Imagine two blocks of masses $m_1$ and $m_2$ joined together by a spring of spring constant $K$.
Now let the spring be stretched by a distance $X$ and then the system is released. suppose during the stretching the block of mass $m_1$ moves towards left by a distance $A$ and the block of mass $m_2$ by a distance B.
Now the centre of mass of the system at any instant will remain at rest. Now let it be present at a point P in between the two blocks. so can I depict the oscillations of the two blocks as two independent oscillations about the centre of mass?
Best Answer
The two oscillations will not be independent as they will share frequency and phase. You can start from Newton's equations of motion.
$$ m_1 \frac{{\rm d}^2\,a(t)}{{\rm d}t^2} = -F(t) $$ $$ \mbox{-}m_2 \frac{{\rm d}^2\,b(t)}{{\rm d}t^2} = F(t) $$ $$ F(t) = k \;\left( a(t)+b(t) \right) $$
Assume simple harmonic motion $a(t) = A \sin(\omega\,t)$, $b(t) = B \sin(\omega\,t)$ which leads to the frequency equation $$ k\,\left(m_1+m_2\right) = m_1 m_2 \omega^2 $$ and the amplitude equation $$ A\, m_1 = B\, m_2 $$ So now you can show that the center of gravity $P$ does not move as long as the amplitudes $A$ and $B$ obey the balance equation above. How? Well what is the equation for the center of gravity? $$ {\rm cg} = \frac{\mbox{-}B\,m_2+A\,m_1}{m_1+m_2} = 0 $$