What is the Period of Oscillation for a Spring Cut to One-Third Its Length?

[LCSphysicist]

Homework Statement

A non-deformed spring whose ends are fixed has a stiffness
x = 13 N/m. A small body of mass m = 25 g is attached at the point
removed from one of the ends by n = 1/3 of the spring's length. Neg-
lecting the mass of the spring, find the period of small longitudinal
oscillations of the body. The force of gravity is assumed to be absent.

Relevant Equations

All below.

I am not sure if i get the problem, but if i understand, we want to know the period of oscillations on a spring with length l/3.

If is this the right interpretation, i would say that the stiffness of the new the spring is k/n, where k is the stiffness of the former spring.
This based on the knowing that the displacement of the points of the spring is proportional to the distance of the end, so to one force :

f = f, kx = k'xn, k' = k/n

T = 2π√(n*m/k)
The answer is, actually
T = 2π√(n*(1-n)m/k)
Where am i wrong?

 

[anuttarasammyak]

The answer you quoted is an imaginary number because $\eta-1=-2/3$ so I doubt it.

 

[LCSphysicist]

The answer you show is an imaginary number so I doubt it.

Oh i made a confusion, actually is 1 - n instead n - 1

 

[haruspex]

we want to know the period of oscillations on a spring with length l/3.

What about the other 2/3 of the spring?