# Proof That The Following System Does Uniform Oscilations

1. Nov 5, 2013

### DorelXD

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

Well, we are given the following system, in which the rope, spring, and pulley is ideal. We know that for the angle $a$ the system is in equilibrium. The question is, to show that if the angle a is small, the system does unfirom oscilations.

I don't know how to tackle this problem, it confuses me a little. We must show that the resultant force on the body is proportional to a constant and the displacement for the equilibrium position at a given time.

2. Relevant equations

Some basic trigonometry, with some tricks for aproximating the trigonometric function for small angles and basic vector addition, I think...

3. The attempt at a solution

Well, the first thing that is obvious is that if the system is at rest, then:

$T\cos{a} = ky_0$ , where y_0 is the initial comprimation of the spring, and k is its spring constant.

Writing the resultant force for the body we get that: $R = T - mg$. How do I continue from here? Can somebody help me, please :D ?

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2. Nov 5, 2013

### haruspex

You mean ".. if the deviations from angle a are small ...", right?
So consider a small deviation in a. Find how that affects the forces and accelerations, and thus how the deviation will change over time.

3. Nov 6, 2013

### DorelXD

Yes, indeed. :D

Well, the forces that act on the body are the tension and the weight. The tention depends on the amount of comprimation of the resort. I drew a picture. The resort compresses with OO'(green) and the body descends with CC'. I should find a relation between the two length. I drew the system in a more "gemoetrcal" way because we are interested in lengths, and relations between plain figures, right ?

Let us say that OO' = x because it seems a more familiar notation, and we are interested specifically in that length. When the system is in the state corespoding to the 'b' angle, the spring is compressed with: $y_0 + x$ and we know the expression for y_0. So if the compression of the spring affects the tension, it affects the motion of the body, because tension acts on it. :D

Is my drawing and my judgement correct:D ? Tell me, please. If yes, how do we proceed? I haven't found a connection between CC' and OO'. :(

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4. Nov 6, 2013

### voko

OO' and AO' are related via the triangle OAO', where OA and angle O are constant. AO' + AC' = constant length of the rope.

Apart from this, I would use conservation of energy.

5. Nov 6, 2013

### DorelXD

Well there is no resitive force so the energy is conserved. Let us say that so CC' = d, beacause it's usefull. Also, in the initial position the potential energy of the body is 0 (for convenience).

$$E_a = E_b => \frac{ky_0^2}{2} = \frac{k(y_0 +x) ^2}{2} - mgd => mgd = \frac{k}{2}x(2y_0+x)$$

But, I don't see how that helps me. :D

Last edited: Nov 6, 2013
6. Nov 6, 2013

### voko

You are forgetting the kinetic energy.

7. Nov 6, 2013

### DorelXD

Right, sorry :D .

$$E_a = E_b => \frac{ky_0^2}{2} = \frac{k(y_0 +x) ^2}{2} - mgd + \frac{mv_x^2}{2}$$

where $v_x$ is the speed corespoding to the elongation x.

Is it ok now? What's next? I haven't done the simplifications yet, but a first thing I notice is that I should connect v_x to x, somehow, right? And this isn't exactly straightforward because we don't have an uniform motion, the speed and acceleration are not constant.

And I wanted to thank you for answering me! So, what do we do next ? :D

8. Nov 6, 2013

### haruspex

I'm not sure that energy will make it easier. We don't need to solve the equations, just show that they approximate SHM.

DorelXD, what is the relationship between T and the force in the spring? Between x and force in the spring? Between x and the height of the mass? Between T and the acceleration of the mass? Put all that together.

9. Nov 7, 2013

### voko

The kinetic energy is in the motion of the mass, not in the motion of the ideal massless spring.

What you need to connect is $x$ and $d$. Then you can differentiate the energy equation, and you will obtain the second derivative of $d$ multiplied with the mass; whatever this is equal to is the force.

10. Nov 7, 2013

### DorelXD

Well, $\frac{T}{cos{\alpha}} = k(y_0 + x )$

$F_e = k(x+y_0)$

That, I don't know. How can I find out ?

Well I can find this out by writing Newton's second law, right? But I didn't do it, because I thought that first we should find a relationship between x and the height of mass.

11. Nov 7, 2013

### DorelXD

I dont understand. I must say that my knowledge of calculus is pretty low. Can you explain a little more detalied, please ? What you want me to do is derivate the whole relationship found for energy? :D

Yes, that's what I meant. Sorry for the bad exprimation...:(

12. Nov 7, 2013

### voko

If you do not understand that, then I suggest we drop this approach. Use the forces.

You still have to find the relationship between $x$ and $d$.

13. Nov 7, 2013

### DorelXD

Ok. That's what I've been trying to do. Nothing seems to work. What do you suggest? :)

14. Nov 7, 2013

### voko

Have a look at post #4.

15. Nov 7, 2013

### voko

#### Attached Files:

• ###### spring.pulley.angle.png
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16. Nov 7, 2013

### DorelXD

Thank you! Now It begins to make sense. Can you tell me what program did you used for drawing this on a pc? I would like to post better pictures, so that people who answer me can help me much easier.

Well, from now on, I'm gonna use the notation on your drawing, because it's very clear and well done.

So, on your drawing , we know that: $c + d = constant$. But I can't see the displacement of the body. And we need to find a connection between the displacement of the body and the comprimation of the resort, right?

17. Nov 7, 2013

### DorelXD

Oh, wait a minute. Your drawing is for a general state. So let's consider the length a,b,c,d,x for state 1 and a', b',c',d' for state 2. Also let's consider alpha and alpha'

We have that: $c + d = c' + d' \to d' - d = y = c - c'$
We can easily see that: $c = \frac{a -x }{cos{\alpha}}$ and $c' = \frac{a - x' }{cos{\alpha'}}$

So, combining the two equations: $y =\frac{a - x}{cos{\alpha}} - \frac{a - x' }{cos{\alpha'}}$

We also know that the difference $\alpha' - \alpha \to 0$

Does anything of what I wrote helps in any way :D ?

Last edited: Nov 7, 2013
18. Nov 7, 2013

### DorelXD

I did it!!!!!!!

So $$y = \frac{a-x}{\cos{\alpha}} - \frac{a-x'}{\cos{\alpha'}} = \frac{a}{\cos{\alpha}} -\frac{x}{\cos{\alpha}} + \frac{a}{\cos{\alpha'}} +\frac{x'}{\cos{\alpha'}}$$

$$y = a(\frac{1}{\cos{\alpha}}- \frac{1}{\cos{\alpha'}}) - \frac{x}{\cos{\alpha}} + \frac{x'}{\cos{\alpha'}}$$

Since alpha is very very close to alpha' we can neglect the term with a and say that:

$$y = - \frac{x}{\cos{\alpha}} + \frac{x'}{\cos{\alpha'}}$$

Now, we are interested in $\Delta x$. We write that: $x' = \Delta x + x$ and we substitute in the last realtion found.

After a little bit of algebra we will get to:

$y\cos{\alpha}\cos{\alpha'} = x ( \cos{\alpha} - \cos{\alpha'} ) + \Delta x \cos{\alpha}$

Again, because alpha is very very close to alpha' then aplying the cosine function we will get two values very very close to each other, so we can say that their difference is zero, and hence we can neglect again the first term and say that:

$y\cos{\alpha}\cos{\alpha'} = \Delta x \cos{\alpha}$ which means that: $y\cos{\alpha'} = \Delta x$

Now, for convenience let's say directly that: $y\cos{\alpha'} = x$ . I know that changing notation in the middle of an exercise isn't a good practice, but in this case, given the fact that I clearly specified what I did, and that I'm not on a contest or test, I'm taking the liberty to do so.

Now, is my reasoning correct? Is this a correct result? I strongly believe that it is. And it's a very beautiful one. Now, please tell me if my work is good, so we can continue with th exercise :D

19. Nov 7, 2013

### haruspex

No, you can't do that. x is close to x' in the same order of magnitude, so both terms matter.
Do you know the derivative of sec?

20. Nov 7, 2013

### voko

I used Inkscape.

No, unfortunately you cannot.

But I have a question to ask. Why did you use $(a - x) = c \cos \alpha$ and not $b = c \sin \alpha$? The latter seems a better choice, because it includes only one variable, the angle.