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Two masses $m_1$ and $m_2$ are joined by a spring of spring constant $k$. I am trying to show that the frequency of vibration of these masses along the line connecting them is: $$\omega = \frac{\sqrt {k (m_1+m_2)}}{\sqrt {m_1m_2}}.$$
I was trying to find the formula using Hooke’s law in the system:
$$(m_1+m_2)a = -kx$$ $$a = \frac{-k}{(m_1+m_2)}x$$
by definition $\sqrt{k/(m_1+m_2)} = \omega$
but this is not what the book is asking me to show.
Your force balance is incorrect. There are two degrees of freedom in this problem: the displacements of the two masses $x_1$ of $m_1$ and $x_2$ of $m_2$, referenced about an equilibrium. Presented below is a general approach to finding the natural frequencies.
Consider a force balance on $m_1$. If $x_1 > 0$ while $m_2$ is fixed at an equilibrium location ($x_2 = 0$), then the spring will push $m_1$ to the left. So we have $m_1 \ddot{x}_1 = -k (x_1 - x_2)$. In the same way, if $x_2 > 0$ while $m_1$ is fixed at an equilibrium location ($x_1 = 0$), the spring will push $m_2$ to the left. So, $m_2 \ddot{x}_2 = -k (x_2 - x_1)$.
From this we obtain the linear system of differential equations
$$\begin{pmatrix} m_1 & 0 \\ 0 & m_2 \end{pmatrix} \begin{pmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{pmatrix} + \begin{pmatrix} k & -k \\ -k & k \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0$$
Multiplying through by the inverse of the "mass" matrix, we get
$$\begin{pmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{pmatrix} + \begin{pmatrix} \frac{k}{m_1} & -\frac{k}{m_1} \\ -\frac{k}{m_2} & \frac{k}{m_2} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0$$
The squared natural frequencies are the eigenvalues of the matrix in the equation above. So, we diagonalize. For eigenvalue $\lambda$, the following is satisfied:
$$0 = |A - \lambda I| = \lambda^2 - k \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \lambda $$
$$\implies 0 = \lambda \left( \lambda - k \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \right)$$
This shows that $\omega^2 = 0$ is one natural frequency and $\omega^2 = k \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \frac{k (m_1 + m_2)}{m_1 m_2}$ is another. The first case indicates the presence of a rigid-body mode - indeed, the center of mass of the system is not fixed, so the system is free to arbitrary translation. The other case is the desired result.
You can also solve this without the eigenvalue approach. Catering to your hint, let $m_1$ and $m_2$ be connected to a spring of spring constant $k$. Suppose $m_1$ is at the origin and $m_2$ is some location to the right. Displace $m_1$ by some distance $x_1$ to the right and $m_2$ some distance $x_2$ to the right with $x_1<x_2$.Then the spring with be stretched by a length of $x_2-x_1$. By Newton's Laws, there is a force to pointed to the right for $m_1$ giving
$$k(x_2-x_1)=m_1a_1$$ Similarly, there is a force to the left for $m_2$ such that
$$-k(m_2-m_1)=m_2a_2$$
$m_1$ and $m_2$ have different accelerations because they do not move together, otherwise $\omega = 0$. We can calculate the center of mass of the system right before releasing the mass. This is
$$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$$
Pick your favorite mass. Mine is $m_1$. So, I will solve for $m_2$ and substitute into the equation involving the acceleration of $m_1$.
$$x_2=\frac{x_{cm}(m_1+m_2)-m_1x_1}{m_2}$$
Substituting, $$k\left(\frac{x_{cm}(m_1+m_2)-m_1x_1}{m_2}-x_1\right)=m_1a_1$$ or $$kx_{cm}(x_1+x_2)=(m_1+m_2)x_1+m_1m_2a_1$$
Here is where you impose your hint that $x_{cm}$ is fixed. What does this mean for the equation above? Also, consider why I needed to do all this algebra and what my goal was.
