How Does a Spring Affect the Oscillation Frequency of a Ball in a Medium?

[alimerzairan]

Homework Statement

The terminal speed of a freely falling ball in a medium is v_t, free fall acceleration is g.
In a differrent setting, the same ball, in the same medium, in the same gravity field is supported
by a light elastic spring. In equilibrium, under the weight of the ball, the spring is compressed
by an amount x₀. Find the frequency of oscillation of the system (ball on a spring) in terms of the given quantities. Assume linear law of resistance. Neglect buoyant forces.

Homework Equations

Within attempt solution part.

The Attempt at a Solution

So far, I have said that well in the freely falling ball case we will have: mg = bv_t. So here I can immediately solve what v_t will be.

Now to the case of the spring system, I took into consideration that only two forces act on this and that is the weight of the ball and the restoring force.

m * $$\frac{dx^2}{d^2t}$$ = mg -kx
But at equilibrium all the forces will cancel meaning that:

mg = kx₀.

Is this logic right?

I then get stuck here because I can say that:

$$\frac{g}{x_0}$$ = $$\frac{k}{m}$$ = $$\omega^2$$

so $$\omega$$ = $$\sqrt{\frac{g}{x_0}}$$.

but I can also say that mg = kx₀ = bv_t

Solving this results into:

$$\frac{k x_0}{m}$$ = $$\frac{b v_t}{m}$$

$$\omega$$ = $$\sqrt{\frac{b v_t}{x_0 m}}$$.

but then, would this spring actually be oscillating? because I say that these two forces at equilibrium will be equal and opposite of each other, would this imply or not imply oscillatory motion.

Also is this the correct approach to this problem

Any response is appreciated and if it is possible before 1 p.m. Tuesday, October 26 (USA-Pacific Time Zone GMT -0800).

Thank you.

 

[betel]

Your equations for equilibrium are correct.
The total equation of motion for the ball, medium spring system is missing the friction from the medium. If you include this you will get the frequency in terms of two unknowns m and b which can be eliminated using the equilibrium equations.