In Seattle, Washington, a seconds pendulum is 0.99382 m long. What is the free fall acceleration in Seattle, Washington? (Use 5 significant figures in your answer.)

A seconds pendulum is a pendulum that moves through its equilibrium position once every 1 second, the pendulum will move from its highest point at one end through its equilibrium position to its highest point on the other end. This means it takes 1 second to move from its highest point on one end to its equilibrium position and another second to move from its equilibrium position to its highest position on its other end. (in this scenario equilibrium position is the position of the pendulum when it is at rest). therefore the period of the pendulum (T) = 1 + 1 = 2 seconds

length of the pendulum (L) = 0.99382 m

what is the free acceleration (g) in Seattle?

period (T) = 2π√(L/g) where π = 3.142

substituting the values of T, π and L into the equation above we have

2 = (2 x 3.142)√(0.99382/g)

2 = 6.283√(0.99382/g)

2/6.283 = √(0.99382/g)

0.318 = √(0.99382/g)

[tex]0.318^{2}=(\sqrt{0.99382/g})^{2}[/tex]

0.101 =0.99382/g

g = 0.99382/0.101

g = 9.8080 m/s^{2}