# Simple Harmonic Motion - amplitude when connected to two springs horizontally.

Image of the full question is here...

https://imgur.com/a/B3PnK

The idea of the question is that a particle is attached to two identical springs, the other end of the springs are attached to points $A$ and $B.$ $A , B$ are 4 meters apart in a horizontal line. The natural lengths of the springs are $1.6$m. The particle is initially held at point $C$ ($AC = 2.5$m) and then released.

The answer says amplitude is $0.4$m, but I think it should be $0.5$m.

I know how to show that the particle will move with SHM. The answers state the amplitude of the motion would be $0.4$m. I think I understand where the $0.4$ meters comes from, it is the extension of the spring in a "natural state". Am I correct here, is this why the amplitude is $0.4$?

I initially thought the amplitude should be $0.5$ m, since the system would oscillate about the midpoint of $AB$ and the particle is initially held at a position such that it is $0.5$m from midpoint of $AB.$

Thanks in advance.## Heading ##

• I think you're right. If it didn't swing back to C, energy was lost, and they don't describe how that happens. – Bill Mar 3 '18 at 21:58
• @Bill That was my second thought. The table is smooth so there is no loss due to friction. – NumberCruncher Mar 3 '18 at 22:05