Answer
a) $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$
b) As T decreases, the clock pendulum will move more frequently
$dg=-0.977 \frac{cm}{s^2}$
$g=979\frac{cm}{s^2}$
Work Step by Step
Given that $T=2\pi(\sqrt{\frac{L}{g}})$
on differentiating the above equation:
a) taking L as constant
${dT}=2\pi\frac{1}{2}\frac{1}{\sqrt{\frac{L}{g}}}\frac{-L}{g^2}dg$
$dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$
b) If g increases then $dg\gt0$ and $dT\lt0$
As T decreases, the clock pendulum will move more frequently.
c) L=100,g=980 and dT=0.001
then
$dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$
$0.001=-\pi\frac{100^{\frac{1}{2}}}{980^{\frac{3}{2}}}dg$
$dg=-0.977 \frac{cm}{s^2}$
$g=980-0.977=979\frac{cm}{s^2}$
so the final answer is
a) $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$
b) As T decreases, the clock pendulum will move more frequently
$dg=-0.977 \frac{cm}{s^2}$
$g=979\frac{cm}{s^2}$
