Finding a spring constant of a spring holding a mass

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To find the spring constant k for a mass attached to a spring on a vertical hoop, the spring force can be expressed as F = -kx + 2Rk, where x is the displacement. The potential energy of the spring is given by V(x) = 0.5kx² - 2Rkx + 2R²k, and the total energy equation combines kinetic and potential energy. The initial conditions are x(0) = 2R and v(0) = 0, which helps establish the energy balance. A simpler method to determine k involves applying the harmonic oscillation equation for regular motion of the mass.
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Homework Statement


I have a vertical hoop of radius R. A spring with spring constant k and relaxed length 0 is attached to the top of the hoop. A block of mass m is attached to the spring and dropped. Assuming the motion is a linear vertical oscillation between the top and bottom of the hoop, what is k?

Homework Equations


Spring force = -kx

Potential = -Integral(force) from a reference point to x

E = KE + PE

The Attempt at a Solution


Set x = 0 to the bottom of the hoop and let x positive point up.
This gives me F = -kx + 2Rk for the force exerted by the spring on my mass.
This gives me a spring potential:

V(x) = .5kx2 - 2Rkx + 2R2k

Gives me energy:

E = .5mv2 + V(x) + mgx

I know that x(0) = 2R and v(0) = 0, so I have an initial energy in terms of known variables but I can't think of anything to compare it with. I've written out the equation of motion for the mass to try to find some other functional x(t), v(t) values, but I'd like to know if there's some simpler method of solution that I'm just missing.

Looking for a value for k.
 
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Hi there,

If the block is going up and down in a regular motion, then you can define the k constant through the harmonic oscillation equation.

Cheers
 

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