Proof That The Following System Does Uniform Oscilations

[DorelXD]

Let's just put together a sufficient set of equations before we worry about which variables to eliminate. Writing
s = relaxed length of spring
x = compression of spring
k = spring constant
α = angle
y = length of hypotenuse
b = vertical distance from pulley to spring
T = tension
m = mass
there are fairly simple equations relating:
1. b, s, x, α
2. T, α, k, x
3. y, b, α
4. T, m, g, $\ddot y$
Four variables, four equations, which in principle can boil down to one variable, one equation. But actually doing that will produce quite a mess, so instead consider perturbations from the equilibrium: ΔT = T - T₀, etc. Note that $\ddot y = \ddot{\Delta y}$.
Even then, you don't need to be too exact. You only need to show that $\ddot{\Delta y}$ ~ $c \Delta y$ for some negative c.

Ok, got it. So, let's roll. The initial state coresponds to the equilibrium state, so we have:

mg = T and T\cos{\alpha} = kx_ 0. Combining the two equations we have that: mg\cos{\alpha} = kx_0, which is useful because now we have an expresion for the initial elongation of the resort.

Now, let's consider another state. We have: F_e = T'\cos{\alpha'} => k(x_0 +x ) = T'\cos{\alpha'}, which gives us an expresion for T', that this the tension at a particular moment:
T' = \frac{k(x+x_0)}{\cos{alpha'}}

Where x_0 is the inital displacement of the resort, and x_0 is its elongation at a given time, i.e. the distance from the equilibrium position. Combining the equations we have that:


T' = \frac{k(x+\frac{mg\cos{\alpha}}{k})}{\cos{\alpha'}}


Hmmm, so let's observe something. The resultant force depends on the tension and weight. The tension is the only variable force. So if the resultant force will be variable, it will be because of the tension. So, I believe we should take a closer look at the expresion for the tension, before writing the second law of dynamics. What dou you say ? Is my work now correct ?

P.S. : I'm deeply sorry for the eventual mistakes of typing. I respect the language, but I do write some words wrong :( .

 

[voko]

mg = T and T\cos{\alpha} = kx_ 0. Combining the two equations we have that: mg\cos{\alpha} = kx_0, which is useful because now we have an expresion for the initial elongation of the resort.

Good.

Now, let's consider another state. We have: F_e = T'\cos{\alpha'} => k(x_0 +x ) = T'\cos{\alpha'}

Not good. The spring force is $kx$, not $kx(x_0 + x)$ (remember, $x$ is measured from the position of the end of the spring when the spring is not stressed).

 

[DorelXD]

Good.
Not good. The spring force is $kx$, not $kx(x_0 + x)$ (remember, $x$ is measured from the position of the end of the spring when the spring is not stressed).

Well, then we have that:

T' = \frac{kx}{\cos{\alpha'}}

I guess I had forgotten how we defined x, when I wrote the post.

 

[voko]

That is correct, but what would you do next? Note that the expression includes two variables, the compression of the spring and the angle.

 

[DorelXD]

Well, I don't like that x. So I would try to get rid of it, by using a little trigonometry, although I'm not exactly sure how to do it yet. I'll take your advice and look at the tangent to see if I get something useful.

What do you think? Is it a good approach ?

 

[voko]

Please see #20.

 

[haruspex]

Please see #20.

... and, equivalently, eqn (1) in post #30.

 

[DorelXD]

Well, from post #20 I used:

b = c\sin{\alpha} and b=c'\sin{\alpha'} , and the fact that the length of the wire remains the same: c + d = c' + d' and I found that : d - d' = b(\frac{1}{\sin{\alpha}}-\frac{1}{\sin{\alpha'}})

 

[haruspex]

Well, from post #20 I used:

b = c\sin{\alpha} and b=c'\sin{\alpha'} , and the fact that the length of the wire remains the same: c + d = c' + d' and I found that : d - d' = b(\frac{1}{\sin{\alpha}}-\frac{1}{\sin{\alpha'}})

Oh, right - it's not equivalent to (1) in post #30.
In terms of Voko's diagram at post #15, my eqn (1) relates a, x, b and α. This has the advantage that it only contains two variables, x and α. (In my post #30 I changed a to s to avoid confusion with angle α, which was labelled as a in the OP.)

 

[DorelXD]

Ok, so what should we do next?

 

[voko]

$d' - d = y$ is the dynamic variable you are interested in; now you have an equation that relates $y$ and $\alpha$, and an equation relating the force and $\alpha$. If you now relate the force and $y$, will that finish your task?

 

[DorelXD]

$d' - d = y$ is the dynamic variable you are interested in; now you have an equation that relates $y$ and $\alpha$, and an equation relating the force and $\alpha$. If you now relate the force and $y$, will that finish your task?

Yes, you're right. Well, we have two main equations:

T' = \frac{kx}{\cos{\alpha'}} (1)

y = b(\frac{1}{\sin{\alpha}} - \frac{1}{\sin{\alpha'}}) (2)

After taking a look at the system of equations, we see that indeed, it comes in handy to consider another equation:

\tan{\alpha'} = \frac{\sin{\alpha'}}{\cos{\alpha'}} = \frac{b}{a-x} (3)

Also, I would like to ask you if, although the problem doesn't specify it directly we can assume the following: x <<a

Combining euations (1) and (3), that is, replace the cos in the first equation with its expresion from the tangent function, I got that:

T' = \frac{kxb}{(a-x)\sin{\alpha'}}

A second move I would to would be combining this equation with equation number 2 (that is replacing sin(alpha') )

Now, have I worked correctly?

 

[voko]

I do not follow your logic. I would expect that you used equation (3) to eliminate $x$ and thus express the force via $\alpha$, which would then enable you to express the force via $y$. What you did instead is, mildly said, puzzling.

 

[DorelXD]

I do not follow your logic. I would expect that you used equation (3) to eliminate $x$ and thus express the force via $\alpha$, which would then enable you to express the force via $y$. What you did instead is, mildly said, puzzling.

You're right. =))))) . I will immediately post the right calculations.

 

[DorelXD]

I got that:

T' = \frac{ka}{\cos{\alpha'}} - b\frac{1}{\sin{\alpha'}}

And, replacing the sin I got that:

T' = \frac{ka}{\cos{\alpha'}} - b(\frac{1}{\sin{\alpha}}-\frac{y}{b})

Now, we know the following: \alpha'= \Delta\alpha + \alpha, so: \cos{\alpha'} = \cos{(\alpha+\Delta\alpha)} = \cos{\alpha}\cos{\Delta\alpha} - \sin{\alpha}\sin{\Delta\alpha}but since \Delta\alpha is very small we can say that:
cos{\alpha'} = \cos{\alpha} -\Delta\alpha\sin{\alpha}

Is it ok now ?

 

[voko]

I got that:

T' = \frac{ka}{\cos{\alpha'}} - b\frac{1}{\sin{\alpha'}}

This is not consistent dimensionally.

 

[DorelXD]

What do you mean? Did I make a mistake in the calculations ?

 

[voko]

I mean that T' = \frac{ka}{\cos{\alpha'}} - b\frac{1}{\sin{\alpha'}} is internally inconsistent. The dimension of $\frac{ka}{\cos{\alpha'}}$ is that of force (Newtons), the dimension of $b\frac{1}{\sin{\alpha'}}$ is of length (meters). It is not possible to subtract meters from Newtons. You made a mistake somewhere.

 

[DorelXD]

I got that:

T' = \frac{ka}{\cos{\alpha'}} - b\frac{1}{\sin{\alpha'}}

And, replacing the sin I got that:

T' = \frac{ka}{\cos{\alpha'}} - b(\frac{1}{\sin{\alpha}}-\frac{y}{b})

Now, we know the following: \alpha'= \Delta\alpha + \alpha, so: \cos{\alpha'} = \cos{(\alpha+\Delta\alpha)} = \cos{\alpha}\cos{\Delta\alpha} - \sin{\alpha}\sin{\Delta\alpha}but since \Delta\alpha is very small we can say that:
cos{\alpha'} = \cos{\alpha} -\Delta\alpha\sin{\alpha}

Is it ok now ?

I found the mistake, it's actually:

T' = \frac{ka}{\cos{\alpha'}} - kb(\frac{1}{\sin{\alpha}}-\frac{y}{b})

I forgot to write that 'k'. What do you think now :D ?

 

[voko]

This is better, but you still have $\alpha '$ in the equation. Express it via $y$ and you are almost there.

 

[DorelXD]

This is better, but you still have $\alpha '$ in the equation. Express it via $y$ and you are almost there.

But how can I do that? I can't see any way.

 

[voko]

How are sine and cosine related?

 

[DorelXD]

How are sine and cosine related?

The function tan relates them.

\tan{a} = \frac{\sin{a}}{\cos{a}}

 

[voko]

No, that's not useful. There is a direct relationship between the sine and the cosine, which is equivalent to Pythagoras's theorem.

 

[DorelXD]

Oooo, well that's simpler:

\cos^{x} + \sin^2{x} = 1 [\tex]

 

[voko]

And the result is?

 

[DorelXD]

<br /> <br /> \cos{\alpha&#039;} = \sqrt{1-\sin^2{\alpha&#039;}}<br />

 

[voko]

Are you going to spoon-feed us, one small thing at a time? You have an equation that relates $\sin \alpha'$ with $y$, so you have everything you need to eliminate $\alpha'$ from the force equation. Do it.

 

[DorelXD]

I'm sorry! I didn't mean to be rude, I just want to be sure that I do everything right..this problem frustrates me...Sorry. I appreciate every second of your time, believe me..

So, here I have that:

\frac{1}{\sin{\alpha&#039;}} = \frac{1}{\sin{\alpha}} - \frac{y}{b} \to \sin{ \alpha&#039; } = \frac{b}{b - y\sin{ \alpha&#039;} } \to \cos{\alpha&#039;} = \sqrt{1- \frac{b^2}{(b - y\sin{ \alpha)^2} }}


And we need to replace in:

T&#039; = \frac{ka}{\cos{\alpha&#039;}} - kb(\frac{1}{\sin{\alpha}}-\frac{y}{b})


This...looks...well..awful. Please, don't get mad for not plugging the expression of cos' in the expresion of force, it's just that...I don't know, looks difficult.

 

[voko]

\frac{1}{\sin{\alpha&#039;}} = \frac{1}{\sin{\alpha}} - \frac{y}{b} \to \sin{ \alpha&#039; } = \frac{b}{b - y\sin{ \alpha&#039;} } \to \cos{\alpha&#039;} = \sqrt{1- \frac{b^2}{(b - y\sin{ \alpha)^2} }}

You made a mistake there. $$

\frac{1}{\sin{\alpha'}} = \frac{1}{\sin{\alpha}} - \frac{y}{b} \to \sin{\alpha'} = \frac {1} {\frac{1}{\sin{\alpha}} - \frac{y}{b}} \to \sin{\alpha'} = \frac {b \sin \alpha} {b - y \sin \alpha}
$$

But it is probably much easier to recall that $$ c = \frac {b} {\sin \alpha} = \text{const} $$ and have $$ \sin \alpha' = \frac b {c - y} $$

And we need to replace in:

T&#039; = \frac{ka}{\cos{\alpha&#039;}} - kb(\frac{1}{\sin{\alpha}}-\frac{y}{b})


This...looks...well..awful. Please, don't get mad for not plugging the expression of cos' in the expresion of force, it's just that...I don't know, looks difficult.

Yeah, the final expression is a mess. But you should not care. You need to approximate the net force (which includes a constant weight) for small values of $y$, so that the result is linear in $y$. You are almost there.