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On a smooth surface, a mass `m_1=2 kg` is attached to the first object by a spring with spring constant `k_1=4 N/m` . Another mass `m_2=1 kg` is...

The force on block 1 from the first spring is


`F_1=-k_1x`


Where `x` is the position to the right of the equilibrium position. Let y be the distance to the right of block 2's equilibrium position. Then the second spring exerts a force `F_2` on block 1 given by


`F_2=k_2(y-x)`


Think about it. If `x=y` there should be no force from spring 2. Also if `ygtx` then there should be a pull to the right (positive) on block 1. If `xlty` then the block should pull to the left (negative).


Since block 2 is only attached to spring 2, from Newton's third law


`F_3=-F2=-k_2(y-x)`


Now apply Newton's second law to both the blocks.


`m_1 (d^2x)/dt^2=F_1+F_2=-k_1x+k_2(y-x)`


`m_2 (d^2y)/dt^2=F_3=-k_2(y-x)`


Plug in the values for `k` and `m` , then move the terms to the left hand side.


`2(d^2x)/dt^2+6x-2y=0`


`(d^2y)/dt^2+2y-2x=0`


I'm going to set let `D:=d/dt` and solve this system of equations by the elimination method.


`(1):-gt (2D^2+6)x-2y=0`


`(2):-gt (D^2+2)y-2x=0`


Now multiply eq. `(1)` by `(D^2+2)` and multiply eq. `(2)` by `2` .


`(1):-gt(D^2+2)(2D^2+6)x-2(D^2+2)y=0`


`(2):-gt 2(D^2+2)y-4x=0`


Add eq. `(1)` and eq. `(2)` together to eliminate `y`. 


`(D^2+2)(2D^2+6)x-2(D^2+2)y+2(D^2+2)y-4x=0 `

`2D^4x+6D^2x+4D^2x+12x-2D^2y-4y+2D^2y+4y-4x=0`


`2D^4x+10D^2x+8x=0`


`(3):-gt 2(d^4x)/(dt^4)+10(d^2x)/(dt^2)+8x=0`


Try a solution of the form `x(t)=e^(rt)` .


Then eq. `(3)` takes the form


`2(r^4+5r^2+4)e^(rt)=0`


Solve for the roots of this characteristic equation.


`(r^4+5r^2+4)=0`


`(r^2+1)(r^2+4)=0`


The roots are `r=i, -i, 2i, -2i` .


Using Euler's formula, it follows that two linearly independent solutions are


`z_1(t)=e^(it)=cos(t)+isin(t)`


`z_2(t)=e^(2it)=cos(2t)+isin(2t)`


The general solution is a superposition of the imaginary and real parts of the linearly independent solutions.



`x(t)=a_1cos(t)+a_2sin(t)+a_3cos(2t)+a_4sin(2t) `

To find `y(t)` use earlier equation to put `y` in terms  of `x` ,


`2(d^2x)/dt^2+6x-2y=0`


`y(t)=(d^2x)/dt^2+3x`


Differentiate the general solution for `x(t)` twice and plug it in. You will find


`y(t)=2a_1cos(t)+2a_2sin(t)-a_3cos(2t)-a_4sin(2t)`


To determine the coefficients apply the initial conditions:


`x(0)=3`


`(dx)/(dt)=0`


`y(0)=3`


`(dy)/(dt)=0`


This will yield a system of 4 equations. You should find that,


`x(t)=2cos(t)+cos(2t)`


`y(t)=4cos(t)-cos(2t)`



x(t) is in black and y(t) in red.

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