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I'm working out how much time a pendulum clock will gain or loose due to change of the length of the pendulum due to temperature. so far I've got, new time period, $$T_2=T_1(1+\frac12\alpha\Delta T)$$

due to $\Delta T$ change in temperature, when the coefficient of linear expansion of the pendulum material is $\alpha$. $T_1$ is the time period when there is no change in temperature.

Now, I can not understand how the time period of the pendulum and the time measured by the clock is related. Please help me with this and also mention if I am going correct or not.

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closed as off-topic by ACuriousMind, Carl Witthoft, Daniel Griscom, Gert, John Rennie Feb 4 '16 at 8:13

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  • $\begingroup$ You don't have enough information. How does your pendulum drive the clock ( how and how far do the hands advance for each swing) $\endgroup$ – Carl Witthoft Feb 3 '16 at 14:51
  • $\begingroup$ it's not provided in the question. let me post a screenshot of the question. $\endgroup$ – Subhranil Sinha Feb 3 '16 at 14:54
  • $\begingroup$ q222. imageshack.com/a/img921/3667/jnCIgr.jpg @CarlWitthoft $\endgroup$ – Subhranil Sinha Feb 3 '16 at 15:02
  • $\begingroup$ My first comment may be misleading. You can, as the answers show, estimate the change in clock-time fractional error, but not the absolute error per cycle. $\endgroup$ – Carl Witthoft Feb 3 '16 at 15:06
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    $\begingroup$ This question makes me wonder about the effect of aerodynamic drag on the pendulum. If the air around the pendulum also heats up, then I'm thinking the air density (and therefore drag) would drop as well. This would tend to speed up the clock. Anybody care to estimate the net effect on the clock speed? $\endgroup$ – James Feb 3 '16 at 15:16
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You are on the right track.

For a mechanical pendulum, the relationship is linear. You don't need to know how many swings of the pendulum corresponds to how many seconds. If the pendulum is x% slower, it will report x% fewer seconds per day.

Now since length goes as $\ell = \ell_0(1+\alpha \Delta T)$ and period of pendulum as

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

We can do the Taylor expansion for $\ell$, noting that $\sqrt{1+x}\approx 1+\frac12 x$ for small $x$, to give us

$$T_2 = T_1 (1 + \frac12 \alpha \Delta T)$$

as you noted.

Now the number of swings per day times the time per swing should equal one day, or $N\cdot T=86400$. It follows that the number of seconds per day, N, is changed by

$$N_2 = N_1 \frac{T_1}{T_2} = \frac{N_1}{1+\frac12\alpha \Delta T} = N_1\left(1-\frac12\alpha T\right)$$

The difference in seconds per day then follows. I will leave that as an exercise.

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You can't get an exact number for this without some assumptions. However, you can develop a relationship. What we do know is that one full period corresponds to some amount of time (my niave guess would be 1 second but I believe I have heard faster clocks which might be half seconds). So we introduce some constant C which is the amount of time reported on the clock per period.

First, we look at the ordinary case without expansion $$ t_1 = CT_1 $$

When the length of the pendulum expands, C does not change as it only cares when a full period occurs. However, we need to calculate the actual time in a different way. $$ t_2 = CT_2 $$

Thus, $$ t_2 = t_1 \frac{T_2}{T_1} $$

implying $$ t_2 = t_1 (1 + \frac{1}{2}\alpha \Delta T) $$

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Assume that each swing advances the second hand by one second on the dial. You have a pendulum which is supposed to have a period of exactly one second.
If the pendulum has got longer in length it will have a longer period and so it will take longer for the pendulum to advance the second hand by one second on the dial.
So the clock will run slow.


Just to add to the possible confusion you may come across the term seconds pendulum. Such a pendulum has a period of two seconds. The reason for such a name is that for a lot of clock mechanisms the advance is done every time the pendulum swings through its equilibrium position ie twice per oscillation.

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I have an old pendulum clock (probably over 100 years old) still in working order. I'm pretty sure that the hands advance linearly with the number of swings of the pendulum. I keep my house about 10 degrees Fahrenheit cooler in the winter than in the summer, but don't really notice a difference in its timekeeping. The bob is suspended by a wire made of copper.

My clock was made by the Gilbert company and has a Normandy chiming mechanism. It has two springs that have to be wound separately. One spring provides energy to the pendulum and moves the hands while the other spring drives the chiming mechanism. They tend to need rewinding together. There is a pin that allows for adjustment of the length of the pendulum, but it tends to slip during adjustment making an accurate adjustment difficult. I've found that I can adjust it much more accurately by attaching a paper clip to the wire that holds the bob and moving the clip up or down to change the effective center of mass of the pendulum.

It is pretty easy to get accurate timing to within 5 min per week with my clock. As the energy in the springs gets depleted it does tend to run a bit slower. The chiming rate slows down as well, so that gives you a reminder that it is time to rewind. I try to rewind once a week, but the clock will continue to run for at least a week and a half before stopping if it doesn't get rewound.

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  • $\begingroup$ Yes, exactly. A pendulum clock is a machine that (a) keeps the pendulum swinging, and (b) counts swings of the pendulum. For any given clock, it takes some number of swings to make the hour hand complete one revolution, and that number can't be changed without re-building the whole clock: It's determined by the ratios of all of the gears. $\endgroup$ – Solomon Slow Feb 3 '16 at 16:07
  • $\begingroup$ Interesting that the spring unwinding causes it to slow down... the amplitude of swing must decrease just enough. $\endgroup$ – Floris Feb 3 '16 at 16:24
  • $\begingroup$ That would be my guess. It is not a big effect, but is noticeable. $\endgroup$ – Lewis Miller Feb 3 '16 at 16:33
  • $\begingroup$ You may be interested in this link of a pendulum clock that is being studied really, really carefully... The page has a lot of links along the left side that lead to fascinating articles. $\endgroup$ – Floris Feb 3 '16 at 16:59

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