consider the linear system in the question and let the particle be displaced through a distance x along the line joining the particles

now the distance between the particles varies

now the distance between the middle particle and the particle particle towards which it is moved is (b-x) and obivously the distance from the other particle will be (b+x)

hence the magnitudes of forces due to the individual particles on the middle particle are

f(1)=q^2/[(4 pie epsilon not)(b-x)^2]

f(2)=q^2/[(4 pie epsilon not)(b+x)^2]

therefore the net force acting on the middle particle is

F=f(1)-F(2)

F=q^2/[(4 pie epsilon not)(b-x)^2]-q^2/[(4 pie epsilon not)(b+x)^2]

F=q^2/[(4 pie epsilon not)[{1/(b+x)^2}-{1/(b-x)^2}]

F=q^2/[(4 pie epsilon not)[(b-x)^2-(b+x)^2/(b+x)^2(b-x)^2]

F=q^2/[(4 pie epsilon not)

                                       [b^2+x^2-2bx-b^2-b^2-2bx/(b^2-x^2)^2]

as b^2,-b^2 and x^2.-x^2 cancels out with each other respectively, we have

F=q^2/[(4 pie epsilon not)[(-4bx)/(b^2-x^2)^2]

as in the question it is given that x<

neglecting x^2 in the denominator, we have

F=q^2/[(4 pie epsilon not)[(-4bx)/b^4]

as 4b in the numerator gets cancelled out with that in the denominator, we have

F=-q^2/[(pie epsilon not)(b^3)]x (solution 4 (A))

here the -ve sign just indicates the direction of the force

B) from the above derivation we have

F=-q^2/[(pie epsilon not)(b^3)]x

according to newtons second law F=ma

ma=-q^2/[(pie epsilon not)(b^3)]x

but a=-(omega)^2(x)

m[-(omega)^2(x)]=-q^2/[(pie epsilon not)(b^3)]x

cancelling the -ve sign, (x) on both sides and taking (m) onto the other side, we have

(omega)^2=q^2/[(m)(pie epsilon not)(b^3)]

hence we obtain the value of omega and now substituting this value in the equation for frequency given in shm, we have

freq=omega/2pie

freq=[{q^2/[(m)(pie epsilon not)(b^3)]}^(1/2)]/2 pie

for the third part(C) its just substituting the values of q,b and m in the above equation