The Force Law for Simple Harmonic Motion — Question 9

Question 9

A free spring pair oscillates with the reduced mass

Two masses $m_1$ and $m_2$ slide on a frictionless line and are joined by a massless spring of constant $k$ and natural length $\ell_0$. There is no external horizontal force. Let $q=x_2-x_1-\ell_0$ be the extension.

1. Derive the equation for $q$ and its angular frequency.

2. For $m_1=m$ and $m_2=3m$, find the ratio of their oscillation amplitudes about the center of mass. What separate motion can the center of mass have?

Question 9 - Solution

A free spring pair oscillates with the reduced mass

(a) Derivation and calculation

$$m_1\ddot x_1=kq,\qquad m_2\ddot x_2=-kq,\qquad q=x_2-x_1-\ell_0.$$$$\ddot q=\ddot x_2-\ddot x_1=-k\left(\frac1{m_1}+\frac1{m_2}\right)q.$$$$\boxed{\mu=\frac{m_1m_2}{m_1+m_2}},\qquad\boxed{\omega=\sqrt{k/\mu}}.$$

(b) Result and interpretation

With $X=(m_1x_1+m_2x_2)/(m_1+m_2)$,$$(m_1+m_2)\ddot X=0\Rightarrow X=X_0+V_0t.$$$$\delta x_1=-\frac{m_2}{m_1+m_2}q,\qquad\delta x_2=\frac{m_1}{m_1+m_2}q.$$$$m_1=m,\ m_2=3m:\quad\boxed{A_1/A_2=3},\quad\mu=3m/4.$$Relative oscillation and uniform center-of-mass translation are independent.

Physical check

As one mass tends to infinity, $\mu$ tends to the finite mass and the fixed-support frequency is recovered.