# Using Hooke's Law with simple harmonic motion

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 $$e = (0.4 \pm x)$$?

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 $$C$$.