I have conceptual doubts on when to use the reduced mass.

This is the situation. I have a spring with two charges of mass $m$, each attached to one end of a spring so that

$$ \mu a = -kx $$

from which (determining $a$ and $\omega$)

$$ T = 2 \pi \sqrt{\frac{m}{2k}} $$

Now the question is:

if one charge is fixed to a certain point of the plane, and the other is allowed to oscillate, does $T$ change? i.e., should I still consider $\mu$ when doing the calculation or $m$?

I would intuitively say that the oscillation is just the same with period $T$ since the spring compresses and stretches uniformly. In this case, oscillation on the side of the fixed charge would simply be absorbed by what is fixing the charge. Does this make sense?

  • 3
    $\begingroup$ Could you please define your variables, including μ and where x is being measured from. $\endgroup$ Dec 1, 2021 at 20:35
  • $\begingroup$ Yes, μ is the reduced mass of the system (in this case m/2) and x is the small variation in length that determine small oscillations. You then assume, from the equation of the harmonic motion, that a= -2k x/m and so that ω=sqrt(2k/m) and T = 2pi/ ω. Anyway, the first part of the problem (determining T with two free charges and a spring) is not really the main focus, the thing is: what happens when I fix the position of one of the charges? If further information is needed, feel free to ask! Thank you :-) $\endgroup$ Dec 1, 2021 at 21:09

2 Answers 2


We can use symmetry to gain insight here. The first case involves a spring with spring constant $k$ and a mass $m$ at either end. We could imagine the spring being fixed at the motionless middle, corresponding to the center of mass. The relevant spring on either side is now half as long, so its spring constant is $2k$. The oscillating mass on either side is $m$. The period is then $T=2\pi\sqrt\frac{m}{2k}$, as you note.

Now imagine the original configuration but with one of the masses fixed. The spring constant is $k$, and the oscillating mass is $m$. The corresponding period is $T=2\pi\sqrt\frac{m}{k}$, which is different from the earlier value. So it does make a difference if you fix one end rather than letting the two-mass assembly oscillate around its center of mass.

  • $\begingroup$ Hello @Chemomechanics and thank you very much for your answer. The case with one or with two masses was known to me, the thing I was not sure about is whether the same equations that you are using for masses are true when those masses are charged $\endgroup$ Dec 2, 2021 at 17:59

to obtain the eigen frequencies $~\omega_i~$ you start with the equations of motion

case I

$$m\,\ddot x_1+k\,(x_1-x_2)=0\\ m\,\ddot x_2-k\,(x_1-x_2)=0$$

or $$ \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}+ \underbrace{\frac{k}{m}\,\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}}_{\mathbf{K}}\,\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}=\mathbf{0}$$ the eigen values of the matrix $~\mathbf K~$ are


with $\omega=\frac{2\,\pi}{T}~$ you obtain $~T~$

case II $$m\,\ddot x_2 +k\,x_2=0\quad \Rightarrow\\ \omega^2=\frac {k}{m}~$$

so if the equations of motion changes the eigen frequencies also changes


that the eigenvalues don't chances if you transfer the equations of motion to the center of mass coordinate $~r_c~$ and the relative coordinate $ r=x_1-x_2 $ the zero eigen value belong to the center of mass coordinate

the EOM's are now

with :

$$\begin{bmatrix} {x}_1 \\ {x}_2 \\ \end{bmatrix}=\begin{bmatrix} 1 & -\frac 12 \\ 1 & -\frac 12 \\ \end{bmatrix} \begin{bmatrix} r_c \\ r \\ \end{bmatrix}$$


$$2\,m\ddot r_c=0\quad \Rightarrow\quad \omega_1^2=0 \\ \frac m2 \ddot r+k\,r=0\quad \Rightarrow\quad \omega_2^2=\frac{2\,k}{m}$$


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