Finding the period of a simple pendulum given angular acceleration

[guss]

Homework Statement

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?

Homework Equations

T = 2π √(L/g)

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!

 

[BruceW]

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

 

[guss]

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

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

Another hint please?

 

[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...

 

[guss]

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...

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

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

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

$T=\sqrt{\pi}$

Which is the answer! Thanks!