I.E Irodov Exercise 4.59

I just used the classical analytical solution that many people used. However, here's something more interesting.

I programmed a time-domain simulation of the motion. The main idea is to use two cartesian coordinates and compute the distance between the two blocks at each timestep, and compute the direction and magnitude of the spring force. It's relatively simple as a simulation.

Basically, the force that the spring imparts on block 1 and 2 are:

$s = (x-y) - l$

Where $x$ and $y$ are the coordinates of block 1 and 2 respectively. In the simulation code, the centre of the spring is at the origin.

$|\bold{\vec{F}}_s| = k(x-y-l)$

Hence:

$\bold{\vec{F}}_1 = k(x-y-l) \hat{i}$

$\bold{\vec{F}}_2 = -k(x-y-l) \hat{i}$

These are the two equations which will be solved numerically. Numerical integration does almost everything for you; it's crazy.

Here's a graph of the motion and the code:

It looks like the blocks oscillate back and forth, but ultimately the whole system moves to the right indefinitely, since there's no friction. I've been told that this is the realistic result assuming there's no friction.

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import math

time = 0

#timestep
deltaT = 10**-5

#variable initialisation

l = 10 #natural length
#masses of the left and right block respectively
m_1 = 5 
m_2 = 10
#spring constant
k = 10

#linear position of the left and right blocks respectively
x = -5
y = 5

#velocities of the left and right blocks
xDot = 10 #initial velocity of left block is 10m/s
yDot = 0

count = 0

#time domain simulation
while time <= 10:
  #Center of mass of system
  CM = (m_1*x + m_2*y)/(m_1 + m_2)

  if count % 1000 == 0:
    print(CM)

  #computing stretch of spring
  stretch = (y-x)-l
  springForce = k*stretch #spring force

  #forces acting on the blocks 
  F1 = k*stretch
  F2 = -k*stretch

  #acceleration and numerical integration
  xDotDot = F1/m_1
  yDotDot = F2/m_2

  xDot += xDotDot*deltaT
  yDot += yDotDot*deltaT

  x += xDot*deltaT
  y += yDot*deltaT

  time += deltaT
  count += 1

This particular problem can be solved just purely by dimensional analysis. For the dimensions of $\omega$ to be $T^{-1}$ , the value of $\alpha$ must be $\boxed{0.5}$ . You can check this yourself. This question can be solved without any challenge.

But solving this rigorously can be done as follows:

Let the coordinate of $m_1$ be $x_1$ and that of $m_2$ be $x_2$ . Let the natural spring length be $L_o$ . Applying Newton's second law to each of the masses gives:

$m_1\ddot{x}_1 = x(x_2-x_1-L_o)$ $m_2\ddot{x}_2 = - x(x_2-x_1-L_o)$

Dividing the first equation by $m_1$ and the second one by $m_2$ and subtracting the 1st equation from the second equation gives:

$\ddot{x}_2 - \ddot{x}_1 + x\left(\frac{1}{m_2}+\frac{1}{m_1}\right)(x_2-x_1-L_o)=0$

Let: $x_2-x_1-L_o = y$ $\implies \ddot{y} + \omega^2 y = 0$

$\omega^2 = \frac{x(m_1+m_2)}{m_1m_2}$