displacement HORIZONTALLY:    If the charge q is displaced slightly to the right side along the line joining the two CHARGES -q, then the force of attraction of the right side -q is more than the force of attraction of the -q on the left side.   So +q moves towards right side. As it moves closer to the -q (right side), the force increases.  Then it will ACCELERATE more and collide with it.================case 2)   displacement of +q along the perpendicular bisector of line joining two charges -q and -q.Let the direction of x be towards the TOP upwards from the center.  Let the origin be at the center of two charges -q.  Let the charge q be displaced by x , where x << d.   The distance between the two charges is 2 d.  Mass of charge q  is m.   The distance between -q and +q = √(d² + x²)Let K = 1/[ 4 πε ].see the diagram.  Let F1 be the force due to -q on the right and F2 be the force due to -q on the left side.   They are equal in magnitude and directions are as shown.   magnitude = F1 = F2 = K q² / (d² + x²)  Components of F1 and F2 along the perpendicular bisector are:            F1 Sin Ф = K q²  x / (x²+d²)³/² The components of these forces parallel to the line joining -q charges are = F1 Cos Ф and will cancel as they are in OPPOSITE directions.Let F and a be the instantaneous resultant force and net acceleration of +q.  Then the resultant force on q due to the two charges:       F =  2 K q² x / (d²+x²)³/²  towards the origin in the direction of - x.     m d² x/ dt²   = - 2 K q² x / [d³ (1 + x²/d²)³/² ]                       =  - 2 K q² x / d³   if  x << d, then we ignore x² as it is << d²   m  d² x / d t² =  - [ 2 q² / 4 π ε d³ ]  x                       = ⁻ω² x  This is the equation of motion of the charge q.  SO q oscillates in a simple harmonic motion with an amplitude of  x₀ that is the displacement at t = 0.     The angular frequency of oscillation =  ω  = q / √ (2π ε d³)