Elastic pendulum

The spring has the rest length ${\displaystyle l_{0}}$ and can be stretched by a length ${\displaystyle x}$. The angle of oscillation of the pendulum is ${\displaystyle \theta }$.

The Lagrangian ${\displaystyle L}$ is:

${\displaystyle L=T-V}$

where ${\displaystyle T}$ is the kinetic energy and ${\displaystyle V}$ is the potential energy.

Hooke's law is the potential energy of the spring itself:

${\displaystyle V_{k}={\frac {1}{2}}kx^{2}}$

where ${\displaystyle k}$ is the spring constant.

The potential energy from gravity, on the other hand, is determined by the height of the mass. For a given angle and displacement, the potential energy is:

${\displaystyle V_{g}=-gm(l_{0}+x)\cos \theta }$

where ${\displaystyle g}$ is the gravitational acceleration.

The kinetic energy is given by:

${\displaystyle T={\frac {1}{2}}mv^{2}}$

where ${\displaystyle v}$ is the velocity of the mass. To relate ${\displaystyle v}$ to the other variables, the velocity is written as a combination of a movement along and perpendicular to the spring:

${\displaystyle T={\frac {1}{2}}m({\dot {x}}^{2}+(l_{0}+x)^{2}{\dot {\theta }}^{2})}$

So the Lagrangian becomes:[1]

${\displaystyle L=T-V_{k}-V_{g}}$

${\displaystyle L[x,{\dot {x}},\theta ,{\dot {\theta }}]={\frac {1}{2}}m({\dot {x}}^{2}+(l_{0}+x)^{2}{\dot {\theta }}^{2})-{\frac {1}{2}}kx^{2}+gm(l_{0}+x)\cos \theta }$

With two degrees of freedom, for ${\displaystyle x}$ and ${\displaystyle \theta }$, the equations of motion can be found using two Euler-Lagrange equations:

${\displaystyle {\partial L \over \partial x}-{\operatorname {d} \over \operatorname {d} t}{\partial L \over \partial {\dot {x}}}=0}$

${\displaystyle {\partial L \over \partial \theta }-{\operatorname {d} \over \operatorname {d} t}{\partial L \over \partial {\dot {\theta }}}=0}$

For ${\displaystyle x}$:[1]

${\displaystyle m(l_{0}+x){\dot {\theta }}^{2}-kx+gm\cos \theta -m{\ddot {x}}=0}$

${\displaystyle {\ddot {x}}}$ isolated:

${\displaystyle {\ddot {x}}=(l_{0}+x){\dot {\theta }}^{2}-{\frac {k}{m}}x+g\cos \theta }$

And for ${\displaystyle \theta }$:[1]

${\displaystyle -gm(l_{0}+x)\sin \theta -m(l_{0}+x)^{2}{\ddot {\theta }}-2m(l_{0}+x){\dot {x}}{\dot {\theta }}=0}$

${\displaystyle {\ddot {\theta }}}$ isolated:

${\displaystyle {\ddot {\theta }}=-{\frac {g}{l_{0}+x}}\sin \theta -{\frac {2{\dot {x}}}{l_{0}+x}}{\dot {\theta }}}$

The elastic pendulum is now described with two coupled ordinary differential equations. These can be solved numerically. Furthermore, one can use analytical methods to study the intriguing phenomenon of order-chaos-order[6] in this system.