Imagine two blocks of masses m1 and m2 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 m1 moves towards left by a distance A and the block of mass m2 by a distance B. Now the center 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 center of masss? I appeal for an answer with strong reasons.
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Traditionally, the "mass on a spring" is analyzed with one end of the spring held fixed. Your problem is equivalent to two such systems back-to-back. You have identified the center of gravity as a fixed point; so you can literally fix it and then cut the problem in two. Each half of the system then behaves like a traditional "mass-on-a-spring". |
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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? $$ {\rm cg} = \frac{\mbox{-}B\,m_2+A\,m_1}{m_1+m_2} = 0 $$ |
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