PX285 - G3 - derivation from examples

note: from lecture
often, mechanical systems are intrinsically coupled, and the displacement degrees of freedom move 'together' in a mode that has a characteristic frequency

particle attached to a spring

the lagrangian:

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

the euler-lagrange equation:

m \ddot{x} = - k x ⟹ \ddot{x} = - ω^{2} x

where, $ω = \sqrt{\frac{k}{m}}$

the solution:

x = A \cos ω t + B \sin ω t = C \exp (i ω t)

two particles

the solutions:

\begin{aligned} x_{1} & = D_{1} \exp (i ω_{1} t) \\ x_{2} & = D_{2} \exp (i ω_{2} t) \end{aligned}

the euler lagrange equations:

\begin{aligned} m_{1} {\ddot{x}}_{1} & = - k_{1} x_{1} \\ m_{2} {\ddot{x}}_{2} & = - k_{2} x_{2} \end{aligned}

in matrix form:

(\begin{matrix} m_{1} {\ddot{x}}_{1} \\ m_{2} {\ddot{x}}_{2} \end{matrix}) = (\begin{matrix} - k_{1} x_{1} \\ - k_{2} x_{2} \end{matrix})

to isolate $\ddot{x}$ an $x :$

(\begin{matrix} m_{1} & 0 \\ 0 & m_{2} \end{matrix}) (\begin{matrix} {\ddot{x}}_{1} \\ {\ddot{x}}_{2} \end{matrix}) = (\begin{matrix} - k_{1} x_{1} \\ - k_{2} x_{2} \end{matrix})

the matrix is called the inertia matrix, $M$
also:

(\begin{matrix} k_{1} & 0 \\ 0 & k_{2} \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} k_{1} x_{1} \\ k_{2} x_{2} \end{matrix})

this matrix is called the stiffness matrix, $K$
therefore:

M \vec{\ddot{x}} = - K \vec{x}

solutions:

\begin{array}{r} \vec{x} (t) = (\begin{array}{c} 1 \\ 0 \end{array}) D_{1} \exp (i ω_{1} t) \\ \vec{x} (t) = (\begin{array}{c} 0 \\ 1 \end{array}) D_{2} \exp (i ω_{2} t) \end{array}

they give the vibrations of the first and the second spring respectively
the general solution is a sum of the two:

\vec{x} (t) = (\begin{matrix} 1 \\ 0 \end{matrix}) D_{1} \exp (i ω_{1} t) + (\begin{matrix} 0 \\ 1 \end{matrix}) D_{2} \exp (i ω_{2} t)