3 Springs and 2 Mass System in a line

[Jwen556]

Homework Statement

I have 2 walls, 3 springs, and two masses. The masses and springs are not necessarily similar to the other(s).
They are connected in a configuration: Wall-Spring1-Mass1-Spring2-Mass2-Spring3-Wall.
I need to find equation for both masses, given that not all springs are in equilibrium when the system is set into motion.
Assume that for this system, there is one scenario in which the system is in its equilibrium state. In this state, x_1 and x_2 (defined below) are 0.

Springs constants: k_1, k_2, k_3
Masses: m_1, m_2
Displacement of mass 1 from equilibrium: x_1
Displacement of mass 2 from equilibrium: x_2

Homework Equations

F = -kx
U = 1/2 * kx^2
K = 1/2 * mv^2
E = U + K
dE/dt = 0 (closed system)

The Attempt at a Solution

I first calculated the forces on each mass for a given position and found that
F_1 = -k_1 * x_1 - k_2 * (x_1 - x_2)
F_2 = -k_3 * x_2 - k_2 * (x_2 - x_1)
Also, E = 1/2(∑kx^2 + ∑mv^2)

(' denote derivative with respect to time)
and dE/dT = k_1*x_1*x_1' + k_3*x_2*x_2' + k_2*(x_1 - x_2)(x_1' - x_2') + m_1*x_1'*x_1'' + m_2*x_2'*x_2'' = 0

But I'm not sure what to do from here.
First, are my equations for force correct? The middle spring (#2) applies the force F=-kx to two objects and I have only ever dealt with springs fixed on one end so I'm not sure.
Also the dE/dT = 0 is a nonlinear differential equation and I have only ever dealt with linear differential equations so I don't know how to solve it.

Would this be better in Physics homework help? I posted here since it is homework for a DE course.

 

[HallsofIvy]

You've started correctly, but I would not use "energy" here. Directly from "F= ma" you have the two equations for the motion of the two objects
m_1\frac{d^2x_1}{dt^2}= -k_1x_1- k_2(x_1- x_2)= -(k_1+ k_2)x_1+ k_2x_2
m_2\frac{d^2x_2}{dt^2}= -k_3x_2- k_2(x_2- x_1)= k_2x_1- (k_2+ k_3)x_3

Depending upon how you want to treat it, that can be reduced to a single fourth order equation or to four first order equations.

 

[Jwen556]

I have only ever worked with single second-order linear differential equations and I'm not sure how to approach reducing. I tried taking the derivative of the second equation with respect to time twice and substituting \frac{d^2x_1}{dt^2} = -\frac{k_1}{m_1}x_1 - \frac{k_2}{m_1}(x_1 - x_2) and get (after substituting, distributing minus signs, and multiplying both sides by m_1) m_1m_2\frac{d^4x_2}{dt^4} = -m_1(k_2 + k_3)\frac{d^2x}{dt^2} - k_2(k_1 + k_2)x_1 + k_x^2x_2 but there is still the x_1 term that I'm not sure how to get rid of.

edit; Ok, I substituted x_1 from the original equation for m_2\frac{d^2x_2}{dt^2} and got fourth order differential equations for both x_1 and x_2. Unsurprisingly, the coefficients were the same. I got
m_1m_2\frac{d^4x}{dt^4} + (m_1k_2 + m_1k_3 + m_2k_1 + m_2k_2)\frac{d^2x}{dt^2} + (k_1k_2 + k_1k_3 + k_2k_3)x = 0 in which x is either x_1 or x_2.

Assuming solution is e^{\lambda t} I get four lambdas so my solution is <br /> x = c_1e^{\lambda_1 t} + c_2e^{\lambda_2 t} + c_3e^{\lambda_3 t} + c_4e^{\lambda_4 t} but I only have two initials equations for each variable:
x(0) = x_i \quad and \quad x&#039;(0) = 0
so I have four variables and two equations which is not enough for me to find the coefficients

 

[haruspex]

Assuming solution is e^{\lambda t} I get four lambdas so my solution is <br /> x = c_1e^{\lambda_1 t} + c_2e^{\lambda_2 t} + c_3e^{\lambda_3 t} + c_4e^{\lambda_4 t} but I only have two initials equations for each variable:
x(0) = x_i \quad and \quad x&#039;(0) = 0
so I have four variables and two equations which is not enough for me to find the coefficients

There are two masses. The motions of each are controlled by the same four unknowns. For each mass you have an initial position and an initial velocity: four equations.

 

[Jwen556]

The coefficients on each exponent are not necessarily the same for x_1 and x_2 right?

 

[haruspex]

The coefficients on each exponent are not necessarily the same for x_1 and x_2 right?

Right, but they must be related. What happens if you substitute your generic x₁ solution back into the original two differential equations?