Harmonic motion through a ring of charge

AI Thread Summary
A ring of radius 18 cm with a uniform positive charge of 6x10^-6 C creates an electric field affecting a negatively charged particle oscillating about its center with an angular frequency of 29 rad/s. The challenge is to determine the new angular frequency when the ring's radius is doubled while maintaining the same linear charge density. The participant initially struggled with the relationship between force and displacement in harmonic motion but realized that the force must be proportional to displacement for small oscillations. By graphically analyzing the force equation and recognizing the need for a linear approximation near the center, they derived the new frequency after adjusting the parameters. The discussion highlights the importance of understanding the underlying physics of harmonic motion in varying conditions.
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Homework Statement



A ring of radius 18 cm that lies in the yz plane carries positive charge 6x10^-6 C uniformly distributed over its length. A particle of mass m that carries a charge of -6 \muC oscillates about the center of the ring with an angular frequency of 29 rad/s.

Find the angular frequency of oscillation of the mass if the radius of the ring is doubled while keeping the linear charge density on the ring constant. Answer in rad/sec.

(I was confused about the phrase "oscillates about the center" so I asked my professor, and he clarified that the charge is moving back and forth along the positive/negative x-axis through the center of the ring)

Homework Equations



The electric field due to a ring with linear charge density \lambda at point x above the radius is:

k\lambda(2*pi*R)x/(R^2 + x^2)^(3/2)

I'm expressing the equation in terms of charge density rather than Q because the problem conserves charge density rather than Q. The charge density is easily obtained: Q/(2*pi*R) = 5.306x10^-6 C/m.

Acceleration in harmonic motion can be described:

a = \omega^2*x

The Attempt at a Solution



(i'm just going to work with magnitudes here to avoid negatives)

F = qE = kq\lambda(2*pi*R)x/(R^2 + x^2)^(3/2)

F = ma = m\omega^2*x

Combining these equations:

kq\lambda(2*pi*R)/(R^2 + x^2)^(3/2) = m\omega^2

Now I'm stuck - I don't know the mass of the particle, and I have no idea how to model the displacement (which varies over time, and I don't know the amplitude). It seems as though with the given charge and frequency, one could come up with variable masses m and displacement x which fits the equation, and the variables aren't related in such a way that I could just double R and obtain a solvable system of equations. Is there another approach to this problem I'm not seeing?
 
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I'm going to bump this up hoping that someone can still help me - just a hint towards the right equation or idea would help tons. I've been trying to relate angular frequency to x - I figured that since the particle travels 4x for every period, it travels distance x every .054 seconds. Still don't know what to do with that... maybe the SHM equation is just a dead end and I should be using something else?
 
Solved! I was so close to giving up but I finally got the problem. For SHM (constant angular frequency) the force must be directly proportional to x a la F = -kx. However, in my problem the force must be proportional to x/(R^2+x^2)^(3/2), which is an entirely different beast - or so I thought.

The breakthrough happened when I just looked at the force equation F(x) = kqQx/(r^2+x^2)^(3/2) graphically - it was curvilinear, but it's slope as it crosses the x-axis (at x=0) was approximately constant! In the stated problem the displacement x must obviously revolve about x=0, and if it is truly moving with constant angular frequency then F(x) must be "linear" - even though the problem doesn't mention that x must be small, it was implicit! Obviously, the spring constant is just F'(0), and using [omega]=sqrt(k/m) it's trivial to simply solve for mass, then double the radius and charge, and then use the new spring constant for the new frequency.

Sorry if I'm just wasting people's time, but I'm sure everyone else knows how good the feeling is when you finally break a problem that's been killing you for a week.
 
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