# Simple Pendulum

1. Jan 19, 2010

In the book "Fundamentals of Physics" by H D Young and Freedman,there is formula about the time period of simple pendulum which uses sine function in a complex way which i think is obtained by series expansion of sine function either by Maclurin or Tayler series.
Can any one please tell me how to derive the formula?

2. Jan 19, 2010

### elibj123

Can you please post the formula here so we'll have a more clear idea of what you are talking about?

3. Jan 19, 2010

### Staff: Mentor

What Young and Freedman show is the period of a simple pendulum as a function of the initial angle (in other words, for angles that do not satisfy the usual small angle approximation). They present a series approximation to the exact solution, which involves solving an elliptic integral. (It's not a simple Taylor series expansion of a sine function.)

Last edited by a moderator: Apr 24, 2017
4. Jan 19, 2010

### Bob S

The time period is

t = sqrt(b/g) ∫oπ/2 dφ/sqrt[1-k2sin2(φ)]

where k= sin(θ/2), b= radius of pendulum, θ= max angle, and the integral is the complete elliptic integral of the first kind..

Bob S

5. Jan 21, 2010

### ftfaaa

According to Rigid-body rotation Law ，you can get a formula
mglsinθ=-ml^2*(d2θ/dt2)
m is the mass of the ball ,l is the length of the rope and θ is the angle between the rope and vertical.
According to Talyor series ,when θ is very small,sinθ=θ.
Then it turns to (d2θ/dt2)+g/l *θ=0,and it is a Second-order differential equations with constant coefficients. The result is θ=Asin(sqr(g/l)*t) A is Amplitude
It's clearly that Vibration cycle is 2π*sqr(l/g)

6. Jan 21, 2010

### Staff: Mentor

We're talking about the case where θ is not small enough for that approximation.

7. Jan 21, 2010

### Bob S

The above answer is actually for a quarter period. The full period is 4 times the above answer:

T = 4·sqrt(b/g) ∫oπ/2 dφ/sqrt[1-k2sin2(φ)]

Bob S