Chapter 7, Problem 7.8P

(a)

To determine

The langrangian for two particles of equal masses.

Answer to Problem 7.8P

The Lagrangian of the two particle system is, $L=\frac{1}{2}m{x}_1^2+\frac{1}{2}m{x}_2^2-\frac{1}{2}k{(x_1-x_2-l)}^2$.

Explanation of Solution

The potential energy of the system is,

  $U=\frac{1}{2}k{x}^2$.

Here $k$ is the spring constant and $x$ is the extension of the spring.

Below figure shows the arrangement of the two particles of equal masses.  These two particles are connected by a spring.

Here mass 1 $(m_1)$ always remains right of the mass 2$(m_2)$.  The masses of the two particles are equal.

  $m_1=m_2=m$.

The kinetic energy of the two masses is’

  $T=\frac{1}{2}{m}_1{x}_1^2+\frac{1}{2}{m}_2{x}_2^2$.

Here, $m_1$ is the mass of the first particle, $m_2$ is the mass of the second particle, $x_1$ is the velocity of the first particle and $x_2$ is the velocity of the second particle.

Substitute $m$ for $m_1$ and $m$ for $m_2$ in the equation $T=\frac{1}{2}m{x}_1^2+\frac{1}{2}m{x}_2^2$ to solve for $T$.

  $T=\frac{1}{2}m{x}_1^2+\frac{1}{2}m{x}_2^2$.

The extension of the spring is,

  $x=x_1-x_2-l$.

Here, $l$ is the spring’s upstretched length.

Substitute $x=x_1-x_2-l$ for $x$ in the equation $U=\frac{1}{2}k{x}^2$ and solve for $U$.

  $U=\frac{1}{2}k{(x_1-x_2-l)}^2$.

The Lagrangian of the system is,

  $L=T-U$.

Conclusion:

Substitute $\frac{1}{2}m{x}_1^2+\frac{1}{2}m{x}_2^2$ for $T$ and $\frac{1}{2}k{(x_1-x_2-l)}^2$ for $U$.

Therefore, the Lagrangian of the two particle system is:

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

(b)

To determine

The Lagrange equations for new variable

Explanation of Solution

The position of the centre of mass of the system is,

  $X=\frac{1}{2}(x_1+x_2)$.

Solve the equation for $x_1+x_2$.

  $(x_1+x_2)=2X$                          (1).

The extension of the spring is,

  $x=x_1-x_2-l$

Solve the equation for $x_1-x_2$.

  $x_1-x_2=x+l$        (2)

Add the equations (1) and (2) to solve for $x_1$.

  $\begin{array}{l}{x}_1+x_2+(x_1-x_2)=2X+(x+l)\\ 2x_1=2X+(x+l)\\ x_1=X+\frac{1}{2}(x+l)\end{array}$.

Substitute $X+\frac{1}{2}(x+l)$ for $x_1$ in the equation to solve for $x_2$.

  $\begin{array}{l}X+\frac{1}{2}(x+l)-x_2=x+1\\ x_2=X+\frac{1}{2}(x+l)-x+1\\ =X-\frac{1}{2}(x+l)\end{array}$.

The velocities of the first and second particle in terms of the position of the centre of mass of the system as follows.

  $\begin{array}{l}{\dot{x}}_1=\dot{X}+\frac{1}{2}\dot{x}\\ {\dot{x}}_2=\dot{X}-\frac{1}{2}\dot{x}\end{array}$

The langrangian of two particle system is,

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

Substitute $\dot{X}+\frac{1}{2}\dot{x}$ for ${\dot{x}}_1$, $\dot{X}-\frac{1}{2}\dot{x}$ for ${\dot{x}}_2$ and $x$  for $x_1-x_2-l$

  $\begin{array}{c}L=\frac{1}{2}m{\left(\dot{X}+\frac{1}{2}\dot{x}\right)}^2+\frac{1}{2}{\left(\dot{X}-\frac{1}{2}\dot{x}\right)}^2-\frac{1}{2}k{(x)}^2\\ =\frac{1}{2}m\left(2{\dot{X}}^2+2{\left(\frac{1}{2}\dot{x}\right)}^2\right)-\frac{1}{2}k{(x)}^2\\ =m{\dot{X}}^2+\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{(x)}^2\end{array}$

Conclusion:

Therefore, the Lagrangian of the system of two particles in terms of centre of mass of the system is

  $L=\frac{1}{2}m{\left(4{\dot{X}}^2+\dot{x}\right)}^2-\frac{1}{2}k{(x)}^2$

The lagrangian equation of motion for the two-particle system in terms of $X$ is as follows,

  $\frac{\partial L}{\partial X}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)$

Determine the values of $\frac{\partial L}{\partial X}$

  $\begin{array}{c}\frac{\partial L}{\partial X}=\frac{\partial }{\partial X}\left(m{\dot{X}}^2+\frac{1}{4}m{\dot{x}}^2-\frac{1}{2}k{x}^2\right)\\ =0\end{array}$

Determine the value of $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)$

  $\begin{array}{c}\frac{\partial L}{\partial \dot{X}}=\frac{\partial }{\partial \dot{X}}\left(m{\dot{X}}^2+\frac{1}{4}m{\dot{x}}^2-\frac{1}{2}k{x}^2\right)\\ =2m\dot{X}\\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)=\frac{d}{dt}\left(2m\dot{X}\right)\\ =2m\ddot{X}\end{array}$

Substitute $0$ for $\frac{\partial L}{\partial X}$ and $2m\ddot{X}$ for $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)$ in equation $\frac{\partial L}{\partial X}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)$ and solve.

  $\begin{array}{c}0=2m\ddot{X}\\ \ddot{X}=0\end{array}$

Thus, the lagrange’s equation of motion for $X$ is $\ddot{X}=0$.

The lagrangian equation motion for the two-particle system in terms of $x$ is a follows.    $\frac{\partial L}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$

Determine the value of $\frac{\partial L}{\partial x}$

  $\begin{array}{c}\frac{\partial L}{\partial x}=\frac{\partial }{\partial \dot{x}}\left(m{\dot{X}}^2+\frac{1}{4}m{\dot{x}}^2-\frac{1}{2}k{x}^2\right)\\ =-\frac{1}{2}\left(2kx\right)\\ =-kx\end{array}$

Determine the value of $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$

  $\begin{array}{c}\frac{\partial L}{\partial \dot{x}}=\frac{\partial }{\partial \dot{x}}\left(m{\dot{X}}^2+\frac{1}{4}m{\dot{x}}^2-\frac{1}{2}k{x}^2\right)\\ =-\frac{1}{4}m\left(2\dot{x}\right)\\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}}\right)=\frac{d}{dt}\left(\frac{1}{4}m\left(2\dot{x}\right)\right)\\ =\frac{1}{2}m\ddot{x}\end{array}$

Substitute $-kx$ for $\frac{\partial L}{\partial x}$ and $\frac{1}{2}m\ddot{x}$ for $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$ in equation $\frac{\partial L}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$ and solve.

  $\begin{array}{c}-kx=\frac{1}{2}m\ddot{x}\\ \dot{x}+\left(\frac{2k}{m}\right)x=0\end{array}$

Thus, the lagrange’s equation of motion for $x$ is $\dot{x}+\left(\frac{2k}{m}\right)x=0$$\ddot{X}=0$.

(c)

To determine

Solve for $X\left(t\right)$ and $x\left(t\right)$

Answer to Problem 7.8P

The value for $X\left(t\right)$ is $X\left(t\right)=v_{0}t+X_c$

Explanation of Solution

Integrate the equation $\ddot{X}=0$ and solve for $\dot{X}$ .

  $\dot{X}\left(t\right)=v_0$.

Here, $v_0$ is constant.

Integrate the equation solve $X\left(t\right)$,

  $X\left(t\right)=v_{0}t+X_c$                          (1).

Here, $X_c$ is constant.

Hence the value for $X\left(t\right)$ is $X\left(t\right)=v_{0}t+X_c$.

Conclusion:

Solve for the solution for the lagrange;s equation for $x$ is $x\left(t\right)=A\cos \left(\omega t+\varphi \right)$

Compare the above equation of motion $\ddot{x}+\left(\frac{2k}{m}\right)x=0$ for $x$ with general equation of motion $\ddot{x}+{\omega }^{2}x=0$ to obtain the angular frequency of oscillation.

So, the relative position oscillates with angular frequency $\omega =\sqrt{\frac{2k}{m}}$.

The centre of mass of the system moves as a free particle. This is because of the no external force acting on the system. Hence, two particles oscillate relative to each other.