Harmonic Motion with Electricity

[nothilaryy]

Homework Statement

A uniform circular ring of charge Q= 4.50 microCoulombs and radius R= 1.30 cm is located in the x-y plane, centered on the origin. A point z is located along the Z axis. If z << R then E is proportional to z. (You should verify this by taking the limit of your expression for E for z << R.) If you place an electron on the z-axis near the origin it experiences a force Fz= -kz, where k is a constant. Obtain a numerical value for k. [I did this and obtained the value 2.946×10-9 N/m which I know is correct] What is the frequency of the small axial oscillations that the electron will undergo if it is released along the z-axis near the origin?

Homework Equations

So far I have looked at:

[tex]
\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}
[/tex]

KE= 1/2mv^2

The Attempt at a Solution

Ok so what haven't I thought about so far?
My first approach was to use Fnet= ma and then substitute dz^2/d^2t for a and try separation of variables and integrate it, but I didn't really know how to make that work for a position dependent force as opposed to a velocity dependent force and I really couldn't get anything useful out of that. Then I thought about trying Potential energy and setting the potential energy at that point to the kinetic energy of the point in the middle of the ring of charge. Then I would have a velocity of that point I couldn't figure out if that would tell me anything, and I don't even think I'm supposed to use potential energy because we haven't even gotten close to learning that chapter yet. I figure I was closer with the Fnet=ma approach because the problem had us find the force and there should be a reason for that, right? So maybe I'm just messing up the calculus.

 

[tiny-tim]

Welcome to PF!

Hi nothilaryy! Welcome to PF!

(have a mu: µ and try using the X² tag just above the Reply box )

… If you place an electron on the z-axis near the origin it experiences a force Fz= -kz, where k is a constant. Obtain a numerical value for k. [I did this and obtained the value 2.946×10-9 N/m which I know is correct] What is the frequency of the small axial oscillations that the electron will undergo if it is released along the z-axis near the origin?
…
My first approach was to use Fnet= ma and then substitute dz^2/d^2t for a and try separation of variables and integrate it, but I didn't really know how to make that work for a position dependent force as opposed to a velocity dependent force and I really couldn't get anything useful out of that …

(I'm sorry nobody replied earlier: I hope you've done it by now, but if not …)

Yes, it's z'' = -(k/m)z, which is a standard simple harmonic motion equation with general solution z = Acosωt + Bsinωt, where ω = … ? (work it out by differentiating twice! )

(if you want to solve it properly, use the trick of putting v = dz/dt, so the chain rule gives you dv/dt = v dv/dz)

 

[tomcue]

Your approach of using Fnet= ma and substituting dz^2/d^2t for a is the correct approach in solving this problem. This is known as the equation of motion for simple harmonic motion, which is applicable in this situation since the force acting on the electron is proportional to its displacement from the origin.

To solve for the frequency of the small axial oscillations, we can use the equation for the period of simple harmonic motion, T= 2π√(m/k), where m is the mass of the electron and k is the spring constant (in this case, the constant of proportionality between the force and displacement).

Since k= 2.946×10^-9 N/m, we can use the mass of an electron (9.11×10^-31 kg) to solve for the period, which is equal to the inverse of the frequency. Thus, the frequency of the small axial oscillations is approximately 6.88×10^12 Hz.

It is also worth mentioning that your approach of using potential energy would also work, as the potential energy at the point where the force is acting is equal to the kinetic energy of the electron at that point. However, since you have not learned about potential energy yet, it is not necessary to use this approach.