Proof That The Following System Does Uniform Oscilations

[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 dosen'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:

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

 

[voko]

And the result is?

 

[DorelXD]

$$\cos{\alpha'} = \sqrt{1-\sin^2{\alpha'}}$$

 

[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:

$$\frac{1}{\sin{\alpha'}} = \frac{1}{\sin{\alpha}} - \frac{y}{b} \to \sin{ \alpha' } = \frac{b}{b - y\sin{ \alpha'} } \to \cos{\alpha'} = \sqrt{1- \frac{b^2}{(b - y\sin{ \alpha)^2} }}$$


And we need to replace in:

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


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]

$$\frac{1}{\sin{\alpha'}} = \frac{1}{\sin{\alpha}} - \frac{y}{b} \to \sin{ \alpha' } = \frac{b}{b - y\sin{ \alpha'} } \to \cos{\alpha'} = \sqrt{1- \frac{b^2}{(b - y\sin{ \alpha)^2} }}$$

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:

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


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.

 

[DorelXD]

You made a mistake there.

Right, I forgot the sin..

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

But I can't see where does that come from.

 

[voko]

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

 

[DorelXD]

Ohhh right, then we have that:

$$\cos{\alpha'} = \sqrt { \frac{ (c-y)^2 - b^2 }{ (c-y)^2 } }$$

Replacing in the force expresion we get that:

$$T' =ka \sqrt { \frac{ (c-y)^2 }{ (c-y)^2 - b^2 } } - kb(\frac{1}{\sin{\alpha}}-\frac{y}{b})= ka \sqrt { \frac{ (c-y)^2 }{ (c-y)^2 - b^2 } } - kb(\frac{c-y}{b})$$


Hmmm, now it looks a little more friendly. And what next? I know that this is the point when we begin to "kill" the problem, but how ? How do we continue?

I really appreciate your patience, and I promise that I will be mor attentive, and that I'll work harder.

If y is very small, then we can say that c- y is always positive, so:


$$T' = ka \frac{c-y}{ \sqrt{ (c-y)^2 - b^2 } } - k(c-y)= k(c-y)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 )$$

 

[voko]

I cannot see any error in what you did, yet I am not happy with the result. When $y = 0$, we should have equilibrium, i.e, $T' = T = mg$. But we get $T' = 0$ instead. There must be a mistake somewhere, I will check all the steps you made to get this, and I advise you do the same thing.

 

[DorelXD]

Sure, I'm checking right now.

 

[DorelXD]

But if y = 0 then $$T' = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 )$$ ,which dosen't seem to be 0 .

 

[voko]

Argh. I have confused myself and misled you. Somehow I imagined that $a^2 + b^2 = c^2$, but that does not follow from my own diagram :)

Anyway, I have checked all the steps and the result is good, except for one thing. Because we measure $d$ downward from the pulley, we ended up the direction downward being positive. But the force of tension is upward, so you must either negate the entire expression for the force of tension, or change the convention and have the upward direction positive, which means changing all the signs at the $y$ variable.

And then, depending on how you fix that, add or subtract $mg$ to obtain the net force. It is the net force that you must cast into the form $-k_e y$ for small values of $y$.

 

[DorelXD]

Argh. I have confused myself and misled you. Somehow I imagined that $a^2 + b^2 = c^2$, but that does not follow from my own diagram :)

Anyway, I have checked all the steps and the result is good, except for one thing. Because we measure $d$ downward from the pulley, we ended up the direction downward being positive. But the force of tension is upward, so you must either negate the entire expression for the force of tension, or change the convention and have the upward direction positive, which means changing all the signs at the $y$ variable.

And then, depending on how you fix that, add or subtract $mg$ to obtain the net force. It is the net force that you must cast into the form $-k_e y$ for small values of $y$.

This sign convention has also put me in trouble in the past. Can you be a little more specific? Indeed, I agree with you, the downward direction is positive. So, why do we need to change the expresion for the tension? This, I don't get :( .

 

[voko]

Take your final formula for tension in #63. Let $y = 0$, then, as you observed in #66, you get $$ T' = kc\left(\frac a {\sqrt{c^2 - b^2}} - 1\right) = kc \left(\frac a {a - x} - 1\right) $$ $a > a - x$ so $a / (a - x) > 1$ and then $T' > 0$. But positive, when downward is positive, means it must be directed downward, yet tension (at the block-end of the rope, which we are considering) is always upward. Hence it is seen the sign must be changed.

Put another way. You started to derive the formula for tension by considering the spring-end of the rope, where tension is downward and to the right, so the resultant formula is correct for the tension at that end of the rope, but we need tension at the other end.

 

[DorelXD]

Now I get it! :) . So, by changing the signes, I get that:

$$T'= k(y-c)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 )$$


So the resultant force, with the positive direction downward, will be:

$$R = -T' + mg = k(c-y)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 ) + mg$$

And now I'm stuck again...