Series Combination of Spring

Finding the equivalent spring and solving the 1DOF equation is a shortcut to a more complex problem. So if you want to know about the more complex approach, then read on.

The problem is a special case of the general problem with two springs and two masses. The special case is when one of the masses is zero.

So the general problem has mass $m_1$ (the black dot) in the end of $k_1$ and $m_2$ (the gray box) in the end of $k_2$.

The equations of motion in terms of the positions of the masses $x_1$ and $x_2$ are as follows:

$$ begin{aligned} m_1 ddot{x}_1 & = -k_1 x_1 + k_2 (x_2-x_1) m_2 ddot{x}_2 & = -k_2 (x_2-x_1) end{aligned} tag{1}$$

_{Note the op uses $x_1$ and $x_2$ for spring extensions, and I am using these variables as mass positions. The extension of the first spring equals $x_1$, but the extension of the second springs equals $x_2-x_1$.}

Any solution of the above is the superposition of the two natural frequency responses of the system. And there is a standard way to solve this problem using eigenvalues and eigenvectors and some linear algebra.

But in this case $m_1=0$ which makes the above equations equal to

$$ begin{aligned} 0 & = -k_1 x_1 + k_2 (x_2-x_1) m_2 ddot{x}_2 & = -k_2 (x_2-x_1) end{aligned} tag{2}$$

which makes them a DAE system (differential algebraic equations). Here you solve the first equation for $x_1$ and plug it into the second equation.

$$ m_2 ddot{x}_2 = -underbrace{left( frac{k_1 k_2}{k_1 + k_2} right)}_{k_{rm eq}} x_2 tag{3} $$

So the solution is the same as solving a 1DOF differential equation in terms of $x_2$ using the equivalent spring $k_{rm eq}$. Note that for each solution of $x_2$ the value of $x_1$ is found from (2) with $$x_1 = frac{k_2}{k_1+k_2} x_2 tag{4}$$

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Appendix I

The two natural frequencies of the general system are

$$ begin{aligned} omega_1^2 & = frac{ left( tfrac{k_1+k_2}{m_1}+tfrac{k_2}{m_2}right) - sqrt{ left( tfrac{k_1+k_2}{m_1}+tfrac{k_2}{m_2}right)^2 - tfrac{4 k_1 k_2}{m_1 m_2}}}{2} omega_2^2 & = frac{ left( tfrac{k_1+k_2}{m_1}+tfrac{k_2}{m_2}right) + sqrt{ left( tfrac{k_1+k_2}{m_1}+tfrac{k_2}{m_2}right)^2 - tfrac{4 k_1 k_2}{m_1 m_2}}}{2} end{aligned} tag{5} $$

For the case when $m_1=0$ then the above becomes

$$ begin{aligned} omega_1^2 & = pm infty omega_2^2 & = frac{tfrac{k_1 k_2}{k_1+k_2}}{m_2} end{aligned} tag{6}$$

Appendix II

The equations of motion in terms of spring extensions $x_1$ and $x_2$ are

$$ begin{aligned} m_1 ddot{x}_1 & = -k_1 x_1 - k_2 x_2 m_1 ddot{x}_1 + m_2 ddot{x}_2 & = -k_2 x_2 end{aligned} tag{7}$$

with the solution for $m_1=0$ as $$x_1 = frac{k_2}{k_1} x_2$$ and

$$ ddot{x}_1+ ddot{x}_2 = - frac{k_2}{m_2} x_2 $$

but since $x_1$ is a function of $x_2$ only, then $ddot{x}_2 = tfrac{k_2}{k_1} ddot{x}_1$ and the 1DOF equation becomes

$$ left( 1 + frac{k_2}{k_1} right) ddot{x}_2 = - frac{k_2}{m_2} x_2 $$

which becomes equivalent to (3) when solved for $ddot{x}_2$.