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Animation of swinging pendulum on spring

Nasser M. Abbasi

January 27, 2025   Compiled on January 27, 2025 at 4:32pm

Let original length of pendulum be \(L\) and extension at instance shown be \(x\). The position vector of bob is

\[ \bar {r}=\left ( L+x\right ) \bar {b}_{1}\]

Hence the velocity vector is

\begin{align*} \left ( \frac {d}{dt}\bar {r}\right ) _{A} & =\left ( \frac {d}{dt}\bar {r}\right ) _{B}+\bar {\omega }\times \bar {r}\\ \bar {v}_{A} & =\left ( \frac {d}{dt}\left ( L+x\right ) \bar {b}_{1}\right ) _{B}+\bar {\omega }\times \left ( \left ( L+x\right ) \bar {b}_{1}\right ) \\ & =\dot {x}\bar {b}_{1}+\dot {\theta }\bar {b}_{3}\times \left ( \left ( L+x\right ) \bar {b}_{1}\right ) \\ & =\dot {x}\bar {b}_{1}+\dot {\theta }\left ( L+x\right ) \bar {b}_{2}\end{align*}

And the acceleration is

\[ \left ( \frac {d}{dt}\bar {v}\right ) _{A}=\left ( \frac {d}{dt}\bar {v}\right ) _{B}+\bar {\omega }\times \bar {v}\]

But

\begin{align} \left ( \frac {d}{dt}\bar {v}\right ) _{B} & =\frac {d}{dt}\left ( \dot {x}\bar {b}_{1}+\dot {\theta }\left ( L+x\right ) \bar {b}_{2}\right ) _{B}\nonumber \\ & =\ddot {x}\bar {b}_{1}+\left ( \ddot {\theta }\left ( L+x\right ) +\dot {\theta }\dot {x}\right ) \bar {b}_{2}\tag {2}\end{align}

And

\begin{align} \bar {\omega }\times \bar {v} & =\bar {\omega }\times \left ( \dot {x}\bar {b}_{1}+\dot {\theta }\left ( L+x\right ) \bar {b}_{2}\right ) \nonumber \\ & =\dot {\theta }\bar {b}_{3}\times \left ( \dot {x}\bar {b}_{1}+\dot {\theta }\left ( L+x\right ) \bar {b}_{2}\right ) \nonumber \\ & =\dot {\theta }\dot {x}\bar {b}_{2}+\dot {\theta }\dot {x}\bar {b}_{2}-\dot {\theta }^{2}\left ( L+x\right ) \bar {b}_{1}\tag {3}\end{align}

Substituting (2,3) into (1) gives the acceleration of the bob in frame \(A\) as

\begin{align*} \bar {a}_{A} & =\ddot {x}\bar {b}_{1}+\ddot {\theta }\left ( L+x\right ) \bar {b}_{2}+\dot {\theta }\dot {x}\bar {b}_{2}+\dot {\theta }\dot {x}\bar {b}_{2}-\dot {\theta }^{2}\left ( L+x\right ) \bar {b}_{1}\\ & =\ddot {x}\bar {b}_{1}+\ddot {\theta }\left ( L+x\right ) \bar {b}_{2}+2\dot {\theta }\dot {x}\bar {b}_{2}-\dot {\theta }^{2}\left ( L+x\right ) \bar {b}_{1}\end{align*}

To obtain \(\bar {F}=m\bar {a}\) we now just need to find \(\bar {F}\). The force on bob is given by just the weight and the spring force acting on it. The spring force is proportional to spring coefficient \(k\) times the extension. \(F_{s}=-kx\). Hence the force vector is

\[ \bar {F}=mg\cos \theta \bar {b}_{1}-mg\sin \theta \bar {b}_{2}-kx\bar {b}_{1}\]

Therefore the equation of motion is

\begin{align*} \bar {F} & =m\bar {a}\\ mg\cos \theta \bar {b}_{1}-mg\sin \theta \bar {b}_{2}-kx\bar {b}_{1} & =m\left ( \ddot {x}\bar {b}_{1}+\ddot {\theta }\left ( L+x\right ) \bar {b}_{2}+2\dot {\theta }\dot {x}\bar {b}_{2}-\dot {\theta }^{2}\left ( L+x\right ) \bar {b}_{1}\right ) \\ g\cos \theta \bar {b}_{1}-g\sin \theta \bar {b}_{2}-\frac {k}{m}x\bar {b}_{1} & =\ddot {x}\bar {b}_{1}+\ddot {\theta }\left ( L+x\right ) \bar {b}_{2}+2\dot {\theta }\dot {x}\bar {b}_{2}-\dot {\theta }^{2}\left ( L+x\right ) \bar {b}_{1}\end{align*}

Or, by equating each vector component

\begin{align*} \ddot {x}-\dot {\theta }^{2}\left ( L+x\right ) +\frac {k}{m}x & =g\cos \theta \\ \ddot {\theta }\left ( L+x\right ) +2\dot {\theta }\dot {x} & =-g\sin \theta \end{align*}

Or

\begin{align*} \ddot {x} & =\dot {\theta }^{2}\left ( L+x\right ) -\frac {k}{m}x+g\cos \theta \\ \ddot {\theta } & =-\frac {2\dot {\theta }\dot {x}}{L+x}-\frac {g}{L+x}\sin \theta \end{align*}

Let initial conditions be \(L=1,x\left ( 0\right ) =0.1,\dot {x}\left ( 0\right ) =0,\theta \left ( 0\right ) =20^{0},\dot {\theta }\left ( 0\right ) =.1\) (rad/sec).

We can now solve for \(x,\theta \) as function of time and make the animations. We can assume values for \(m,k\) and \(g=9.81\).

The solution will give \(x\left ( t\right ) ,\theta \left ( t\right ) \). To plot the path, we need to express \(x\left ( t\right ) \) in frame \(A\) coordinates of course which is the fixed frame. But this is easy, since the pendulum frame \(B\) is fixed at the base on the frame A origin.

The following is plot of the solution using \(k=0.99,m=0.09\)

The following is a quick animation of the above

PIC

The following is the code used 

(*Nasser M. Abbasi, Jan 25, 2025*) 
 
SetDirectory[NotebookDirectory[]] 
 
L = 1; 
k = .99; 
m = .09; 
g = 9.81; 
ode1 = x''[t] == z'[t]^2*(L + x[t]) + g*Cos[z[t]] - k/m*x[t]; 
ode2 = z''[t] == -2 z'[t]*x'[t]/(L + x[t]) - g*Sin[z[t]]/(L + x[t]); 
IC = {x[0] == .1, x'[0] == 0, z[0] == 20 Degree, z'[0] == .1}; 
sol = NDSolve[{ode1, ode2, IC}, {x, z}, {t, 0, 100}] 
 
Manipulate[ 
 Module[{currentX, currentY}, 
  currentX = (L + Evaluate[x[t0] /. sol])*Sin[Evaluate[z[t0] /. sol]]; 
  currentY = (L + Evaluate[x[t0] /. sol])*Cos[Evaluate[z[t0] /. sol]]; 
  Grid[{ 
    {Graphics[{ 
       Line[{{0, 0}, {First@currentX, -First@currentY}}], 
       {Blue, Disk[{First@currentX, -First@currentY}, .1]} 
       }, 
      PlotRange -> {{-2, 2}, {-4, 1}}, GridLines -> Automatic, 
      GridLinesStyle -> LightGray 
      ]}}] 
  ], 
 {{t0, 0, "time"}, 0, 10, .01}, 
 TrackedSymbols :> {t0} 
 ]