Oscillation of a cutted spring

In summary, the conversation discusses finding the period of oscillations on a spring with length l/3. The stiffness of the new spring is k/n, based on the displacement of the points being proportional to the distance. The formula for the period is T = 2π√(n*(1-n)m/k), where n is the stiffness of the former spring and m is the mass. There is confusion about the answer being an imaginary number and a correction is made regarding the value of n.
  • #1
LCSphysicist
645
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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?

1594524961396.png
 
Last edited:
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  • #2
The answer you quoted is an imaginary number because ##\eta-1=-2/3## so I doubt it.
 
  • #3
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
 
  • #4
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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Likes anuttarasammyak and LCSphysicist

1. What is the definition of oscillation?

Oscillation refers to the repetitive back-and-forth movement or fluctuation of a system around its equilibrium point.

2. How does a cutted spring affect the oscillation of a system?

A cutted spring can change the natural frequency of a system, which is the rate at which the system oscillates without any external forces. This can impact the amplitude and period of oscillation.

3. What factors can affect the oscillation of a cutted spring?

The length and thickness of the spring, the weight attached to it, and the angle at which it is cut can all affect the oscillation of a cutted spring.

4. What is the equation for calculating the period of oscillation for a cutted spring?

The equation is T = 2π√(m/k), where T is the period, m is the mass attached to the spring, and k is the spring constant.

5. How can the oscillation of a cutted spring be applied in real-life situations?

Oscillation of a cutted spring can be seen in many everyday objects, such as pendulum clocks, guitar strings, and car suspensions. It is also used in scientific instruments, such as seismographs, to measure vibrations and movements.

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