One Dimensional motion of particle in a potential field

[sabanboy]

A classical particle constrained to move in one dimension (x) is in the potential field V(x) = V0(x – a)(x –b)/(x – c)^2, 0 < a < b < c < ∞.
a. Make a sketch of V
b. Discuss the possible motions, forbidden domains, and turning points. Specifically, if the
particle is known to be at x → ∞ with E = 3V0(b – 4a + 3c)/(c – b), at which value of x
does it reflect?

I'm not sure how to approach this problem any tips or advice would be greatly appreciated.

 

[BvU]

Hello Saban, and welcome to PF.
Here's my tip:
Start with a)

 

[sabanboy]

Thanks I have it drawn and reviewed it with my professor who told me the sketch is correct. I'm still not sure as to how to complete part b though.

 

[BvU]

I don't have your sketch, but I can kind of telepatically pick it up
So you see a big peak at x=c. And a minimum between a and b. Right ?

Good that you already did the first discussion in part b) and now want to complete it.

In the completion of part b the particle comes from the right, with a given E.
Is it clear to you that it doesn't have enough energy to get to x < c ?

So what you know about the turning point is that at that point there is no more kinetic energy, just potential energy.
Since the given E(x≈∞) = E_kin(x≈∞) + V(x≈∞) is the same as E at the turning point x, you obtain an equation in x.

 

[HalfLostAndSearching]

I would approach this problem by first understanding the potential field V(x) and its behavior. From the given equation, we can see that V(x) has three distinct terms: V0, (x-a)(x-b), and (x-c)^2. The term V0 represents a constant potential, while the other two terms involve the position of the particle (x) and the constants a, b, and c.

a. To make a sketch of V, we can plot the potential as a function of x. We know that the potential is zero when x = a, b, or c, and it increases as x moves away from these points. We also know that the potential is at a maximum when x is halfway between a and b, and at a minimum when x is halfway between b and c. Therefore, our sketch of V(x) would look like a "W" shape, with the minimum points at x = (a+b)/2 and x = (b+c)/2, and the maximum point at x = (a+c)/2.

b. Now, let's discuss the possible motions of a particle in this potential field. The particle is constrained to move in one dimension, so it can only move along the x-axis. The particle will experience a force given by F(x) = -dV(x)/dx, where dV(x)/dx is the derivative of the potential with respect to x. This force will cause the particle to move towards points of lower potential, and away from points of higher potential.

The forbidden domains in this potential field would be the regions where the potential is infinite, which would occur at x = a, b, or c. This means that the particle cannot exist at these points, and if it were to reach these points, it would bounce back due to the infinite force.

The turning points in this potential field would be the points where the potential changes from increasing to decreasing, or vice versa. These points would be the minimum and maximum points we identified in our sketch of V(x).

Now, let's consider the specific scenario given in the question. The particle is known to be at x → ∞ with E = 3V0(b – 4a + 3c)/(c – b). At this point, the particle has a certain energy, E, which is related to its velocity and position through the equation E = (1/2)