This is a doozy. First off Merideth, you are right that the period (and frequency) remain constant as a damped pendulum system swings. But that is not saying that it is equal to the period (or frequency) of the same pendulum experiencing a different magnitude of dampening, which is what the original question is asking. I'll weigh in as best I can, but have to be up front that the PhD guy who wrote this is a heck of a lot smarter than I am, and there's an element of blind faith I put in the passages he wrote.
I think one of the best ways to address questions like this is to consider an extreme case I can visualize. Let's say you have a pendulum that you lift to an inital angle of 30 degrees (either side) and then release. In air, it would swing quickly and reach an angle really close to 30 degrees on the opposite side and then return quickly to an angle very close to 30 degrees. If you were in water, then it would go notably slower and not reach a value that close to 30 degrees and then swing back slowly to an angle even further reduced from 30 degrees. It will definitely move more slowly in water, in part because of drag and in part because of the bouyant force offsetting the effect of gravity to some extent. But it will travel less distance.
The question is asking you to make a decision about the time it takes for a complete cycle in those two scenarios. I really like the reasoning so far in terms of thinking at the MCAT level. A complex equation is not their target, although it supports an increased period (as mentioned in a previous post). It comes down to the dilemma of whether reduced speed or reduced distance players a bigger role.
If you take the simple period equation where period (T) is inversely proportional to g, then the impact of the buoyant force and drag will reduce the acceleration and lead us to a situation where g is essentially less (which is the more difficult equation referred to in previous posts). Let's simplify the thought to say that g is reduced and refer to the ideal period equation. If g goes down, then T goes up, but by a smaller percentage than g changes. T should increase slightly as the viscosity and density of the medium increase. I would probably stop at that point figuring it fit the force equation I had memorized.
You can also consider it from an energetics perspective, but that one wasn't my first inclination. In water, it is going to lose a certain amount of energy to work done by resistive forces. That will manifest itself as less speed at the bottom and less height at the top during ech subsequent cycle. Although those two points are not directly comparable given that as it goes from its lowest (fastest) point to its highest point, it is losing energy, it's still valid that somehow we are looking at h versus v^2. If we consider the total energy of the system, it will be PEinitial = PEt + KEt + Wdrag t. If h drops, then d traveled drops by a proportional amount. If time were to remain constant (no change in period), then v would have dropped by the exact same percentage as h, which means that deltah would not offset deltav^2 in the energetics equation. So the time must change, but by a different factor than h and therefore a different factor than distance. I'm assuming the math works out in the same as it did when looked at forces, because of the square root term, but I'm too lazy to work it out.
I hope this helps, but I'm not sure I did an adequate job explaining my perspective.
And milski, while I can't say whether or not they care what you think, I care. 😳 🙂 You post a great deal of helpful information at this site!
