Assume it is displacd by x straight wrt to one spring...then correspondingly the extensions in the other two springs will be xcos(60) each

The force eqn (which is not much different,would be)

 

F = kx+(k²x²cos²(60)+k²x²cos²(60)+2k²x²cos²(60)cos(120))

(Using the formula of vectors .a+b(vectors)(magnitude) = (a²+b²+2abcos(alpha)

where alpha is the angle between the vectors a and b

 

Thus F = kx+kx/2 =3kx/2

T = 2(m/k_eff) where k_eff in this case is 3k/2

Thus T = 2(2m/3k)