# Decreased period of pendulum

I'm doing an experiment with a physical pendulum and as time passes, the time taken for n cycles is shorter than would be predicted from the period (i.e. the time taken for the first n cycles is less than n*(period)). I'm not sure why this is as I thought damping and air resistance were supposed to increase the period. Is there any possible explanation for why the period is actually decreasing with time?

• You're saying that the pendulum is starting to oscillate slower and slower with time? That is indeed very odd. Are you sure that it isn't precessing a little, and you're observing it from outside the plane? What sort of increase in time-period are we talking about? Could you include your data and (if possible, of course) a short video? Even a photo of your setup would help. – Philip May 23 at 20:22
• The pendulum seems to be oscillating with a decreased period. I counted the time taken for 15 periods and this was less than the 15*(period). This was consistent for all the trials. Also, the time taken for the first half cycle was very accurate so I am unsure as to why this is occuring. I tried to add a video but I can't seem to find a way to include it. Is there any way I could email it to you or is there some way to include a video here? – planckton May 23 at 20:35
• Apologies, I misunderstood the question slightly. (There are ways to convert videos to gifs on imgur for example.) However, the most important thing would be your actual data points. For eg. when you say "I counted the time taken for 15 periods and this was less than the 15*(period)", how well do you know "(period)", how was it measured? Furthermore, the most standard mistake that people make when timing a simple pendulum is to count one extra oscillation by accident. It's an obligatory error that all undergraduates I've encountered have made at least once. – Philip May 23 at 20:39
• I have checked multiple times and I've counted the right number of cycles. Also counting the wrong number of cycles would result in the period being off by multiply seconds whereas here, the error is about 1 s. The period is calculated using the theory for a uniform rigid rod (this is pretty well established so I don't think the theoretical period should be a problem – planckton May 23 at 20:43
• Perhaps. I will restate my most basic point here once again: until we know what sort of numbers you are dealing with, no one will be able to give you a satisfactory answer. Do you mean that the time for 15 oscillations is off by 1 s from what you'd expect "theoretically"? That would not be unreasonable, in most cases (how did you time this? A stopwatch? A photogate?). Again, one would need to see your data points to be sure. Also, when you say "uniform rigid rod" are you considering a simple pendulum or a compound pendulum? I think the question needs a lot more details to be answerable. – Philip May 23 at 20:51

Not sure if this would help, but in reality, the oscillation period of a pendulum increases with the oscillation amplitude. A second order expansion for period as a function of amplitude is $$T \approx 2\pi \sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16}\right).$$ See this Wikipedia article for more details. The period only approaches $$2\pi\sqrt{\frac{L}{g}}$$ as the oscillation amplitude approaches zero, and it is larger than that for finite amplitude oscillations. Therefore, the oscillation period of your pendulum might be decreasing due to damping.

• Yes. OP: what was the starting amplitude of your oscillations? – Chet Miller May 23 at 23:10

The period of oscillation of a simple, small angle pendulum is given by:

$$T=2\pi\sqrt{\frac{L}{g}}$$

where $$L$$ is the length of the pendulum.

[...] the time taken for n cycles is shorter than would be predicted for the period [...]

So it could be caused by shortening of the rod which may in turn be caused by lowering temperature.

Temperature compensated pendulums certainly do exist.

• The period for the first couple cycles is very accurate but as time passes the time taken for the first n cycles becomes significantly less n periods. Would the lowering temperature allow for the first couple of cycles to be accurate yet decrease for later cycles? – planckton May 23 at 20:21
• The larger the deviation of temperature $\Delta T$, the larger the 'shrinkage' $\Delta L$ will be. But it's hard to evaluate w/o quantifying things. – Gert May 23 at 20:25
• @planckton See my comment above. To quantify any such statements, we really need to know what "significantly less n periods" means. Being able to look at your data would help. Also, while I believe Gert's answer is not strictly wrong, for most table-top experiments I would seriously doubt that lowered temperatures could cause "significant" changes in the time period (or rather, changes more significant than the multitude of other uncertainties that inevitably creep in). – Philip May 23 at 20:26