Answer: A pendulum of length 1m and period 2.01s is placed at the top of Mount Everest having an altitude of 8849m. Calculate the value of 'g' at that point.
The period of this pendulum can be solved using the formula shown in the picture below. T is the period in seconds; L is the length in meters and g is the acceleration due to gravity which is equal to 9.81 m/s²near the equator.
T = 2 × π × (square root of 1.00 m / 9.81 m/s²)
T = 2 * 3.1416 * (square root of 0.101936799)
T = 6.2832 * 0.319275428
T = 2.00 seconds
The period of this pendulum correct to 3 significant figures is 2.00 seconds.
The frequency of this pendulum is solved by the formula f = 1/T
f = 1/2.00 seconds
f = 0.500 hertz
Answer: 9.76 m/s^2
Explanation: