Harmonic motion through a ring of charge

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SUMMARY

The discussion centers on calculating the angular frequency of a particle oscillating about the center of a charged ring. The ring has a radius of 18 cm and carries a uniform positive charge of 6x10^-6 C. The particle, with a charge of -6 µC, oscillates with an initial angular frequency of 29 rad/s. The solution involves recognizing that the force acting on the particle is proportional to the displacement divided by the cube of the distance, leading to a modified approach for calculating the new angular frequency when the radius is doubled while maintaining constant linear charge density.

PREREQUISITES
  • Understanding of harmonic motion and angular frequency
  • Familiarity with electric fields generated by charged rings
  • Knowledge of force equations in simple harmonic motion (SHM)
  • Ability to manipulate equations involving charge density and displacement
NEXT STEPS
  • Explore the derivation of electric fields from charged ring configurations
  • Study the relationship between force and displacement in SHM
  • Investigate the effects of varying radius on oscillatory motion in charged systems
  • Learn about the implications of linear charge density in electrostatics
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Students and educators in physics, particularly those focusing on electromagnetism and oscillatory motion, as well as anyone solving problems related to charged particles in electric fields.

king vitamin
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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 [tex]\mu[/tex]C 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 [tex]\lambda[/tex] at point x above the radius is:

k[tex]\lambda[/tex](2*pi*R)x[tex]/[/tex](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 = [tex]\omega[/tex]^2*x

The Attempt at a Solution



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

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

F = ma = m[tex]\omega[/tex]^2*x

Combining these equations:

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