How Can I Analyze Particle Movement Given Potential Energy and Force Function?

[RubroCP]

Homework Statement

The graph in the figure below shows the potential energy of a particle expressed by the equation $U(x)=3x-4x^2-0.45x^3+0.6x^4$, in SI units. If the energy of the particle be $E=0\,\mathrm{J}$, what are the possible motions? If the particle energy be $E=-5\,\mathrm{J}$, what are the return points?

Relevant Equations

$F(x)=-\mathrm{d}V(x)/\mathrm{d}x$,
$E=U+T$, being $T$ the kinetic energy.

Knowing that $F(x)=-\mathrm{d}V(x)/\mathrm{d}x$, I found that $F(x)=-2.4x^3+1.35x^2+8x-3$. But it was the only thing I could find. How can I analyze what will be the type of movement with the information presented by the question statement?

 

[kuruman]

What are the available positions of the particle when its energy is -5 J?

 

[RubroCP]

What are the available positions of the particle when its energy is -5 J?

Okay, but this $E(x)$ is not the potential energy, it's the total energy, which is obtained by adding the potential and the kinetics. That's what I can't understand.

 

[kuruman]

The plot shows the potential energy as a function of $x$. It is indeed $U(x)$. The total energy is constant and equal to $-5~\text{J}$ no matter where the particle is because mechanical energy is conserved. The total energy cannot be a function of $x$.

 

[RubroCP]

The plot shows the potential energy as a function of $x$. It is indeed $U(x)$. The total energy is constant and equal to $-5~\text{J}$ no matter where the particle is because mechanical energy is conserved. The total energy cannot be a function of $x$.

Thank you very much, now I understand.

 

[kuruman]

To complete the picture, if you solve the equation $U(x)=-5~\text{J}$, the roots will give the $x$-values at which the potential energy is equal to the total energy. At these points, the kinetic energy is equal to zero. They are called the "turning" points because the particle stops instantaneously and goes back to where it came from.

 

[Orodruin]

Note to OP: You should also use these ideas to solve the first part of the problem where energy is zero.

 

[Delta2]

You can do the question that involves total energy $E=0J$ with two ways:

1. As you suggested by finding first the force and then solving the differential equation $F(x)=m\frac{dv}{dt}$ for solution expressed as $v(x)$. you might need to do one trick here to set $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$. You will also need the initial condition v(0)=0 (it follows from E=0) in order to solve this ODE fully
2. from U+T=E=0 you can solve for T and then solve (easily) for v(x).

If all goes well you should get the same $v(x)$ from both ways.

 

[Orodruin]

You can do the question that involves total energy $E=0J$ with two ways:

1. As you suggested by finding first the force and then solving the differential equation $F(x)=m\frac{dv}{dt}$ for solution expressed as $v(x)$. you might need to do one trick here to set $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$. You will also need the initial condition v(0)=0 (it follows from E=0) in order to solve this ODE fully
2. from U+T=E=0 you can solve for T and then solve (easily) for v(x).

If all goes well you should get the same $v(x)$ from both ways.

You do not need to solve for v(x) to give a basic description of the possible motions. Also, there is no requirement that v(0) = 0 is even part of the solution.

 

[Delta2]

You do not need to solve for v(x) to give a basic description of the possible motions. Also, there is no requirement that v(0) = 0 is even part of the solution.

I don't understand how we give a basic description of the possible motions if not to solve for v(x).
Also from U(0)+T(0)=0 and U(0)=0 doesn't follow that T(0)=0 i.e. v(0)=0?

 

[Delta2]

I guess from the graph of U(x) we can determine where it accelerates or decelerates and in what range of x values the particle is constricted. Is that it?

 

[hilbert2]

That function $V(x)$ describes a double potential well with a barrier in between. The barrier is higher than 0 Joules, which is the maximal value of potential energy the particle can have at any part of its trajectory. So the particle will remain either in the left of right part of that double well and doesn't cross the barrier. In addition to this feature, what would you expect the motion to look like?

 

[haruspex]

I guess from the graph of U(x) we can determine where it accelerates or decelerates and in what range of x values the particle is constricted. Is that it?

If the total energy is E then PE is at most E and v(x) =0 corresponds to PE(x)=E.
However, I'm not sure what is expected for "possible motions". If a ball bounces between floor and ceiling it may never reach v(x)=0.

 

[Orodruin]

I don't understand how we give a basic description of the possible motions if not to solve for v(x).
Also from U(0)+T(0)=0 and U(0)=0 doesn't follow that T(0)=0 i.e. v(0)=0?

A basic description would be to say: The particle moves back and forth between points A and B.

The point is that x = 0 does not need to be part of the motion and so v(0) as stated may be undefined. For example, if $0 < x_A < x_B$, then the particle will never pass $x = 0$ and $v(0)$ has no meaning.

 

[Delta2]

Well ok, I can see now, the particle with total energy 0 can be "trapped" either between $x_A=-2.6$ and $x_B=0$ or between $x_A=0.8$ and $x_B=2.6$. In the second case $T(0)$ (or $v(0)$) has no meaning indeed.

For some reason I focused only on the left part of the graph .