Period of vibration with springs

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Homework Statement



Two springs, each with unstretched length 0.200 m, but with different force constants k1 and k2, are attached to opposite ends of a block with mass m on a level, frictionless surface. The outer ends of the springs are now attached to two pins P1 and P2, 0.100 m from the original positions of the ends of the springs (the figure ). Let k1 = 2.00 N/m , k2 = 6.40 N/m, and m = 0.100 kg .

Part A
Find the length of each spring when the block is in its new equilibrium position after the springs have been attached to the pins.

I got this part, L1=.352 m and L2=.248 m

Part B
Find the period of vibration of the block if it is slightly displaced from its new equilibrium position and released.


Homework Equations



T=2pi*sqrt{m/k}

The Attempt at a Solution


I was given m, but I did not know what to use for k. I know how to find the k for springs in a series, but not when they are separated and acting in opposite directions. Any help is greatly appreciated!
 

Answers and Replies

  • #2
Delphi51
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This looks interesting! The method for deriving the springs in series formulas is here: http://sepwww.stanford.edu/public/docs/sep77/francis2/paper_html/node2.html

Perhaps the same approach can be taken here. You can write things like F2 - F1 = ma, F1 = k1*x1.
Would you say x1 = -x2?
Try to get something that looks like F = kx so you can find the value of the combined k. That should allow the period to be calculated.
 
  • #3
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Okay, I have tried messing around with it a little bit, but am not getting anywhere. I can't set x1=x2 can I since their k's are different? Are the forces of both sides equal?
 
  • #4
Delphi51
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It is tricky, isn't it? Seems to me as it pulls one way, say x1 gets smaller, then x2 must get larger. I take the wording to mean that x1 + x2 = 0.2 initially. Wouldn't that remain true as one gets smaller and the other larger?

For the forces, the free body diagram gives F2 - F1 = ma
(depending on how you defined the directions of the forces)
 
  • #5
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okay, plugging in .2-x2 for x1 gives me 8.4x1-1.28=ma But where do I go from here?
 
  • #6
Delphi51
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I'm getting 0.2*k2 - (k1+k2)*x1 = ma.
It doesn't look quite like F = -kx. If it was, ω = sqrt(k/m). If I had to guess about the double spring I would say
ω = sqrt((k1+k2)/m)
but it sure isn't clear.
Upon reading the question again, I'm confused about what it says - maybe because I haven't seen the diagram that apparently goes with it.
 
  • #7
vela
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It is tricky, isn't it? Seems to me as it pulls one way, say x1 gets smaller, then x2 must get larger. I take the wording to mean that x1 + x2 = 0.2 initially. Wouldn't that remain true as one gets smaller and the other larger?
From Julie's answer to part (a), the springs appear to be stretched so that the total distance between the two pins is 0.600 m.

Intuitively, I think you should find that the spring constants just add.
 
  • #8
Delphi51
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Maybe so, vela. That would mean they are stretched .2 beyond their unstretched length as I have been thinking. It would be nice to see the diagram to make sure.

Julie, I've asked for more helpers on this.
 
  • #9
tiny-tim
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hi everyone! :smile:

am i missing something?

this looks very simple to me …

force is proportional to displacement (doesn't matter where we start from) …

and displacement is equal (and opposite) for the two spings …

and we can ignore the "equilibrium forces" …

so isn't mx'' = -(k1 + k2)x ? :confused:
 
  • #10
Delphi51
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Thanks for coming to our rescue, tiny-tim!
But it seems to me your F = -(k1 + k2)x would only be true if the x was the same for both springs. That is, if both springs were on the same side of the mass. With one on each side, if one side stretches x, then the other side will be compressed . . . . 0.2-x if the vision that x1+x2 = 0.2 is right.
 
  • #11
tiny-tim
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Hi Delphi51! :smile:

But the extension is x on one side and minus x on the other …

an extra force pushing from the left, say, and a reduced force pushing from the right …

overall a "double" extra force pushing from the left

(the 0.2 is a constant which is absorbed into the equilibrium position and thereafter doesn't affect the motion).

And surely it would be the same whether the massive block is in the middle or on either side?
 
  • #12
Delphi51
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But the extension is x on one side and minus x on the other
I picture it like this:
springs.jpg

Initially they are stretched a total of 0.2, so x1 + x2 = 0.2. As x1 is increased, x2 is decreased but always x1 + x2 = 0.2.
ma = F2 - F1 = k2*x2 - k1*x1
Using x1 + x2 = 0.2 to eliminate x2 we have
ma = k2(0.2 - x1) - k1*x1
ma = 0.2*k2 - (k1+k2)x1
(we had this in post 5)
If the 0.2*k2 term was missing, it would be straightforward as you say.
The question is what affect does that extra term have on the frequency. I guessed none, but I don't know.
 
  • #13
tiny-tim
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Hi Delphi51! :smile:
ma = 0.2*k2 - (k1+k2)x1
(we had this in post 5)
If the 0.2*k2 term was missing, it would be straightforward as you say.
The question is what affect does that extra term have on the frequency. I guessed none, but I don't know.
if we put x3 = x1 - 0.2k2/(k1 + k2), then we have:

ma = -(k1 + k2)x3

x3 is therefore the equilibrium position (which is what i meant by absorbing the constant into the equilibrium position) :wink:
 
  • #14
Delphi51
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Wow, very clever substitution! So the frequency will indeed be
1/(2π)*sqrt[(k1+k2)/m]
Thank you very much.
 

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