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.

3 Answers
Apr 27, 2017

#\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}}#

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.

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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.

Apr 27, 2017

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)#