Answer:
[tex]\Delta T=-\pi\sqrt{\frac{l}{g}}\frac{\Delta g}{g}[/tex]
Explanation:
The period of a pendulum is given by the following formula:
[tex]T=2\pi\sqrt{\frac{l}{g}}[/tex]
l: length of pendulum
g: gravity constant
If there is a little increase in g to g+Δg you have:
ΔT + [tex]T=2\pi\sqrt{\frac{l}{g+\Delta g}}=2\pi\frac{(l)^{1/2}}{(g+\Delta g)^{1/2}}=2\pi(l)^{1/2}(g+\Delta g)^{-1/2}[/tex]
by using the expansion for (g+Δg)^{-1/2} term you obtain:
[tex](g+\Delta g)^{-1/2}=(g)^{-1/2}(1+\frac{\Delta g}{g})^{-1/2}\\\\= (g)^{-1/2}[1-\frac{1}{2}\frac{\Delta g}{g}+\frac{3}{8}(\frac{\Delta g}{g})^2+...][/tex]
However, due to Δg is very lower in comparison with g, you can consider that (Δg/g)^2 ≈ 0 and also for higher exponents. Thus, you obtain for T:
[tex]T+\Delta T=2\pi(l)^{1/2}(g)^{-1/2}[1-\frac{1}{2}\frac{\Delta g}{g}]=2\pi \sqrt{\frac{l}{g}}[1-\frac{1}{2}\frac{\Delta g}{g}]=2\pi\sqrt{\frac{l}{g}}-\pi\sqrt{\frac{l}{g}}\frac{\Delta g}{g}[/tex]
from the previous result you can conclude the following:
[tex]\Delta T=-\pi\sqrt{\frac{l}{g}}\frac{\Delta g}{g}[/tex]
