if a simple pendulum with length 1.50 m makes 72 oscillations in 3 minutes. what is the acceleration of the gravity at its location ?

Answer 1

let I = m L^2
theta = A sin wt
torque = -m g L sin theta = approx -m g L theta
dTheta/dt = A w cos w t
d^2Theta/dt^2 = -Aw^2 sin wt = - w^2 Theta
I d^2Theta/dt ^2 = m g L theta
m L^2 w^2 = m g L
w^2 = g/L
(YOU MIGHT START HERE IF FORMULA IN YOUR BOOK)
2 pi f = sqrt(g/L)
f = 72/180
2 pi (72/180) = sqrt (g/1.5)
6.317 = g/1.5
g = 9.47 m/s^2