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I was reading longitudinal oscillations of two masses from Crawford's Waves.

longitudinal oscillation
(source: rochester.edu)

The displacement of $m_1$ is given by $\psi_a$ & that of $m_2$ is given by $\psi_b$. The differential equations that Crawford's writes is: $$\begin{align}M\frac{d^2 \psi_a}{dt^2} = -K\psi_a \color{red}{+} K(\psi_b - \psi_a) \\M\frac{d^2 \psi_b}{dt^2} = -K\psi_b \color\red{-} K(\psi_b - \psi_a) \end{align}$$ . Look at the signs- the former has $+$ but the later has $-$. I was expecting $-$ sign in the former equation also but everywhere it is given as $-$ sign.

I then read this in which it is written as when $\psi_a\lt \psi_b$, the coupling spring stretches more than the leftmost spring- that is quite reasonable to me. But then it abruptly ends writing that the coupling spring doesn't always provide restoring force adding that the force on $m_1$ would be along the direction of motion.

Now can anyone explain me why doesn't the coupling spring provide restoring force? Why is there $-$ sign in the later but $+$ in the former equation?

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  • $\begingroup$ Sometimes it's useful to write your differential equations as a block diagram - to help illustrate, reveal structure. For these equations the block diagram shows 2 distinct negative feedback loops with cross coupling. $\endgroup$
    – docscience
    Commented Sep 6, 2015 at 14:47

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If $\psi_b-\psi_a \gt 0$, the center spring is stretched longer than its normal length. It will then pull to the right on mass $a$ and to the left on mass $b$. The difference in signs reflects this. If $\psi_a$ had been measured positive to the left, making the diagram symmetric, the sign would be minus.

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  • $\begingroup$ Sir, I want to know why it is so. Why doesn't the former equation contain the $-$ sign? After all, so far I've been taught restoring force acts in the opposite direction of displacement, but here it is otherwise. The link above tells that the coupled spring doesn't provide restoring force; my question is why it is so; doesn't a stretched spring impart restoring force?? $\endgroup$
    – user36790
    Commented Sep 6, 2015 at 14:48
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    $\begingroup$ The force acts opposite to displacement when the other end of the spring is fixed. Here if only mass $a$ is displaced, the force will be opposite the displacement-the minus sign in the $(\psi_b-\psi_a)$ term takes care of that. But if mass $b$ is displaced to the right, the force on mass $a$ will be directed to the right, so the acceleration is positive. $\endgroup$ Commented Sep 6, 2015 at 15:50

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