Answer:
The g at the top of the mountain is 9.820 m/s².
Explanation:
The period of simple pendulum is given as;
[tex]T = 2\pi \sqrt{\frac{l}{g} }[/tex]
where;
T is period of the oscillation
g is acceleration due to gravity
l is length of the pendulum
[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} }\\\\\frac{T^2}{4 \pi ^2} = \frac{l}{g}\\\\T^2 g = 4 \pi ^2 l\\\\let \ 4 \pi ^2 l = k\\\\T^2_{sea \ level} \ \times \ g_{sea \ level} = T^2_{top \ of \ mountain} \ \times \ g_{top \ of \ mountain}\\\\ g_{top \ of \ mountain} = \frac{T^2_{sea \ level} \ \times \ g_{sea \ level}}{T^2_{top \ of \ mountain} } \\\\ g_{top \ of \ mountain} = \frac{1^2 \ \times \ 9.8}{0.999^2} = 9.820 \ m/s^2[/tex]
Therefore, the g at the top of the mountain is 9.820 m/s².
