Symmetry behind charged spring-mass system in Electric field

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

If we assume that x = 0 is where the spring connects to the wall, then the rest position of the mass-spring-electric field position is x = EQ/k and the max position is x = 2EQ/k. Is there a reason for the symmetry between the rest position and max position? (The symmetry being: max position = rest position + EQ/k)

Many thanks!

 

[kuruman]

The reason is that in this case EQ/k is the amplitude of the motion. Note that this problem is equivalent to a vertical spring mass system with EQ replacing mg.

 

[ChiralSuperfields]

The reason is that in this case EQ/k is the amplitude of the motion. Note that this problem is equivalent to a vertical spring mass system with EQ replacing mg.

Thanks for your reply @kuruman! Whoops, forgot amplitude was max position - rest position, why was I thinking about symmetry??!