Here is one way.

T = 2pi*sqrt(L/g)

Differentiate both sides,

Since g is constant,

T = [2pi / sqrt(g)]*sqrt(L)

dT = [2pi / sqrt(g)][(1/2) / sqrt(L)]dL

dT = [pi / sqrt(gL)]dL -------------------(1)

When dL = (1/2)percent of L, or 0.005L, the L becomes (L +0.005L) = (1.005L).

Plug those into (1), to get the change in T,

dT = [pi / sqrt(g*1.005L)](0.005L)

dT = [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] ------(2)

So the percent change in T is

(dT / T) *100percent

Where dT /T =

= [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] / [2pi*sqrt(L/g)]

= [(pi)sqrt(g) / sqrt(g)*2pi]*[(0.005L) / sqrt(1.005L)*sqrt(L)]

= [(1/2)]*[0.005 / sqrt(1.005)]

= 0.0024938

or,

= 0.0025

And that is 1/4)percent.

Therefore, when the L was increased by (1/2) percent, the period T was increased by (1/4) percent ------------answer.

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b) Using the result in part (a), find the approximate error in this pendulum clock in one day.

T is in seconds.

1 day = 24hrs*(60min/hr)(60sec/min) = 24*3600 = 86,400 sec

(0.0025)(86,400) = 216 sec = 3min and 36sec.

Therefore, in one day the pendulum clock is faster by 3 minutes and 36 seconds than the correct time. ----------answer.