# Frequency of vibration of two masses connected by a spring?

## Two masses ${m}_{1}$ and ${m}_{2}$ are joined by a spring of spring constant $k$. Show that the frequency of vibration of these masses along the line connecting them is: $\setminus \omega = \setminus \sqrt{\setminus \frac{k \left({m}_{1} + {m}_{2}\right)}{{m}_{1} {m}_{2}}}$ So I have that the distance traveled by ${m}_{1}$ can be represented by the function ${x}_{1} \left(t\right) = A \cos \left(\setminus \omega t\right)$ and similarly for the distance traveled by ${m}_{2}$ is ${x}_{2} \left(t\right) = B \cos \left(\setminus \omega t\right)$. The force the spring exerts on these two masses is $- k {x}_{n} \left(t\right) = {m}_{n} \setminus \frac{{d}^{2} {x}_{n}}{{\mathrm{dt}}^{2}}$. But after, simplifying I have no idea how to combine these two functions.

Apr 27, 2017

$\setminus \omega = \setminus \sqrt{\setminus \frac{k \left({m}_{1} + {m}_{2}\right)}{{m}_{1} {m}_{2}}}$

#### Explanation:

The distance each mass is displaced can be represented by the following
${x}_{1} \left(t\right) = A \cos \left(\setminus \omega t\right)$
x_2(t)=Bcos(\omega t

Using the formula for simple harmonic oscillation $- k x = m \setminus \frac{{d}^{2} x}{{\mathrm{dt}}^{2}}$, you get the following
$k \left(B \cos \left(\setminus \omega t\right) - A \cos \left(\setminus \omega t\right)\right) = - {m}_{1} A \setminus {\omega}^{2} \cos \left(\setminus \omega t\right)$
$k \left(B \cos \left(\setminus \omega t\right) - A \cos \left(\setminus \omega t\right)\right) = {m}_{2} B \setminus {\omega}^{2} \cos \left(\setminus \omega t\right)$

Solving we get
$- {m}_{1} A \setminus {\omega}^{2} \cos \left(\setminus \omega t\right) = {m}_{2} B \setminus {\omega}^{2} \cos \left(\setminus \omega t\right)$
$- {m}_{1} A = {m}_{2} B$

Plugging this into the first equation for simple harmonic oscillation we get
$k \left(B \cos \left(\setminus \omega t\right) + \setminus \frac{{m}_{2} B}{{m}_{1}} \cos \left(\setminus \omega t\right)\right) = {m}_{2} B \setminus {\omega}^{2} \cos \left(\setminus \omega t\right)$

Simplify to get
$k \left(1 + \setminus \frac{{m}_{2}}{{m}_{1}}\right) = {m}_{2} \setminus {\omega}^{2}$
$k \left({m}_{1} + {m}_{2}\right) = {m}_{1} {m}_{2} \setminus {\omega}^{2}$
$\setminus {\omega}^{2} = \setminus \frac{k \left({m}_{1} + {m}_{2}\right)}{{m}_{1} {m}_{2}}$
$\setminus \omega = \setminus \sqrt{\setminus \frac{k \left({m}_{1} + {m}_{2}\right)}{{m}_{1} {m}_{2}}}$

Apr 27, 2017

This problem has only 1 dimension, so it seems simple as it is motion along the number line. But there are 2 variables (degrees of freedom).

The Lagrangian is often a safe (lazy) way to extract the DE's.

If ${x}_{1} \left(t\right)$ and ${x}_{2} \left(t\right)$ are the displacements of the masses from the equilibrium (no tension) position, then we have:

$T = \frac{1}{2} {m}_{1} {\dot{x}}_{1}^{2} + \frac{1}{2} {m}_{2} {\dot{x}}_{2}^{2}$

$V = \frac{1}{2} k {\left({x}_{2} - {x}_{1}\right)}^{2}$

Lagrangian $L = T - V$; and Euler-Lagrange: $\frac{d}{\mathrm{dt}} \left(\frac{\partial L}{\partial {\dot{x}}_{i}}\right) = \frac{\partial L}{\partial {x}_{i}}$

$\implies {m}_{1} {\ddot{x}}_{1} = k \left({x}_{2} - {x}_{1}\right) \text{ and } {m}_{2} {\ddot{x}}_{2} = - k \left({x}_{2} - {x}_{1}\right)$

Or:

$\left(\begin{matrix}{\ddot{x}}_{1} \\ {\ddot{x}}_{2}\end{matrix}\right) = \left(\begin{matrix}- \frac{k}{m} _ 1 & \frac{k}{m} _ 1 \\ \frac{k}{m} _ 2 & - \frac{k}{m} _ 2\end{matrix}\right) \left(\begin{matrix}{x}_{1} \\ {x}_{2}\end{matrix}\right) \implies \ddot{m} a t h b f x = M m a t h b f x$

We can solve that system based upon matrix $M$, but we can go further than that in terms of understanding the physics, because matrix $M$ has 2 discrete eigenvectors and can be diagonalised:

• ${\lambda}_{1} = 0 , m a t h b f {e}_{1} = \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

• ${\lambda}_{2} = - k \frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}} , m a t h b f {e}_{2} = \left(\begin{matrix}- {m}_{2} / {m}_{1} \\ 1\end{matrix}\right)$

This is the diagonalisation trick. Because:

$M = S \Lambda {S}^{- 1}$ where:

• S is the matrix of eigenvectors = $\left(\begin{matrix}1 & - {m}_{2} / {m}_{1} \\ 1 & 1\end{matrix}\right)$

• $\Lambda$ is the diagonal matrix of eigenvalues $= \left(\begin{matrix}0 & 0 \\ 0 & - k \frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}}\end{matrix}\right)$

Then:

$\ddot{m} a t h b f x = M m a t h b f x \implies \ddot{m} a t h b f x = S \Lambda {S}^{- 1} m a t h b f x$

Pre-multiply both sides by ${S}^{- 1}$:

$\implies {S}^{- 1} \ddot{m} a t h b f x = \Lambda {S}^{- 1} m a t h b f x$

$\implies {d}^{2} / \left({\mathrm{dt}}^{2}\right) \left(\textcolor{b l u e}{{S}^{- 1} m a t h b f x}\right) = \Lambda \textcolor{b l u e}{{S}^{- 1} m a t h b f x}$

And if we make linear transform: $m a t h b f u = {S}^{- 1} m a t h b f x \setminus \left(\star\right)$, the we have:

$\ddot{m} a t h b f u = \Lambda m a t h b f u \implies \left(\begin{matrix}{\ddot{u}}_{1} \\ {\ddot{u}}_{2}\end{matrix}\right) = \left(\begin{matrix}0 & 0 \\ 0 & - k \frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}}\end{matrix}\right) \left(\begin{matrix}{u}_{1} \\ {u}_{2}\end{matrix}\right)$, or:

${\ddot{u}}_{1} = 0 \setminus \triangle \text{ and } {\ddot{u}}_{2} = - k \frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}} {u}_{2} \setminus \square$

• $\triangle \implies {u}_{1} \left(t\right) = \alpha t + \beta$, with $\alpha , \beta$ as some sundry constants

• and with ${\omega}^{2} = k \frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}}$, then $\square \implies {u}_{2} \left(t\right) = A \cos \left(\omega t + \phi\right)$

To understand what this really means, we go back to $\star$. If we invert $S$, we find that:

$u = {S}^{- 1} m a t h b f x \implies \left(\begin{matrix}{u}_{1} \\ {u}_{2}\end{matrix}\right) = \frac{1}{{m}_{1} + {m}_{2}} \left(\begin{matrix}{m}_{1} & {m}_{2} \\ - {m}_{1} & {m}_{1}\end{matrix}\right) \left(\begin{matrix}{x}_{1} \\ {x}_{2}\end{matrix}\right)$

So:

• ${u}_{1} = \frac{{m}_{1} {x}_{1} + {m}_{2} {x}_{2}}{{m}_{1} + {m}_{2}}$; and so the dimension on the number line that ${u}_{1}$ measures is either stationary or moving at constant velocity. That makes sense. There are no external forces on this system. And ${u}_{1}$ is a recipe for the Centre of mass of the system. It's just minding it's own business.

• ${u}_{2} = \frac{{m}_{1}}{{m}_{1} + {m}_{2}} \left({x}_{2} - {x}_{1}\right)$. This is a recipe for the actual oscillation within the system, that is based upon the relative displacement of the masses, and ergo the spring constant and the particles masses.

Apr 27, 2017

See below.

#### Explanation:

For each mass we have

${m}_{1} {\ddot{x}}_{1} = k \left({x}_{2} - {x}_{1}\right)$
${m}_{2} {\ddot{x}}_{2} = - k \left({x}_{1} - {x}_{2}\right)$

The movement is one dimensional so multiplying by ${m}_{2}$ the first equation and by ${m}_{1}$ de second equation we have

${m}_{2} {m}_{1} {\ddot{x}}_{1} = {m}_{2} k \left({x}_{2} - {x}_{1}\right)$
${m}_{1} {m}_{2} {\ddot{x}}_{2} = - {m}_{2} k \left({x}_{2} - {x}_{1}\right)$

now subtracting the second from the first

${m}_{1} {m}_{2} \left({\ddot{x}}_{2} - {\ddot{x}}_{1}\right) = - \left({m}_{2} + {m}_{1}\right) k \left({x}_{2} - {x}_{1}\right)$

now if

${x}_{2} - {x}_{1} = {c}_{1} \cos \omega t + {c}_{2} \sin \omega t$ we have

${\ddot{x}}_{2} - {\ddot{x}}_{1} = - {\omega}^{2} \left({c}_{1} \cos \omega t + {c}_{2} \sin \omega t\right)$

then after substitution and simplification

$- {\omega}^{2} {m}_{1} {m}_{2} = - \left({m}_{1} + {m}_{2}\right) k$

so

$\omega = \sqrt{\frac{{m}_{1} + {m}_{2}}{{m}_{1} {m}_{2}} k}$