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In according with Landau's Mechanics the number of independent integrals of motion for a closed mechanical system with $s$ degrees of freedom is $2s-1$. We can express the $2s-1$ arbitrary constants $C_1,C_2,...,C_{2s-1}$ as functions of $q$ and $\dot{q}$, and these functions will be integrals of the motion.

Thus for a mass with a spring we have only 1 integral (the energy), but I do not understand how to obtained it, indeed for $m\ddot{x}+kx=0$ I can write: $x=A\cos{\omega t}+B\sin{\omega t}$ and $\dot{x}=-A\omega \sin{\omega t}+B\omega\cos{\omega t}$. How can I express $A=f_1(q,\dot{q})$ and $B=f_2(q,\dot{q})$ if, for istance, $A=x_0$ and $B=0$?

There is a similar question in What is the standard procedure to form the first integrals of a motion? but I don't understand the answer.

Qmechanic
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Paolo Secchi
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1 Answers1

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your example:

$$\ddot x+\omega_0^2\,x=0$$

with $~x(0)=A~,\dot x(0)=B~$ you obtain the solution

$$x={\frac {B\sin \left( \omega_{{0}}t \right) }{\omega_{{0}}}}+A\cos \left( \omega_{{0}}t \right) \quad \Rightarrow\\ v=\dot x=B\cos \left( \omega_{{0}}t \right) -A\sin \left( \omega_{{0}}t \right) \omega_{{0}} $$

solve those two equations for $~A~,B~$

$$A=-{\frac {v\sin \left( \omega_{{0}}t \right) -x\omega_{{0}}\cos \left( \omega_{{0}}t \right) }{\omega_{{0}}}} \\ B=\sin \left( \omega_{{0}}t \right) x\omega_{{0}}+v\cos \left( \omega_{{0 }}t \right) $$ thus $~A=f_A\,(q,\dot q)~$ and $~B=f_B\,(q,\dot q)~$ where $~x=q~$ and $~v=\dot q$

Eli
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