Question

a pendulum-driven clock gains 5.00 s/day, what fractional change
in pendulum length must be made for it to keep perfect time?

Answer 1

T = 2π√(L/g)

or, in terms of L:

L = g(T/2π)²

To cut down on writing, let's define a constant k = g/(2π)². Then the above is:

L = kT²

So a small change in length (dL) is related to a small change in period (dT) like so (via derviative):

dL = 2kTdt

But notice (from previous equation) that kT = L/T. Substitute:

dL = 2(L/T)dt

The fractional change in pendulum length is (by definition) (dL)/L. By the above equation, this is:

(dL)/L = 2(dt)/T

That is, fractional change in pendulum length is twice the fractional change in period. Now, we know the clock is ticking too fast, and we want to slow it down by 5.90s per day. That is accomplished by increasing the period (making it tick slower) by a fraction of (5.90 s per day). That is:

(dt)/T = 5.90s per day = 5.90s / 86400s = 6.83×10^-5

So the fractional change in pendulum length would be:

(dL)/L = 2(dt)/T = 2(6.83×10^-5) = 1.36×10^-4