Answer
a. $-\pi\sqrt L(g)^{-3/2}dg$
b. speed up.
c. $-0.98cm/s^2$, $979.02cm/s^2$
Work Step by Step
a. With the given equation $T=2\pi(L/g)^{1/2}$, we have $dT=2\pi(L)^{1/2}(-\frac{1}{2}(g)^{-3/2})dg=-\pi\sqrt L(g)^{-3/2}dg$
b. With the relation $dT=-\pi\sqrt L(g)^{-3/2}dg$, an increase in $g$ will give a negative change or decrease in $T$. Thus, when $g$ increases, the period $T$ will decrease and a shorter period means that the pendulum clock will move faster or speed up.
c. Given $L=100cm, g=980cm/s^2, dT=0.001sec$, we have $0.001=-\pi\sqrt {100}(980)^{-3/2}dg$ and $dg\approx-0.98$. Thus the new $g_1=980-0.98=979.02cm/s^2$
