2.4. Reduced mass

Here, we derive the reduced mass for a system of two point-like masses connected by a force that obeys Hooke’s law.

We start out with the two point-like masses $m_{1}$ and $m_{2}$ at positions $x_{1}$ and $x_{2}$ , assuming $x_{2} > x_{1}$ . The bond length is the difference $r = x_{2} - x_{1}$ . The equilibrium bond length is $r_{e}$ .

If the two masses are connected by a spring that follows Hooke’s law with force constant $k$ , the forces acting on the two are

(2.72)

\begin{array}{r} \begin{matrix} F_{1} = m_{1} \frac{d^{2}}{d t^{2}} x_{1} = + k (r - r_{e}) \\ F_{2} = m_{2} \frac{d^{2}}{d t^{2}} x_{2} = - k (r - r_{e}) \end{matrix} \end{array}

Dividing the first equation by $m_{1}$ and the second equation by $m_{2}$ , and then subtracting the first from the second equation gives

(2.73)

\frac{d^{2}}{d t^{2}} (x_{2} - x_{1}) = \frac{d^{2}}{d t^{2}} r = - k (\frac{1}{m_{1}} + \frac{1}{m_{2}}) (r - r_{e})

The expression in parentheses can be abbreviated by defining the reduced mass $μ$

(2.74)

\frac{1}{μ} = \frac{1}{m_{1}} + \frac{1}{m_{2}} μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

This gives finally

(2.75)

μ \frac{d^{2}}{d t^{2}} r = - k (r - r_{e})

This equation is Hooke’s law for a single effective particle of mass $μ$ on a spring with force constant $k$ and equilibrium length $r_{e}$ .

This shows that the two spring-connected masses can be modeled as a single mass connected to a spring of the same force constant.