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I'm trying to understand the way my teacher found the lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equalibrium point is $(0,0,\frac{-mg}{k})$

The length of the relaxed spring is $0$.

The lagrangian with respect to the equilibrium point is

 $$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2. $$

I don't understand why the gravitational potential energy wasn't taken into account.

Can it be related to the given initial conditions? the mass starts at the equilibrium point with velocity $v$ to the right ($x$ direction).

I'm trying to understand the way my teacher found the Lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equilibrium point is $(0,0,\frac{-mg}{k})$.

The length of the relaxed spring is $0$.

The Lagrangian with respect to the equilibrium point is  $$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2. $$

I don't understand why the gravitational potential energy wasn't taken into account.

Can it be related to the given initial conditions? The mass starts at the equilibrium point with velocity $v$ to the right ($x$ direction).

I'm trying to understand the way my teacher found the lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equalibrium point is $(0,0,\frac{-mg}{k})$

The length of the relaxed spring is $0$.

The lagrangian with respect to the equalibrium point is

$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2 $

I don't understand why the gravitational potential energy was'nt taken into account.

Can it be realted to the given initial conditions? the mass starts at the equalibrium point with velocity $v$ to the right ($x$ direction).

I'm trying to understand the way my teacher found the lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equalibrium point is $(0,0,\frac{-mg}{k})$

The length of the relaxed spring is $0$.

The lagrangian with respect to the equilibrium point is

$$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2. $$

I don't understand why the gravitational potential energy wasn't taken into account.

Can it be related to the given initial conditions? the mass starts at the equilibrium point with velocity $v$ to the right ($x$ direction).

Im trying to understand the way my teacher found the lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equalibrium point is $(0,0,\frac{-mg}{k})$

The length of the relaxed spring is $0$.

The lagrangian with respect to the equalibrium point is

$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2 $

I dont understand why the gravitational potential energy was not taken into accouunt.

Can it be realted to the given initial conditions? the mass starts at the equalibrium point with velocity $v$ to the right ($x$ direction).

I'm trying to understand the way my teacher found the lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equalibrium point is $(0,0,\frac{-mg}{k})$

The length of the relaxed spring is $0$.

The lagrangian with respect to the equalibrium point is

$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2 $

I don't understand why the gravitational potential energy was'nt taken into account.

Can it be realted to the given initial conditions? the mass starts at the equalibrium point with velocity $v$ to the right ($x$ direction).