Is this oscillator driven?

Question

A mass $m$ is attached to a vertical massless spring or a spring constant $k$. Originally, the spring was relaxed because the mass was held by a clip. Suddenly the clip was released. THe mass dropped down and the maximal elongation of the spring was recorded as $l$. What is the resonant frequency of the system? The gravitational constant is $g = 9.8 \frac{m}{s}$.

I assume (based on instructions from the professor) the intention of the question is that there is some damping going on here (damping constant $b$). What I'm confused about is if the gravitational force here makes this a driven damped oscillator or not. The "driving" force is a constant, so it isn't changing the oscillations at all (I think?). Meaning, is my equation (for $\omega = \sqrt{\frac{k}{m}}$ and $\beta = \frac{b}{2m}$):

$$ \ddot{x} + 2\beta\dot{x} +\omega^2x = 0 $$

or

$$ \ddot{x} + 2\beta\dot{x} +\omega^2x = g $$

If it is driven, is my resonant frequency $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$? What else would resonant frequency mean in this case?

Answer 1

Your equation is Newton's second law. That's always true for any system in Newtonian mechanics. So it's very straightforward to figure out which equation of motion applies to this system: write out $\sum F = ma$, plug in the forces, and simplify.

Now to address the other issue: does a constant force qualify as a driving force? I'd say there are two ways to think about this:

• Intuitively, a "driving force" is meant to be some kind of force that will sustain oscillation even without the natural response of the oscillator. Imagine what this system would do without a spring. Gravity wouldn't make it oscillate; it would just fall, and that's not really what most people would consider "driving." Strictly speaking, yes it is the limit of an oscillatory driving force as the frequency goes to zero, but in this case, some of the properties that characterize a driven oscillator, as we normally think of it, don't carry over to zero driving frequency.
• Mathematically, any simple harmonic oscillator (driven or not, damped or not) has an equilibrium position, and it's conventional to choose a coordinate $q$ such that the equilibrium position is at $q = 0$. With that in mind, consider the coordinate transformation $q = x - \frac{g}{\omega^2}$. I'll let you work out the implications of that. :-) (Fun fact: this is mathematically equivalent to the Higgs mechanism.)

The point to take away from this, either way, is that no, a constant force is not a driving force, but that's really a matter of terminology, specifically what people usually understand "driving force" to mean.