Calculate the length of a seconds pendulum at a place where the gravity is 9.83 metre per second

Question

Calculate the length of a seconds pendulum at a place where the gravity is 9.83 metre per second

Answer 1

period T of pendulum is given by 2pi * sqrt(L/g) where L is length and g is acceleration due to gravity (9.83 in this case)

rearrange in terms of L

T = 2pi* sqrt(L/g)

T / 2pi = sqrt(L/g)

(T/2pi)^2 = L/g

L = g * ( T / 2pi )^2

T is given as 1second, substitute this and g =9.83 m/s^2 and you will have the answer

Answer 2

Pendulum period in seconds

T ≈ 2π√(L/G)

L is length of pendulum in meters

G is gravitational acceleration = 9.8 m/s²

Period is inversely proportional to the square root of gravity.

So gravity goes down by 9.83/9.8 = 1.00306

Square root of that is 1.00153

so period is 1/1.00153 = 0.99847 seconds

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