# Transverse Oscillation of horizontal Mass-Springs system

1. Nov 11, 2015

### omegasquared

1. The problem statement, all variables and given/known data

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:

2. Relevant equations

F = kx
sinθ = o/h

3. The attempt at a solution

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

2. Nov 11, 2015

### Chandra Prayaga

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. Nov 11, 2015

### omegasquared

The x is to do with the extension of the spring which will occur when you displace the ball by y upwards.

4. Nov 11, 2015

### haruspex

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. Nov 11, 2015

### omegasquared

Additional extension beyond what it was (L) at y=0.

6. Nov 11, 2015

### haruspex

Then reconsider your equation for Fr.

7. Nov 11, 2015

### omegasquared

Ah, good point.

So it should be: Fr = 2k(L+x)sinθ

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

Agreed?

8. Nov 11, 2015

### haruspex

Yes.

9. Nov 11, 2015

### omegasquared

I'm still unsure how to get from there to simply ky however.

10. Nov 11, 2015

### haruspex

It says small displacements. If y is small, x is very small. What approximations does that allow you to make?

11. Nov 11, 2015

### omegasquared

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. Nov 11, 2015

### haruspex

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.

13. Nov 11, 2015

### Chandra Prayaga

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.

14. Nov 11, 2015

### omegasquared

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!