math
posted by Gabi .
The formula T=2*pi*sqrt[L/32] gives the period of a pendulum of length l feet. The period P is the number of seconds it takes for the pendulum to swing back and forth once. Suppose we want a pendulum to complete three periods in 2 seconds. How long should the pendulum be? Thank you so much for helping. The solution must be sbout 0.36 ft. I tried everything but I'm still wrong. Thanks again!!!!

To complete three periods in 2 seconds, the period must be
T = 2/3 = 0.6667 s
I will use your formula although physics says that the denominator in the sqrt[L/32] should be 32.2 ft/s^2, not 32. That is the value of g, the acceleration due to gravity.
Anyway, using 0.6667 = 2*pi*sqrt(L/32),
sqrt(L/32) = 0.10611
L/32 = (0.10611)^2 = 0.01127
L = 0.3603 feet
With the correct formula,
L/32.2 = (0.10611)^2
L = 0.3625 ft