If a spring is cut in half, does the period change?

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SUMMARY

The period of a simple harmonic oscillator changes when a spring is cut in half. The formula for the period T is given by T = 2π√(m/k), where k is the spring constant. When the spring is halved, the spring constant doubles, resulting in a new period T' = T/√2. This definitive conclusion indicates that the correct answer to the problem is option C: T/√2.

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  • Understanding of simple harmonic motion
  • Familiarity with the spring constant (k) and its relationship to spring length
  • Knowledge of the formula for the period of oscillation (T = 2π√(m/k))
  • Basic concepts of forces and mass in physics
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  • Study the effects of spring constants in series and parallel configurations
  • Explore the relationship between amplitude and period in simple harmonic motion
  • Investigate the impact of mass changes on the period of oscillation
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to spring dynamics and harmonic oscillators.

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Homework Statement



A simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. Particle oscillates as shown in (i) with preriod T. If the spring is cut in half and used with the same particle, as shown in (ii), the period will be.

A) 2T
B) \sqrt{2}T
C) T/\sqrt{2}
D) T
E) T/2

Homework Equations



T = 2\pi\sqrt{m/k}

The Attempt at a Solution



I figured that since the length of the spring isn't relevant to the period, the period would stay the same. The answer provided is C. I'm not sure why.
 

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Hm, a thought, does cutting the spring in half alter the constant in some way?
 
This is all I can think of, because the mass stays the same, therefore the only thing that would change is the K, however why would cutting the spring in half change the K?
 
I read this problem in an archive, however, the answers given there don't seem to make sense. They start talking about the amplitude changing, however, isn't the amplitude independent of the period?
 
Consider a spring which extends by length x under a load of mass m.
Connect two such springs in series to form one long spring. What would be the new extension of this spring under a load of mass m? What is the effective new spring constant of this long spring?
Can you extend this idea to cutting a spring in half?
 
The extension would be longer, but how does that affect the spring constant?
 
If you take the full spring and hang a weight W from it, it will stretch an amount Δx. So the spring constant is k = W/Δx.

Now consider the two halves of that spring. How much stretch does each half get? The force (W) is the same, so what must be the spring constant for each half?
 
Oh I see. So the first spring:
k = mg/x

Thus, T = 2\pi\sqrt{x/g}

So in the second spring:

k = 2mg/x

Thus, T' = 2\pi\sqrt{x/2g} = T/\sqrt{2}

Is this correct?
 
Yes, that's right. The key thing is that the spring constant of half a spring is twice the spring constant of the full spring. (This should make a bit of sense. You'd expect it to be harder to pull a shorter spring to the same extension.) And since the period depends on the spring constant, it will change accordingly.
 

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