Pendulum

Pendulum period in seconds
T ≈ 2π√(L/g)
or, rearranging:
g ≈ 4π²L/T²
L ≈ T²g/4π²
L is length of pendulum in meters
g is gravitational acceleration
    (on earth, nominal of 9.8 m/s²)
Length for ½ second = 0.062 m
Length for 1 second = 0.248 m
Length for 2 second = 0.993 m
Length for 4 second = 3.97 m

Uniform beam pendulum
center of oscillation is 2/3 of the length of the uniform
beam L from the pivoted end.
T ≈ 2π√((2L/3g)

Compound pendulum
Any swinging rigid body free to rotate about a fixed
horizontal axis is called a compound pendulum or physical
pendulum. The appropriate equivalent length L for
calculating the period of any such pendulum is the distance
from the pivot to the center of oscillation. This point is
located under the center of mass at a distance from the
pivot traditionally called the radius of oscillation, which
depends on the mass distribution of the pendulum. If most of
the mass is concentrated in a relatively small bob compared
to the pendulum length, the center of oscillation is close
to the center of mass. The radius of oscillation or
equivalent length L of any physical pendulum can be shown to
be:
L = I/mR
where I is the moment of inertia of the pendulum about the
pivot point,
m is the mass of the pendulum, and
R is the distance between the pivot point and the center of mass.

The period of a compound pendulum is given by
T = 2π√(I/mgR)
for a uniform rod pivoted at one end, I = mL²/3
T = 2π√(2L/3g)

Speed of bob at lowest point
O is the pivot point of the pendulum.
A is the release point
B is the point of lowest travel
θ is the deflection angle at release
φ is angle ABO
α is angle BAC
L = OA = OB = length of string

OC = OA cos θ
h = BC = OA – OC = OA – OA cos θ = OA (1 – cos θ)

Now use KE = PE to get the speed
½mV² = mgh
V = √(2gh)

Pendulum period in sec for larger angle, θ in radians
T ≈ [2π√(L/G)] [1 + (1/16)θ² + (11/3071)θ⁴ +
     (173/737280)θ⁶ + (22931/1321205760)θ⁸ + ... ]
T ≈ [2π√(L/G)] [1 + (1/2)²(sin²(θ/2) +
     ((1·3)/(2·4))²(sin⁴(θ/2) +
     ((1·3·5)/(2·4·6))²(sin⁶(θ/2) + .. ]

Conical Pendulum period
T = 2π√(r/(gtanθ))
r is radius of the rotation
θ is angle of bob with vertical
v = 2πr/T
v is velocity


Change in simple pendulum period due to length change
dT/dL = (1/2)2π√(1/g)√1/L) = π√(1/Lg)
dT = (π/√(gL))dL

for a 1% change in length, what is time change for 1
day, 86400 seconds?
dL/L = 0.01 (ie, 1%)
dL = 0.01L
dT = π/√(gL) (0.01)L
dT = (0.01)(π/√g) L/√L
dT = (0.01)(π/√g)√L
dT = (0.01)(π√(L/g))
dT = (0.01)(1/2)(2π√(L/g))
dT = (0.005)T
dT = (0.005)(86400) = 432 seconds

Conical Pendulum where the bob moves in a circle.
T = tension in string of length L
m = mass of bob
θ = angle of string with vertical
t = 2π√(Lcosθ/g)

Compound Pendulum where the rod is not massless
or the bob is extended.
T ≈ 2π√(I/mgL)
g is gravitational acceleration = 9.8 m/s²
I is moment of inertia around the pivot in kg·m²
m is the mass of the pendulum in kg
L is the distance from the pivot to the center of
   mass of the pendulum in meters

Moment of inertia, I, in kg·m²
I = cMR²
    M is mass (kg), R is radius (meters)
    c = 1 for a ring or cylinder
    c = 2/5 solid sphere
    c = ½ solid cylinder or disk around its center
    c = 1/12 rod around its center, R = length