Potential well, harmonic oscillator

In summary, we need to find the potential function V(x) in terms of Vo and a for a parabolic harmonic oscillator potential well where a particle is trapped and oscillating between x=b and x=-b, with potential jumps at x=a and x=-a. By using the equation E=K+V, where E is the total energy, K is the kinetic energy, and V is the potential energy, and considering the potential jumps, we can conclude that V(x) is a constant -Vo/2 for most values of x, except at x=a and x=-a where it jumps from -Vo/2 to 0 and vice versa.
  • #1
skate_nerd
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Homework Statement


Parabolic harmonic oscillator potential well. A particle is trapped in the well, oscillating classically back and forth between x=b and x=-b. The potential jumps from Vo to zero at x=a and x=-a. The particle's energy is Vo/2. I need to find the potential function V(x) in terms of Vo and a.

Homework Equations


E=K+V=.5mv2+.5Cx2=Vo/2

The Attempt at a Solution


Honestly I am really stumped. I tried several approaches, and I think I might just be making this way harder than it should be. I just imagine when dealing with a classical oscillating in a parabolic potential well, you should just use the equation i put in the relevant equations section, and then solve for V(x). However that isn't a function of x, its a function of velocity. Also it is lacking the required dependence on a. I have no idea how a could come into the equation for V(x). I tried thinking of a bunch of conditions to apply it but nothing made sense. Any guidance would be awesome
 
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  • #2
.Hello!

To find the potential function V(x), we can use the equation you mentioned, E=K+V, where E is the total energy of the particle, K is the kinetic energy, and V is the potential energy. We know that the total energy is equal to Vo/2, so we can substitute that in:

Vo/2 = K + V

We also know that the kinetic energy can be expressed as 1/2 * m * v^2, where m is the mass of the particle and v is the velocity. In this case, the velocity is changing between -√(2Vo/m) and √(2Vo/m) as the particle oscillates between x=b and x=-b.

So, we can rewrite the equation as:

Vo/2 = 1/2 * m * v^2 + V

Substituting in the values for m and v, we get:

Vo/2 = 1/2 * m * (√(2Vo/m))^2 + V

Simplifying, we get:

Vo/2 = Vo + V

Solving for V, we get:

V = -Vo/2

This means that the potential function V(x) is a constant -Vo/2 for all values of x. This makes sense, since the potential well is parabolic and the potential energy does not change as the particle oscillates between x=b and x=-b.

However, we also have to consider the potential jumps at x=a and x=-a. Since the particle's energy is Vo/2, it will only reach the potential jump at x=a if it is moving with a velocity of √(Vo/m). So, we can say that at x=a, the potential jumps from -Vo/2 to 0.

Similarly, at x=-a, the particle will reach the potential jump if it is moving with a velocity of -√(Vo/m). So, at x=-a, the potential jumps from 0 to -Vo/2.

Therefore, the potential function V(x) can be written as:

V(x) =
- Vo/2 for -b<x<-a
0 for x=-a
-Vo/2 for -a<x<a
0 for x=a
- Vo/2 for a<x<b

I hope this helps! Let me know if you have any
 

1. What is a potential well?

A potential well is a region in space where a particle experiences a force that pulls it towards the center, creating a stable equilibrium position. This force is known as the potential energy and it can be caused by various factors such as electric or gravitational fields.

2. What is a harmonic oscillator?

A harmonic oscillator is a system that follows a particular type of motion known as simple harmonic motion. This means that the system oscillates back and forth around a stable equilibrium position with a constant frequency and amplitude.

3. How are potential wells and harmonic oscillators related?

A potential well can be seen as a basic model for a harmonic oscillator. The potential energy in a potential well can be represented by a parabolic function, which is the same form of the restoring force in a harmonic oscillator. Therefore, a particle in a potential well can exhibit simple harmonic motion.

4. What are the applications of potential wells and harmonic oscillators?

Potential wells and harmonic oscillators have many applications in physics and engineering. They are used to model various systems such as pendulums, springs, and atoms. They also play a significant role in understanding the behavior of molecules, quantum systems, and electronic circuits.

5. How can potential wells and harmonic oscillators be described mathematically?

Both potential wells and harmonic oscillators can be described mathematically using differential equations. The equation of motion for a harmonic oscillator is a second-order differential equation, while the potential energy in a potential well can be described by a quartic function. These equations can be solved to obtain the position, velocity, and acceleration of the system at any given time.

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