Simple Harmonic Motion: Pendulum

[TJC747]

The Earth's acceleration due the gravity varies from 9.78 m/s2 at the equator to 9.83 m/s2 at the poles. A pendulum whose length is precisely 1.000 m can be used to measure g. Such a device is called a gravimeter.
(a) How long do 100 oscillations take at the equator?
(in sec)

(b) How long do 100 oscillations take at the north pole?
(in sec)

(b) Suppose you take your gravimeter to the top of a high mountain peak near the equator. There you find that 100 oscillations take 201 seconds. What is g on the mountain top?
( in m/s^2)

I guess I'd use T = 2*pi*sqrt(L/g) in conjunction with other Hooke formulae. Help would be appreciated. Thanks.

 

[rock.freak667]

right, so use

T= 2 \pi \sqrt{\frac{L}{g}}

to find the time for one oscillation