Originally Posted by

**HallsofIvy** Yes, that is correct.

It is interesting that, although you are given the 'natural length' of the two springs, you don't need it.

Suppose we have two springs, each with modulus of elasticity k N/m (NOT Newtons) and natural length l. The two springs are attached to points distance L m apart and two an object of mass m. Let x be the distance from the object to the "left" point of attachment. Then the spring on the left has length x and is extended to distance x- l beyond its natural length and so applies force -k(x- l) to the object. The spriing on the right has length L- x and so is extended L- x- l beyond its natural length and applies force k(L- x- l) to the object. The total force on the object is -k(x- l)+ k(L- x- l)= -kx+ kl+ kL- kx- kl= kL- 2kx.

"Force= mass times acceleration" gives, as the "equation of motion",

$\displaystyle m\frac{d^2x}{dt^2}= kL- 2kx$

so the motion is independent of the natural length "l".