) is "often" used for potential, potential energy is commonly denoted U ... for example: gravitational potential, as mentioned in the Wikipedia article. This is not to say one cannot use any letter for anything you like.

But I was not going by the letter - I was going by the relation [itex]F_x = -\frac{d}{dx}V(x)[/itex] which is the relation between potential and force.

Still need to identify what "a" means in the expression for V(x).
 
  • #14
turin
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But I was not going by the letter - I was going by the relation [itex]F_x = -\frac{d}{dx}V(x)[/itex] which is the relation between potential and force.
That is the relationship between potential energy and force, as vela suggests. You are perhaps confusing the relationship between electric field and electric potential. And vela is also correct that V is quite often used for potential energy. Indeed, V is confusingingly sometimes used for (electrostatic) potential. Wikipedia is great, but it is certainly not a definitive nor authoritative resource.
 
  • #15
vela
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That's not correct. The force is given by the negative gradient of the potential energy, not the potential energy per unit mass.

http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html

The page you linked to says the force field is given by the negative gradient of the potential. For instance, the electric field is the negative gradient of the electric potential, but the actual force is obtained by multiplying E by the charge.
 
  • #16
Simon Bridge
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Oh that's interesting ... in the link in post #15, force is given by [itex]F=-\frac{d}{dx}U[/itex] with U for potential energy - not V.

But I'll concede the relationship.
Vela has correctly identified my confusion - thanks.
So my question amounts to asking if the F stands for the actual force or the the field ... since V is usually potential and U energy, and OP has continually made notational errors, I'd still maintain this is unclear... (hoping for the former from the reference to F=ma).

Anyone got any objection to the enumerated steps?
Time to hear from OP.
 
  • #17
turin
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I agree with your point, Simon. Without clearly and consistently identifying the letters F, V, etc., we're all just guessing ...
 

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