Question

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:
`\omega=\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}`
So I have that the distance traveled by `m_1` can be represented by the function `x_1(t)=Acos(\omega t)` and similarly for the distance traveled by `m_2` is `x_2(t)=Bcos(\omega t)`. The force the spring exerts on these two masses is `-kx_n(t)=m_n\frac{d^2x_n}{dt^2}`. But after, simplifying I have no idea how to combine these two functions.

Answer 1

`\omega =\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}`

Explanation:

The distance each mass is displaced can be represented by the following
`x_1(t)=Acos(\omega t)`
`x_2(t)=Bcos(\omega t`

Using the formula for simple harmonic oscillation `-kx=m\frac{d^2x}{dt^2}`, you get the following
`k(Bcos(\omega t)-Acos(\omega t))=-m_1A\omega ^2cos(\omega t)`
`k(Bcos(\omega t)-Acos(\omega t))=m_2B\omega ^2cos(\omega t)`

Solving we get
`-m_1A\omega ^2cos(\omega t)=m_2B\omega ^2cos(\omega t)`
`-m_1A=m_2B`

Plugging this into the first equation for simple harmonic oscillation we get
`k(Bcos(\omega t)+\frac{m_2B}{m_1}cos(\omega t))=m_2B\omega ^2cos(\omega t)`

Simplify to get
`k(1+\frac{m_2}{m_1})=m_2\omega ^2`
`k(m_1+m_2)=m_1m_2\omega ^2`
`\omega ^2=\frac{k(m_1+m_2)}{m_1m_2}`
`\omega =\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}`

Answer 2

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(t)` and `x_2(t)` are the displacements of the masses from the equilibrium (no tension) position, then we have:

`T = 1/2 m_1 dot x_1^2 + 1/2 m_2 dot x_2^2`

`V = 1/2 k (x_2 - x_1)^2`

Lagrangian `L = T - V`; and Euler-Lagrange: `d/(dt) ((partial L)/(partial dot x_i)) = (partial L)/(partial x_i)`

`implies m_1 ddot x_1 = k (x_2 - x_1) " and " m_2 ddot x_2 = - k (x_2 - x_1)`

Or:

`((ddot x_1),(ddot x_2)) = ((-k/m_1, k/m_1),(k/m_2, - k/m_2)) ((x_1),(x_2)) implies ddot mathbf x = M mathbf 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, mathbf e_1 = ((1),(1))`

• `lambda_2 = - k(m_1 + m_2)/(m_1m_2), mathbf e_2 = ((- m_2/m_1),(1))`

This is the diagonalisation trick. Because:

`M = S Lambda S^(-1)` where:

• S is the matrix of eigenvectors = `((1, -m_2/m_1),(1, 1))`

• `Lambda` is the diagonal matrix of eigenvalues `= ((0,0),(0, - k(m_1 + m_2)/(m_1m_2)))`

Then:

`ddot mathbf x = M mathbf x implies ddot mathbf x = S Lambda S^(-1) mathbf x`

Pre-multiply both sides by `S^(-1)`:

`implies S^(-1) ddot mathbf x = Lambda S^(-1) mathbf x`

`implies d^2/(dt^2) ( color(blue)(S^(-1) mathbf x)) = Lambda color(blue)(S^(-1) mathbf x)`

And if we make linear transform: `mathbf u = S^(-1) mathbf x \ (star)`, the we have:

`ddot mathbf u = Lambda mathbfu implies ((ddot u_1),(ddot u_2)) = ((0,0),(0, - k(m_1 + m_2)/(m_1m_2))) ((u_1),(u_2))`, or:

`ddot u_1 = 0 \ triangle " and " ddot u_2 = - k(m_1 + m_2)/(m_1m_2) u_2 \ square`

• `triangle implies u_1(t) = alpha t + beta`, with `alpha, beta` as some sundry constants

• and with `omega^2 = k(m_1 + m_2)/(m_1m_2)`, then `square implies u_2(t) = A cos ( omega t + phi)`

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

`u = S^(-1) mathbf x implies ((u_1),(u_2)) = 1/(m_1 + m_2)((m_1, m_2),(- m_1, m_1))((x_1),(x_2))`

So:

• `u_1 = (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 = ( m_1)/(m_1 + m_2)( x_2 - x_1)`. 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.

Answer 3

See below.

Explanation:

For each mass we have

`m_1 ddot x_1=k(x_2-x_1)`
`m_2 ddot x_2=-k(x_1-x_2)`

The movement is one dimensional so multiplying by `m_2` the first equation and by `m_1` de second equation we have

`m_2m_1 ddot x_1=m_2k(x_2-x_1)`
`m_1m_2 ddot x_2=-m_2k(x_2-x_1)`

now subtracting the second from the first

`m_1m_2(ddot x_2-ddot x_1)=-(m_2+m_1)k(x_2-x_1)`

now if

`x_2-x_1 = c_1cos omegat+c_2 sin omega t` we have

`ddot x_2-ddot x_1 = -omega^2(c_1cos omegat+c_2 sin omega t)`

then after substitution and simplification

`-omega^2m_1m_2=-(m_1+m_2)k`

so

`omega = sqrt((m_1+m_2)/(m_1m_2)k)`