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

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The discussion centers on calculating the period of oscillation for a spring cut to one-third of its original length. The original poster initially misinterprets the problem but later clarifies that the correct formula for the period is T = 2π√((1-n)m/k), where n represents the ratio of the new length to the original length. There is confusion regarding the stiffness of the spring after cutting, with the poster suggesting that the new stiffness is k/n. The conversation highlights the importance of understanding the relationship between spring length, stiffness, and oscillation period. The final question raised concerns the implications for the remaining two-thirds of the spring.
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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?

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The answer you quoted is an imaginary number because ##\eta-1=-2/3## so I doubt it.
 
anuttarasammyak said:
The answer you show is an imaginary number so I doubt it.
Oh i made a confusion, actually is 1 - n instead n - 1
 
LCSphysicist said:
we want to know the period of oscillations on a spring with length l/3.
What about the other 2/3 of the spring?
 
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