2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ (Hint: The center of mass remains at rest.) $\endgroup$ Commented May 27, 2023 at 14:37
  • $\begingroup$ seems like my equations for hooks law is wrong $\endgroup$ Commented May 27, 2023 at 14:56

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ why did you multiply by the inverse of the mass matrix? $\endgroup$ Commented May 27, 2023 at 15:24
  • $\begingroup$ We want an equation of the form $\vec{\ddot{X}} + [\omega^2] \vec{X} = 0$. Starting from $[M] \vec{\ddot{X}} + [K] \vec{X} = 0$, multiplying by $[M]^{-1}$ gives the desired form. $\endgroup$
    – K. Jiang
    Commented May 27, 2023 at 15:30
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .