Derive the equation T = 2pi(l/g)^1/2.

Question

That's period of motion.

Answer 1

hang mass m (will not matter) from string of length L

deflect angle A from up and down
Tension T in string
m g = T cos A
horizontal restoring force = -T sin A
so
F =-( m g/cosA) sin A = - m g tan A
if A is small
F = - m g A

distance x from centered
x = L sin A
for small angle
x = L A
if x = s sin (2 pi t/T)
v = s ( 2 pi/T) cos (2 pi t/T)
a = - s (2 pi/T)^2 sin (2 pi t/T)
or
a = - (2 pi /T)^2 x

now F = m a
-m g A = -m(2 pi/T)^2 L A

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

2 pi/T = (g/L)^.5

T = 2 pi (L/g)^.5