Advanced spring mechanics

I will going to use the Newtonian model for finding out the forces acting on the block, so let's first draw a free body diagram for one of the springs:

Let's call the force the spring applies on the block directly $\bold{F_s}$ . The vertical and horizontal components respectively are:

$\bold{F_y} = \bold{F_s} \sin(\theta)$ and $\bold{F_x} = \bold{F_s} \cos(\theta)$

The vertical component of the forces exerted by both springs are opposite in direction and cancel each other out.

Where $\displaystyle \cos(\theta) = \frac{L}{ \sqrt{L^2+x^2}}$

A single spring's stretch is equal to $\sqrt{L^2+x^2} - L$ , so the force a single spring exerts on the block is $k(\sqrt{L^2+x^2} - L)$

So the force both springs exert on the object is $m \ddot{x} = -2kx \left( 1 - \Large{\frac{1}{\sqrt{L^2+x^2}}} \right)$

Since the block is of unit mass, the initial acceleration is also $-2kx \left( 1 - \Large{\frac{1}{\sqrt{L^2+x^2}}} \right) = 4.10266$ m/ $s^2$

@Karan Chatrath analytically found out the integral which gives the time it takes for the block to reach the origin by integrating $\Large \frac{1}{\dot{x}}$ over distance. He did it by using energy conservation.

Here is my numerical simulation which uses Explicit Euler numerical integration:

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import math

#double spring system

time = 0
deltaT = 10**-7 #timestep 1e-07

x = 3 #initialisation of position
xDot = 0 #initialisation of velocity

while x >= 0: #simulating till block hits the origin

    #Explicit Euler numerical integration

    xDotDot = -2*x*(1-(1/math.hypot(1,x))) #differential equation
    xDot += xDotDot*deltaT #numerical extrapolation of velocity
    x += xDot*deltaT #numerical extrapolation of position

    time +=deltaT #updating value of time


print("numerical simulation timestep: " +str(deltaT))
print("time taken for block to reach the origin: " + str(time))

#time taken for block to reach the origin: 1.413782

I agree with @Steven Chase . A nice problem indeed. Also posting my solution, which is a semi-analytical approach.

Consider the block at a general time instant $t$ . Its kinetic energy is:

$T = \frac{1}{2}m\dot{x}^2$

The system is symmetrical about the X-axis and the springs are identical. The potential energy of each spring is the same and the total potential energy of the system is therefore:

$V = 2\left(\frac{1}{2}K\left(\sqrt{x^2+1}-1\right)^2\right)$ $V = K\left(\sqrt{x^2+1}-1\right)^2$

The total energy of the system is a constant which is equal to the initial potential energy of the system. Note that initially the particle is at $x=3$ and is released from rest. This gives:

$T + V = K\left(\sqrt{10}-1\right)^2$ $\implies \frac{1}{2}m\dot{x}^2 + K\left(\sqrt{x^2+1}-1\right)^2=K\left(\sqrt{10}-1\right)^2$ $\implies \frac{1}{2}\dot{x}^2 + \left(\sqrt{x^2+1}-1\right)^2=\left(\sqrt{10}-1\right)^2$ Re-arranging gives:

$\implies \dot{x} = -\sqrt{2\left(\left(\sqrt{10}-1\right)^2-\left(\sqrt{x^2+1}-1\right)^2\right)}$

The additional negative sign is included to imply that as time progresses, $x$ decreases.

Separating the variables and integrating gives the required time period:

$\int_{0}^{3} \frac{dx}{\sqrt{2\left(\left(\sqrt{10}-1\right)^2-\left(\sqrt{x^2+1}-1\right)^2\right)}} = T \approx 1.4138$

To find the initial acceleration, consider the energy conservation equation:

$\implies \frac{1}{2}\dot{x}^2 + \left(\sqrt{x^2+1}-1\right)^2=\left(\sqrt{10}-1\right)^2$

Differentiating the above with respect to time gives:

$\left(\ddot{x} + 2x\left(1 - \frac{1}{\sqrt{x^2+1}}\right)\right)\dot{x} = 0$

Since $\dot{x}$ cannot be zero, the term:

$\ddot{x} + 2x\left(1 - \frac{1}{\sqrt{x^2+1}}\right)=0$

At the initial instant when the block is just released, the magnitude of the acceleration of the block is obtained by substituting $x=3$ above, which gives:

$\lvert \ddot{x}(0) \rvert \approx 4.102633$

The required answer is, therefore:

$\boxed{\lvert \ddot{x}(0) \rvert + T \approx 5.5164}$

Nice problem. Simulation code attached below (with comments)

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import math

# Constants

m = 1.0  # mass
L0 = 1.0 # natural length
k = 1.0  # spring constant

dt = 10.0**(-6.0)  # time step

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

# Initialize time, position, velocity

t = 0.0

x = 3.0
xd = 0.0

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

# Initialize horizontal acceleration

L = math.hypot(L0,x)  # spring length
S = L - L0 # spring stretch
Fs = k*S   # individual spring force

Fx = -2.0*Fs*(x/L)  # total horizontal force

xdd = Fx/m  # horizontal acceleration

a = abs(xdd)  # magnitude of initial horizontal acceleration

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

while x >= 0.0:

    x = x + xd*dt        # numerical integration
    xd = xd + xdd*dt

    L = math.hypot(L0,x)  # spring length
    S = L - L0  # spring stretch
    Fs = k*S  # individual spring force

    Fx = -2.0*Fs*(x/L)  # total horizontal force

    xdd = Fx/m  # horizontal acceleration

    t = t + dt

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

print "time step"
print dt
print ""
print "initial acceleration"
print a
print ""
print "final time"
print t
print ""
print "initial acceleration + final time"
print (t+a)
print ""

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

# Results

#time step
#1e-05

#initial acceleration
#4.1026334039

#final time
#1.41378

#initial acceleration + final time
#5.5164134039

#>>> ================================ RESTART ================================
#>>> 
#time step
#1e-06

#initial acceleration
#4.1026334039

#final time
#1.41378199997

#initial acceleration + final time
#5.51641540387

>>>