Wedge, Block, and Spring

Let the horizontal coordinate of the bottom left vertex of the wedge be $x_0$ . Let the distance of the block up the ramp be $\alpha$ . The coordinates of the block are:

$x = x_0 + \alpha \cos \theta \\ y = \alpha \sin \theta$

Let $x_0 = 0$ be the initial location of the wedge. The kinetic energy, potential energy, and lagrangian are:

$T = \frac{1}{2} m (\dot{x}_0^2 + 2 \cos \theta \, \dot{x}_0 \dot{\alpha} + \dot{\alpha}^2) + \frac{1}{2} M \dot{x}_0^2 \\ V = m g \alpha \sin \theta + \frac{1}{2} k x_0^2 \\ L = T - V = \frac{1}{2} m (\dot{x}_0^2 + 2 \cos \theta \, \dot{x}_0 \dot{\alpha} + \dot{\alpha}^2) + \frac{1}{2} M \dot{x}_0^2 - m g \alpha \sin \theta - \frac{1}{2} k x_0^2$

Equations of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}_0}} = \frac{\partial{L}}{\partial{x}_0} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\alpha}}} = \frac{\partial{L}}{\partial{\alpha}}$

Evaluating the equations of motion results in:

$(m+M) \ddot{x}_0 + m \cos \theta \, \ddot{\alpha} = - k x_0 \\ m \cos \theta \, \ddot{x}_0 + m \ddot{\alpha} = - m g \sin \theta$

Solve for both second derivatives on each time step of a numerical simulation routine. The initial conditions are:

$x_0 = 0 \\ \dot{x}_0 = 0 \\ \alpha = \frac{h}{\sin \theta} \\ \dot{\alpha } = 0$

Plots of $x_0$ and $\alpha$ over time are given below. The simulation is run until $\alpha = 0$ .

Simulation code:

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import math
import numpy as np

h = 10.0
M = 1.0
theta = math.pi/6.0

m = 1.0
g = 10.0
k = 4.0

dt = 10.0**(-5.0)

#############################

x0 = 0.0

# alpha*math.sin(theta) = h
alpha = h/math.sin(theta)

U0 = m*g*h

x0d = 0.0
alphad = 0.0

x0dd = 0.0
alphadd = 0.0

t = 0.0
count = 0

#############################

print "t x0 alpha"

while alpha >= 0.0:

    x0 = x0 + x0d*dt
    alpha = alpha + alphad*dt

    x0d = x0d + x0dd*dt
    alphad = alphad + alphadd*dt

    J11 = m + M
    J12 = m*math.cos(theta)

    J21 = m*math.cos(theta)
    J22 = m

    J = np.array([[J11,J12],[J21,J22]])
    vec = np.array([-k*x0,-m*g*math.sin(theta)])

    Sol = np.linalg.solve(J, vec)

    x0dd = Sol[0]
    alphadd = Sol[1]

    t = t + dt
    count = count + 1

    if count % 100 == 0:
        print t,x0,alpha

#############################

print ""
print ""

vx = x0d + alphad*math.cos(theta)
vy = alphad*math.sin(theta)

print dt
print t
print ""
print x0
print x0d
print ""
print vx
print vy
print ""

E = 0.5*m*(vx**2.0 + vy**2.0) + 0.5*M*(x0d**2.0) + 0.5*k*(x0**2.0)
print (E/U0)
print ""
print ""

#############################

I considered the X-axis to be placed horizontally to the left and the Y-axis vertically upwards. The origin is at the initial position of the wedge.

The coordinates of the wedge at any instant of time are $(x_1,y_1)=(x,0)$

Let the distance moved downward by the block along the hypotenuse be $s$ , then the coordinates of the block are:

$(x_2,y_2)=(h - s \sin{\theta}, x + s \cos{\theta})$

The kinetic and potential energies of the system are:

$T = \frac{1}{2}M\left(\dot{x}_1^2 + \dot{y}_1^2\right) + \frac{1}{2}m\left(\dot{x}_2^2 + \dot{y}_2^2\right)$ $V = mgy_2 + \frac{k}{2}x^2$

The Lagrangé equations give two linear differential equations as such:

$\ddot{x} = -\frac{16x}{5}-2\sqrt{3}$ $\ddot{s} = \frac{8\sqrt{3}x}{5} +8$

Now:

$x(0)=s(0)=\dot{x}(0)=\dot{s}(0)=0$

Applying and solving:

$\boxed{x = \frac{5\sqrt{3}}{8} \left(\cos \left(\frac{4t}{\sqrt{5}}\right) -1\right)}$

The solution for $s$ can be obtained by substituting $x$ into the equation for $\ddot{s}$ and double integrating twice. This gives:

$\boxed{s = -\frac{15 \cos \left(\frac{4t}{\sqrt{5}}\right) -40t^2-15}{16}}$

A plot of $s$ vs. $x$ as plotted by @Steven Chase will give the required answer. However, the plot will be scaled by -1 in this case as I chose the axes to be differently oriented.