A pendulum which has a period of 1s in London, where g=9.8118m s-2, is taken to Paris where it is found to lose 20s per day. What is the value of g in Paris?

I like this question but can't figure it out. I have the length of the pendulum correct, I think. I know that there are 86400s in one day; minus 20s for the pendulum period in Paris. Can't figure it out.
Any clues please?

It loses 20s per day, means that when the pendulum is taken to Paris, it swings twenty fewer times in 24hrs than when it was in London. (You would say in common parlance that your clock had 'lost time' in such circs, not having ticked enough ticks as it were)

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If the pendulum makes n oscillations per day,
the period T1 in London is 86400/n
the period T2 in London is 86380/n
Substitute these values in the relevant equation and then find T1/T2.

ok. so i may have the answer. hopefully.



how's that?

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Your calculation is wrong, because you have not squared T1/T2
If you simplify the equations you can get the following relation.
T(L)^2*g(L) = T(P)^2*g(P)
Find g(P).

So, I don't need to calculate the length at all?

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Your answer is correct.

Yipeeee!!!

Thank you very much

why are they all T^2, when they are not squared?

Formula for the period of the pendulum is given by
T^2 = 4*π^2*L/g. So
T^2*g is a constant.