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A particle, of mass $9$kg, is attached to two identical springs. The other ends of the springs are attached to fixed points, $A$ and $B$, which are $1.2$ metres apart on a smooth horizontal surface. The springs have natural length $0.4$m and the magnitude of the tension in each spring is given by $112.5e$, where $e$ is the extension of the spring. The particle is released from rest at a distance of $0.5$ metres from $B$ and moves on the line $AB$. The midpoint of $AB$ is $C$. At time $t$ seconds after release, the displacement of the particle from $C$ is $x$ metres, where the direction from $A$ to $B$ is taken to be positive.

Show that the resultant force on the particle, at time $t$, is $-225x$ newtons.

Why is $e = (0.2 \pm x)$ (depending on which side of the particle the spring is), why is it not $e = (0.4 \pm x)$?

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Because $x$ is not measured from $A$ or $B$, but from $C$. The midpoint of $AB$ is at $0.6$ meters from $A$ (or $B$). The position of the end of the unextended spring is $0.4$ meters from $A$, which means that is $0.6-0.4=0.2$ meters from $C$.

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