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

In summary, when a simple harmonic oscillator consisting of a mass m and an ideal spring with spring constant k is cut in half and used with the same particle, the period will decrease to T/\sqrt{2}. This is because cutting the spring in half doubles the spring constant, resulting in a shorter period.
  • #1
clairez93
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0

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) [tex]\sqrt{2}[/tex]T
C) T/[tex]\sqrt{2}[/tex]
D) T
E) T/2

Homework Equations



T = 2[tex]\pi[/tex][tex]\sqrt{m/k}[/tex]

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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  • #2
Hm, a thought, does cutting the spring in half alter the constant in some way?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
The extension would be longer, but how does that affect the spring constant?
 
  • #7
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?
 
  • #8
Oh I see. So the first spring:
[tex]k = mg/x[/tex]

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

So in the second spring:

[tex]k = 2mg/x[/tex]

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

Is this correct?
 
  • #9
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.
 

1. What is a spring's period?

The period of a spring is the time it takes for one complete oscillation, or back-and-forth motion, to occur.

2. How is a spring's period affected by cutting it in half?

If a spring is cut in half, its period will decrease.

3. Why does cutting a spring in half affect its period?

When a spring is cut in half, its stiffness or spring constant increases, which causes it to oscillate at a faster rate and therefore have a shorter period.

4. Does the mass of the spring affect its period?

Yes, the mass of the spring does affect its period. The greater the mass, the longer the period will be.

5. Is there a mathematical equation that shows the relationship between a spring's period and its length?

Yes, the equation is T = 2π√(m/k), where T is the period, m is the mass of the spring, and k is the spring constant. This equation shows that as the length of the spring decreases, the period also decreases.

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