Question

derive hamiltonians equation of motion and used it to obtain time period of a simple pendulum.

Answer 1

coordinate that varies in this problem is θθ (rr is fixed). The mass, on a circular path, has speed v=rθ˙v=rθ˙, hence we can write down the kinetic energy immediately,

T=12mr2θ˙2T=12mr2θ˙2

The potential is proportional to the height, h=rcosθh=rcos⁡θ, and if we set

U(π2)=0U(π2)=0

then we get

U(θ)=−mgrcosθU(θ)=−mgrcos⁡θ

So now the Lagrangian is

L(θ,θ˙)=T−U=12mr2θ˙2+mgrcosθL(θ,θ˙)=T−U=12mr2θ˙2+mgrcos⁡θ

The momentum conjugate to θθ is

πθ=∂L∂θ˙=mr2θ˙πθ=∂L∂θ˙=mr2θ˙

and

∂L∂θ=−mgrsinθ∂L∂θ=−mgrsin⁡θ

Thus, the equations of motion are

0=dπθdt−∂L∂θ=mr2θ¨+mgrsinθ0=dπθdt−∂L∂θ=mr2θ¨+mgrsin⁡θ

or

θ¨=−grsinθθ¨=−grsin⁡θ

Now, the Hamiltonian for the pendulum is

H=πθθ˙−L=12mr2θ˙2−mgrcosθ=T+UH=πθθ˙−L=12mr2θ˙2−mgrcos⁡θ=T+U

which is the energy, as usual.