Simple harmonic motion of two horizontal springs

[scoutfai]

Figure 1 shows a mass m in equilibrium between two stretched helical springs on a smooth horizontal surface. k_1 and k_2 are the force constants of the springs and e_1 , e_2 are their respective equilibrium extensions.
The mass m is displaced X to one side ( X is smaller than e_1 and e_2 ), and then released. how you show that the acceleration (a) of the mass when its displacement from the equilibrium position is x obey the following equation :
a = - [ (k_1) + (k_2) ]*x / m

Please show me step by step ! thanks you.

 

[Tide]

The force is
[tex]F(x) = k_1 (x - x_{1E}) + k_2 (x - x_{2E})[/itex]<br /> Simply note that F(0) = 0 and your desired result follows.[/tex]