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Transverse Oscillation of horizontal Mass-Springs system

  • #1

Homework Statement



Mass M is supported by a smooth table and connected by two light horizontal springs to two fixed blocks. Each spring is of natural length L. Both springs are initially extended by L to get a total width between blocks of 4L. The spring constant of both springs is k.

When the mass is given a small transverse displacement, y, show that the restoring force is approximately ky.

See photo below:

Photo on 11-11-2015 at 15.54.jpg


Homework Equations



F = kx
sinθ = o/h[/B]

The Attempt at a Solution



Restoring force, Fr = 2kxsinθ
sinθ = y / (2L + x)
x = (y / sinθ) - 2L
 

Answers and Replies

  • #2
Chandra Prayaga
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First you should draw a diagram showing a small transverse displacement y of the ball, and draw a coordinate system. Then write your equations. For example, I don't understand why there is an x in your equations at all.
 
  • #3
First you should draw a diagram showing a small transverse displacement y of the ball, and draw a coordinate system. Then write your equations. For example, I don't understand why there is an x in your equations at all.
The x is to do with the extension of the spring which will occur when you displace the ball by y upwards.
 
  • #4
haruspex
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Restoring force, Fr = 2kxsinθ
sinθ = y / (2L + x)
Are you being consistent about the meaning of x? Is it the total extension of the spring, or the additional extension beyond its extension when y=0?
 
  • #5
Are you being consistent about the meaning of x? Is it the total extension of the spring, or the additional extension beyond its extension when y=0?
Additional extension beyond what it was (L) at y=0.
 
  • #6
haruspex
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Additional extension beyond what it was (L) at y=0.
Then reconsider your equation for Fr.
 
  • #7
Then reconsider your equation for Fr.
Ah, good point.

So it should be: Fr = 2k(L+x)sinθ

Which would then be: = 2ky(L+x) / (2L+x)

Agreed?
 
  • #8
haruspex
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Ah, good point.

So it should be: Fr = 2k(L+x)sinθ

Which would then be: = 2ky(L+x) / (2L+x)

Agreed?
Yes.
 
  • #10
haruspex
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I'm still unsure how to get from there to simply ky however.
It says small displacements. If y is small, x is very small. What approximations does that allow you to make?
 
  • #11
It says small displacements. If y is small, x is very small. What approximations does that allow you to make?
Expanding previously, Fr = (2kL+2kx)y / (2L+x), so if x is small then 2kx ≈ 0 and 2L + x ≈ 2L? Which would leave 2kLy/2L = ky but then surely the y on the top would make the whole thing ≈ 0?
 
  • #12
haruspex
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Expanding previously, Fr = (2kL+2kx)y / (2L+x), so if x is small then 2kx ≈ 0 and 2L + x ≈ 2L? Which would leave 2kLy/2L = ky but then surely the y on the top would make the whole thing ≈ 0?
What we're doing here is finding the first order approximation. You can think of it as writing the full form as the sum of a series of terms in which, as the displacement tends to zero, each term tends to zero faster than the one before it: a power series, say. The first order approximation throws away all except the first term.
 
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  • #13
Chandra Prayaga
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Well, certainly, if you put y = 0, then you don't have to do anything. There are no oscillations! You are making an approximation, where y, and x are small compared to L. So expressions like L + x, where L is large, and x is small, are nearly equal to L. e.g., 1000 + 1 ~ 1000. If you look at the y on top, you cannot neglect it because you don't have anything to compare it with. After all, while 1 << 1000, 1 is not small compared to 0.
 
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  • #14
What we're doing here is finding the first order approximation. You can think of it as writing the full form as the sum of a series of terms in which, as the displacement tends to zero, each term tends to zero faster than the one before it: a power series, say. The first order approximation throws away all except the first term.
Well, certainly, if you put y = 0, then you don't have to do anything. There are no oscillations! You are making an approximation, where y, and x are small compared to L. So expressions like L + x, where L is large, and x is small, are nearly equal to L. e.g., 1000 + 1 ~ 1000. If you look at the y on top, you cannot neglect it because you don't have anything to compare it with. After all, while 1 << 1000, 1 is not small compared to 0.
Thank you both for your help, in combination I believe I understand this much better now but I'll continue to ponder this idea of approximations (in terms of power series) over. Much appreciated!
 

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