Applied Mathematics: Two particles of mass $m$ moving smoothly along the $x$-axis connected by a spring?

Question

Two particles of mass $m$ can move smoothly along the $x$-axis and are connected by a spring of natural length $l$ and spring constant $k$.

Here are the questions and my proposed answers. I'm stuck on the last question. I really hope you can help me.

1) If the coordinates of the particles are $x_1$ and $x_2$, show that the potential energy stored in the spring is $\frac{1}{2}k(\space x_2-x_1-l)^2$.

Answer: $PE_{spring}=\frac{1}{2}kx^2$ where $x$ is the extension of the spring relative to equilibrium.

$$x=(x_2-x_1-l)$$ $$\implies \space V=\frac{1}{2}k(\space x_2-x_1-l)^2 .$$

2) Hence deduce the Lagrangian of the system.

Answer: $L=T-V$ where $T$= kinetic energy. $$T=\frac{1}{2}m(\dot {x}_1^2 +\dot {x}_2^2)$$

Substituting into $L=T-V$ we have

$$L=\frac{1}{2}m(\dot {x}_1^2 +\dot {x}_2^2)-\frac{1}{2}k(x_2-x_1-l)^2.$$

3) Write down the Lagrange equations of motion and show that the general solution satisfies the condition $x_1+x_2=Ct +D, x_1-x_2=-l+A\cos(\omega t)+B\sin(\omega t)$ where $\omega =\sqrt{\frac{2k}{m}}$. And hence describe the motion of the system.

I'm really stuck on this question. Any help would be much appreciated.

Answer 1

You should first obtain the Lagrange equations: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=m\ddot{x}_1-k(x_2-x_1-l)=0 $$ $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_2}-\frac{\partial L}{\partial x_2}=m\ddot{x}_2+k(x_2-x_1-l)=0 $$ By summing the two equations we obtain: $$ m(\ddot{x}_1+\ddot{x}_2)=0\implies x_1+x_2=Ct+D $$ By subtracting the two equations we obtain $$ m(\ddot{x}_1-\ddot{x}_2)-2k(x_2-x_1-l)=0 $$ and by setting $x=x_1-x_2$ we have $$ m\ddot{x}+2kx=2kl $$ which is a simple differential equation whose solution is an translated harmonic motion.

Answer 2

The Lagrange equation of motion can be found at wikipedia: $$ \frac{d}{dt} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) = \frac {\partial L}{\partial q_j} $$ In your case you have: $$ q_j = \{ x_1,x_2 \} $$ so by applying the Lagrange eq. to each $x_1$, $x_2$ we get:

$$ \frac{d}{dt} \left (m \dot{x_1}\right) = m \ddot{x_1} = k(x_2-x_1-l)$$ and $$ \frac{d}{dt} \left (m \dot{x_2}\right) = m \ddot{x_2} = -k(x_2-x_1-l) $$

So, if we add the two equations:

$$ m \ddot{x_1} + m \ddot{x_2} = k(x_2-x_1-l) + -k(x_2-x_1-l) $$ thus $$ m \ddot{x_1} + m \ddot{x_2} = m\frac{d^2}{d^2t}\left (x_1 + x_2\right) = 0$$

Which solve to the first condition by integrating twice (think $x_1 + x_2$ as a single variable): $$x_1+x_2=Ct +D$$

On the other hand, if we subtract both equations we get: $$ m \ddot{x_1} - m \ddot{x_2} = k(x_2-x_1-l) - -k(x_2-x_1-l) $$ thus $$ \ddot{x_1} - \ddot{x_2} = \frac{d^2}{d^2t}\left (x_1 - x_2\right) = -2\frac{k}{m}(x_2-x_1) -2\frac{k}{m}l $$

Which can be solved in the standard way as you would solve the equation $\ddot{y} = -ay - b$ by treating $x_1 - x_2$ as a single variable, and so you would arrive at the second condition. (You actually could just substitute the condition to prove it's validity instead of solving the differential equation.)

Hope this helps.