Two masses $m_1$ and $m_2$, constrained to the $x$-axis, are coupled by a light spring of stiffness $s$ and natural length $l$. If $x$ is the extension of the spring and $x_1$ and $x_2$ indicate the positions of the two masses ($x_2 > x_1$), show that the equations of motion of the masses along the $x$-axis are:

$$m_1\ddot{x}_1 = sx$$ and $$m_2\ddot{x}_2 = −sx$$

and combine these to show that the system oscillates with a frequency:

$$ω = \sqrt{\frac{s}{\mu}}$$

where $\mu = \frac{m_1m_2}{m_1 + m_2}$ is the reduced mass.

Attempt: So I know $F = ma = -kx$ and acceleration is the second derivative of $x$ so $a_1 = \ddot{x}_1$ and $a_2 = \ddot{x}_2$. So that gives the first two equations: $m_1\ddot{x}_1 = sx$ and $m_2\ddot{x}_2 = −sx$ and they have different signs because they the force acts in opposite directions.

Then I know the extension of the spring $x = x_2 - x_1 - l$, so $m_1\ddot{x}_1 = s(x_2 - x_1 - l)$ and $m_2\ddot{x}_2 = -s(x_2 - x_1 - l)$, but then combining these all I seem to get is $m_1\ddot{x}_1 + m_2\ddot{x}_2 = 0$, so I'm not sure where to go from there?

So any help would be appreciated.


1 Answer 1


You have found the two differential equations $$\begin{align} m_1\ddot{x}_1&=s(x_2-x_1-l) \\ m_2\ddot{x}_2&=-s(x_2-x_1-l) \end{align} \tag{1}$$ for the unknown functions $x_1(t)$ and $x_2(t)$.

Because you already guess the solution will be an oscillatory motion, you can make the approach $$\begin{align} x_1(t)&=A_1\sin(\omega t) \\ x_2(t)&=l+A_2\sin(\omega t) \end{align} \tag{2}$$ where $A_1$, $A_2$ and $\omega$ are some still unknown constants.

Inserting the approach (2) into the differential equations (1) you get $$\begin{align} -m_1A_1\omega^2&=s(A_2-A_1) \\ -m_2A_2\omega^2&=-s(A_2-A_1) \end{align} \tag{3}$$

There are many different ways (basic and sophisticated) to solve this. A basic way is:

  • resolve one equation of (3) to $A_2$,
  • insert this $A_2$ into the other equation of (3),
  • resolve the resulting equation for $\omega^2$.

You will find the result to be $$\omega^2=s\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$$ and the constant $A_1$ will still be arbitrary.


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