# Time lost by pendulum

1. Feb 6, 2016

### Vibhor

1. The problem statement, all variables and given/known data

Q. What is the time lost in time 't' by a pendulum clock whose actual time period is T and the changed time period at some higher temperature is T' ? (T' is Time period at some higher temperature , l' is the increased length , Δθ is the difference in temperature, α is coefficient of linear expansion)

2. Relevant equations

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

3. The attempt at a solution

Time period of a simple pendulum is given by T = $2\pi \sqrt{\frac{l}{g}}$ . As the tempertaure is increased , length of the pendulum and hence time period gets increased . the pendulum clock becomes slow and hence loses time .

If T' is Time period at some higher temperature , l' is the increased length , Δθ is the difference in temperature, α is coefficient of linear expansion .

$\frac{T'}{T} = \sqrt{\frac{l'}{l}}$ = $\sqrt{\frac{l+Δl}{l}}$ = $\sqrt{\frac{l+lαΔθ}{l}}$ = $(1+αΔθ)^{\frac{1}{2}}$

$T' ≈ T(1+\frac{1}{2} αΔθ)$

$ΔT = T'-T = \frac{1}{2} αΔθ$

Time lost/sec = $\frac{ΔT}{T}$

Time lost in time 't' = $\frac{ΔT}{T}t$

But the answer given is $\frac{ΔT}{T'}t$ . This is what I do not understand . When we calculate Time lost/sec , we should divide ΔT by T or $T'$ ( new/increased time period ) . I think it should be T , but according to the book it should be $T'$

Many Thanks

Last edited: Feb 6, 2016
2. Feb 6, 2016

### Nathanael

I'm not sure why you considered the lengths, or the coefficients of expansion, or the change in temperature; none of those are known, so it's no use in considering them.
In fact, your relevant equation is not important either; even if it swung through large enough angles so that the small angle approximation failed, the answer would be the same. (It doesn't even matter if they are pendulums or another device.)

All that matters is, we have two clocks, both of which are calibrated such that T seconds pass on the clock per cycle of the pendulum. One clock's period really is T (hence it keeps time correctly) and the other clock's pendulum has a period of T'. Suppose a (true) time t goes by; the number of oscillations of the pendulums will then be t/T and t/T'. Thus one clock will say (t/T)*T = t has passed, and the other will say (t/T')*T has passed. The difference between the clocks is then t-t(T/T') = t(T'-T)/T'.

3. Feb 6, 2016

### Vibhor

Sorry . They are part of the problem . I will edit the question . It is a standard problem in thermal expansion chapter.

4. Feb 6, 2016

### Nathanael

Oh. Well all of that information is a red herring (unless there are other parts to the question).

5. Feb 6, 2016

### Vibhor

The problem is part of the theory as well as similar to a worked example given in the book . Could you have a relook at the OP ?

6. Feb 6, 2016

### Nathanael

Ok, but as I said, all of that information is a red herring (none of it matters; it's a distraction to get you thinking about other things).

Have you understood the (second paragraph of) post #2? If you are still confused, can you better explain why you think it should be over T instead of over T' ?

7. Feb 6, 2016

### Vibhor

Because it is the actual time period over which the pendulum has gained time . Whenever we are looking for fractional change in a quantity , the division is made by the true/original quantity . Isn't it ??

8. Feb 6, 2016

### Nathanael

I don't understand what you're saying here.
Yeah, if we want the fractional change in something, then we do it that way. If the problem had said, "find the fractional change in the period" then the answer would be (T'-T)/T. That's not what it asked though. It is asking this: "after time t, how far off from the true time is the T'-clock?" You see, it's a more subtle question than just finding a fractional change.
(Paragraph 2 of post #2 has the precise explanation.)

9. Feb 7, 2016

### Vibhor

Last edited: Feb 7, 2016
10. Feb 7, 2016

### ehild

You (and the link) determined the relative error of time with the linear approximation. Then $\frac {\Delta T}{T'}= \frac{0.5 \alpha \Delta \theta}{1+0.5 \alpha \Delta \theta}$ which is the same as $0.5 \alpha \Delta \theta$ in the linear approximtion. It does not matter if you divide by T or by T'.
In principle, you see the beats of the clock, and assign the time to the number of beats. During t time, there are N=t/T' beats of the wrong clock, what you think is NT time. The time difference is( t/T')*T-t=T(T/T' - 1), as Nathanael has shown.

11. Feb 7, 2016

### Nathanael

To be clear, most of the work in that post is to find T' which we are given at once.
Towards the end, they did not explain their calculation of "$t_{day}$," but it does appear that they (effectively) used t(T'-T)/T instead of t(T'-T)/T'. I say they are wrong in doing that (although to the accuracy involved in that thread it changes nothing).
Well I don't understand your reasoning in the first place so I'm not sure how to help. If you have another way of explaining your reasoning, please do.

12. Feb 7, 2016

### ehild

You calculated the relative increase of time elapsed during one period of the pendulum to the time of one period of a punctual pendulum.
But the time shown by the clock was the question. It is less then the real time.

13. Feb 7, 2016

### Vibhor

This is the key issue . I was not paying enough attention to how the clock works . The number of oscillations in the incorrect clock is multiplied by the correct time period T to get how much time has elapsed as per the incorrect clock . Please correct me if I am wrong.

Thank you very much .

14. Feb 7, 2016

### Vibhor

Spot on

Thanks !

15. Feb 7, 2016

### Vibhor

@ehild ,@Nathanael Please see the attached image . It has a worked out example from the textbook dealing with the same concept . I think it is quite misleading in light of the discussion we had in this thread . The author is using the same approach as in the OP i.e fractional loss of time .

But interestingly ,the answer comes out to be same using both the approaches , the flawed as in the OP and the correct approach as in post#2 .

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16. Feb 7, 2016

### Vibhor

17. Feb 7, 2016

### Nathanael

Exactly.
You are absolutely right, they are all doing it wrong! I guess it's quite a prevalent misconception! Note though, that although they are theoretically wrong, the answers involved are virtually the same either way. This is because T and T' are so close to each other (which in turn is because thermal expansion is so minuscule).
You can see, algebraically, to get from their answer to ours, you must multiply by T/T'. In your textbook's example, this factor is 0.99988, which is very nearly 1 (hence why it seems to give the same answer).
If T and T' were not so nearly the same, though, then their answers would be off by more than that.

Thanks for bringing those examples up, as I did not realize how common that misunderstanding is...
No one seems to pay attention to the finer details!

18. Feb 7, 2016

### ehild

I join to Nathanael's opinion.