If you know that the period of a pendulum is 1.87sec. What is the length of that pendulum? (gravity 9.81m/sec squared)

F = m A

m g sin theta = m a = -m d^2 x/dt^2
g sin theta = -d^x/dt^2
for small Theta sin Theta = theta = x/L
g x/L = -d^x/dt^2
let x = c sin 2 pi f t
then v = dx/dt = 2 pif c cos 2 pi f t
and a = d^2x/dt^2 = - (2 pi f)^2 c sin 2 pi f t
so
g /L = (2 pi f)^2
2 pi f = sqrt (g/L)
but T = period = 1/f
so
T = 2 pi sqrt (L/g)
1.87 = 2 pi sqrt(L/9.81)
sqrt(L/9.81) = .2976
L/9.81 = .088577
L = .869 meter

To find the length of a pendulum when you know its period, you can use the formula for the period of a pendulum:

T = 2π√(L/g)

Where:
T = Period of the pendulum
L = Length of the pendulum
g = Acceleration due to gravity

In this case, you know the period (T = 1.87 sec) and the acceleration due to gravity (g = 9.81 m/s^2). You can rearrange the formula to solve for the length (L):

L = (T^2 * g) / (4π^2)

Now let's substitute the known values into the formula:

L = (1.87^2 * 9.81) / (4π^2)

Calculating this expression will give us the length of the pendulum.