pendulum and air resistance

[Meredith92]

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How does air resistance affect the period of a pendulum?

TBR says " In any environment where the pendulum swings in a medium, there exists drag. the pendulum bob would be slowed by the medium as it traveled its pathway, increasing the time required to complete a cycle. this means the period would be larger."

I may be over-thinking this, but if there is air resistance wouldnt that make the distance traveled less? So the period might not even change since although its "slower" it doesnt have to travel as far to complete one cycle.

This problem also seems to contradict another example in TBR when the length was doubled- they said "longer pendula have a greater distance to traverse per cycle so it takes longer to complete that cycle" so shouldnt we apply the same logic here? Air resistance means shorter distance to traverse so it takes LESS time to complete the cycle?

 

[milski]

Once you have air resistance, the energy of the pendulum will start dissipating with each swing - each swing will have smaller and smaller amplitude. There will also be a very minimal increase to the period, compared to the period of the same pendulum in vacuum. Consider a swing from the equilibrium point all the way out and back to it. In vacuum, it will be completely symmetric - the speeds at each point on the swing out and swing back will be the same. Essentially, you have some amount, ΔE being changed from KE to PE and back to KE. Now let's think about the case with air resistance. Let's say that again you've converted ΔE to potential energy on the upswing. The final speed for that much of a swing corresponding to the "vacuum period" of this pendulum is v=sqrt(ΔE/2m). Since some of that energy will be lost to air resistance, the final speed will be only sqrt((ΔE-&#949😉/2m) and the pendulum will need just a bit more time to get to the equilibrium position. You can apply similar logic to the upswing as well. The final result is that you have an increase in the period and continuously decreasing amplitude.

Actually, disregard that. You cannot apply the same logic for the upswing, the time will be shorter than the vacuum period for that, leaving you with having to decide which has larger magnitude - the decrease on the upswing or the increase on the downswing. Only way out is the solve the equation for that motion and that's pretty hard. At that point I'm willing to take the TBR's statement as a fact. :-/

 

[slz1900]

I think tbr is wrong here, or at least it would be situation dependent. Determining whether decreased amplitude or decreased velocity of the bob affects period more would be very complicated. For our purposes, the equation for a simple pendulum's period is 2 pi sqrt (L/g). Velocity isn't in there, and length and gravitational force don't change with air resistance. There are also way more complicated formulas for periodic motion that might give you a different answer (ones that don't assume the pendulum is in a vacuum like the one I gave does), but for mcat purposes, I think that you should recognize that nothing affected by air resistance is included in your equation for period, so it is unchanged.

 

[Meredith92]

Ok because none of the answers had that as an option :/ I feel weird saying tbr is wrong ... But it makes
More sense that air resistance doesn't change the period

 

[milski]

Ok because none of the answers had that as an option :/ I feel weird saying tbr is wrong ... But it makes
More sense that air resistance doesn't change the period

It sure does. It's not part of 2π.sqrt(L/g) because it explicitly is not taken into account in the derivation of the formula. The formula with the air resistance will be much more complicated but they sort of expect you to be able to make qualitative statements for that case.

In a similar way h=gt^/2 applies only for free fall in vacuum. You're expected to know that and the fact that in air, it takes a bit more time for the same fall but you're not expected to know exactly how much more time that is.

I would take TBR's conclusion as fact and move on - reasoning why that is the case does not work too well and the math behind it is certainly not MCAT territory.

 

[slz1900]

It sure does. It's not part of 2π.sqrt(L/g) because it explicitly is not taken into account in the derivation of the formula. The formula with the air resistance will be much more complicated but they sort of expect you to be able to make qualitative statements for that case.

In a similar way h=gt^/2 applies only for free fall in vacuum. You're expected to know that and the fact that in air, it takes a bit more time for the same fall but you're not expected to know exactly how much more time that is.

I would take TBR's conclusion as fact and move on - reasoning why that is the case does not work too well and the math behind it is certainly not MCAT territory.

How would you make a qualitative statement? Even if TBR is right it seems to me like it goes beyond intuition. For a falling block it is easy to see that it would take longer with air resistance. For a swinging pendulum, you have to weigh two competing factors.

 

[milski]

How would you make a qualitative statement? Even if TBR is right it seems to me like it goes beyond intuition. For a falling block it is easy to see that it would take longer with air resistance. For a swinging pendulum, you have to weigh two competing factors.

Yes, it does go beyond simple logic/intuition. Based on the explanation, they expect you do some rather arbitrary assumptions. I don't think this is a good question but I doubt that TBR cares much about what I think.😎

 

[Meredith92]

Hate to bring this up again, but on rereading the chapter on my review day I really think TBR is wrong with this question. I think its a really important point and would love if a BerkeleyReviewTech could help us out on this.

In the chapter, on example 5.1b it talks about dampened harmonic oscillation: "dampened harmonic oscillation refers to a cyclic process that loses energy as it moves. this loss in energy is generally attributed to friction... the period remains constant. dampening reduces the amplitude, but it does not reduce or increase the time necessary to complete a cycle. over time, the object will cover less distance in a given cycle, but because its doing it at a reduced average speed, the time to complete a cycle remains constant."

This seems to directly go against the question i originally posted. Air resistance is equivalent to friction, so wouldnt the pendulum have the same period??

 

[slz1900]

TBR is actually right on air resistance increasing period. Here is a paper that talks about it (pg 114):

http://fy.chalmers.se/~f7xiz/TIF080/pendulum.pdf

The paper also says that increased drag increased the period, so tbr is wrong on the second statement.

Notice though that it is a difference in a fraction of a MICROsecond, so I highly doubt that you'll be asked directly about it.

 

[milski]

The paper uses material well beyond what's on the topic list for physical sciences. I don't think TBR's expectations for the doing estimates here are very appropriate.

 

[BerkReviewTeach]

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!

 

[slz1900]

BRTeach, I respect your explanation, but I disagree that you could reasonably be expected to use your intuition to figure this one out. The difference in period is literally less than half of a microsecond, which to me seems smaller than the lower bound of human guesstimation ability given the number of variables here. I don't think your water example works either because, per the paper, the math will change substantially depending on the buoyant force and drag of the fluid the pendulum is in. It only has to become 0.03 microseconds shorter to invalidate BR's statement.

 

[BryanNextStep]

I respect your explanation, but I disagree that you could reasonably be expected to use your intuition to figure this one out.

Gotta agree with slz1900 on this one. We just got what was essentially a *fantastic* science answer and a kinda "wha?" MCAT answer.

From everything I've read in the thread so far, the most important thing said was when milski pointed out the following: "It's not part of 2π.sqrt(L/g) because it explicitly is not taken into account in the derivation of the formula." This seems to be one of those things the MCAT really wants us to know - when stuff doesn't matter. If something isn't in the equation, then it doesn't matter. The classic example here would be changing the mass at the end of the bob.

Gotta remember when solving any individual question to stay focused on the basic question, "Okay this question is irrelevant. What matters is: what is this question teaching me that I can use on test day?"

When it comes to that, I think milski's got the most important point in the thread.

 

[Meredith92]

( Wrote this before i saw the posts above but would still appreciate these questions answered thanks!!)

Thanks so much for your help BRtech. It's slowly starting to make sense,

A few follow up questions- in your water example would you say once it is in water and experiencing dampened harmonic motion that the period would remain the same? It seems odd that this could be true because the pendulum will stop rather quickly.

Also ( this is probably a dumb question and separate from this topic a bit) how does the distance the mass travels even play into our formula for period? T=2pi sqrt(L/g) I get how you used the smaller value of g to support the problem under water but does distance fit into the period equation other than knowing v=d/t. Tbr earlier in the chapter (example5.1b) used the concept that both v and d decrease to explain why in dampened motion period stays the same... But how can they do that when both v and d are not present in the period equation. I apologize if this is a really stupid question!

 

[Meredith92]

Oh wait my second question may have already been answered by milski - we can just ignore air resistances effect on v and since it's not in the equation?

I don't know if that makes intuitive sense to me... Tension is not In the period equation T=2pi sqrt L/g but it does have an effect on period (through another equation)