system 2-mass 1-spring 1-d

System of two mass and one spring

Two frictionless and free moving masses connected by a spring will have two natural frequencies, where one is zero ¹. The system can be considered as similar system with two springs ², but with the second spring is unattached to a fix point. Simpler system with only one mass but with massful spring is still interesting to study ³.

Here the formulation for a system consist of 2 masses $m_1$, $m_2$ and 1 spring with constant $k$ is given.

Mass $m_i$ is located at $x_i$ with $i = 1, 2$, where $x_1 < x_2$. Normal length of the spring is $l_0$. Force on each mass are

$$\tag{1} F_{k,1} = k[ (x_2 - x_1) - l_0 ]$$

and

$$\tag{2} F_{k,2} = -k[ (x_2 - x_1) - l_0 ].$$

Using Newton’s second law of motion will turn Eqns (1) and (2) into

$$\tag{3} m_1 \frac{d^2 x_1}{dt^2} = k[ (x_2 - x_1) - l_0 ]$$

and

$$\tag{4} m_2 \frac{d^2 x_2}{dt^2} = -k[ (x_2 - x_1) - l_0 ].$$

Sum of Eqns (3) and (4) gives

$$\tag{5} \frac{d^2 X_{\rm com}}{dt^2} = 0, \ \ \omega_1 = 0$$

with

$$\tag{6} X_{\rm com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

is system center of mass.

Multiply Eqn (3) with $m_2$ and Eqn (4) with $m_1$ will produce

$$\tag{7} m_1 m_2 \frac{d^2 x_1}{dt^2} = k m_2 [ (x_2 - x_1) - l_0 ]$$

and

$$\tag{8} m_1 m_2 \frac{d^2 x_2}{dt^2} = - k m_1[ (x_2 - x_1) - l_0 ].$$

Substract Eqn (4) with Eqn (3) will give

$$\tag{9} \omega_2 = \sqrt{\frac{k}{\mu}}$$

and

$$\tag{10} \mu = \frac{m_1 m_2}{m_1 + m_2}$$

is effective mass, whose details are in system 2-mass 1-spring.