# Finding the period of a simple pendulum given angular acceleration

1. Dec 23, 2011

### guss

1. The problem statement, all variables and given/known data
A physical pendulum oscillates through small angles about the vertical with the angle, measured in radians, obeying the differential equation d2θ/dt2 = -4πθ. What is the period of the oscillation?

2. Relevant equations
T = 2π √(L/g)

3. The attempt at a solution
I attempted to integrate twice which should give me θ. This did not work, though, because the acceleration is defined by θ.

Thanks!

2. Dec 23, 2011

### BruceW

try to think of a function which you could use for theta. (i.e. a substitution)

3. Dec 23, 2011

### guss

I am trying to think...
θ = acos(H/L)
But I don't think that's what you're thinking of.

4. Dec 23, 2011

### I like Serena

The differential equation for a pendulum (that belongs to your relevant equation) is:
$${d^2\theta \over dt^2} + {g \over L}\theta=0$$
(See http://en.wikipedia.org/wiki/Pendulum_(mathematics).)

Perhaps you can match that with your differential equation?

Otherwise you would need to learn how to solve the DE...

5. Dec 23, 2011

### guss

So from there,
$-4\pi\theta + {g \over L}\theta=0$

${L \over g}={1 \over 4\pi}$

$T=\sqrt{4\pi^{2} \over 4\pi}$

$T=\sqrt{\pi}$