Proof That The Following System Does Uniform Oscilations

[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.

 

[DorelXD]

You made a mistake there.

Right, I forgot the sin..

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

Replacing in the force expresion we get that:

T&#039; =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&#039; = 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&#039; = 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&#039;= 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&#039; + mg = k(c-y)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 ) + mg

And now I'm stuck again...

 

[voko]

You changed the sign in $T'$, and then used $-T'$ in the resultant, huh? Think again.

As to what to do next, if you have a function f(x) and know the value of the function and its derivative at some value of x, how can you approximate the values of the function in the neighborhood of that point?

 

[DorelXD]

Yes, I was a little confused. By doing that, I still have the tension pointing downward. So, I don't change the sign in T' and I have that:

R = -T&#039; + mg = k(y-c)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 ) + mg

As to what to do next, if you have a function f(x) and know the value of the function and its derivative at some value of x, how can you approximate the values of the function in the neighborhood of that point?

Agh...this is beyond my knowledge. If before I wasn't able to answer corectly at the questions you asked me, because I was nervous and frustrated, now I simply don't know the answer to that. I haven't studied the derivate yet. I just know the concept, that's all.

 

[voko]

Honestly, I do not know how you are supposed to perform the final step without using some calculus.

What has been explained to you in your class?

 

[DorelXD]

Well, I haven't studied it in school yet, but I took a closer look by myself, as I was curious. Well, I understood the concept, and the graphic interpretation. I also know the derivate of a few elementary functions, and and some basic derivation rules. I'm sorry for my late response. So, what do you suggest? You could show me what you mean, and then I could ask you what I don't understand.

P.S. : I edited my post because it was full of typing mistakes. I don't know why I made so many.

 

[voko]

You have the resultant function as a function of $y$: $R(y)$.

Now, for any function $R(y)$ its derivative $R'(y)$ at point $y_0$ is defined as the limit of $$ \frac {R(y_0 + \Delta y) - R(y)} {\Delta y} $$ when $\Delta y \to 0$. That means that when $\Delta y$ is small, we have $$ R'(y_0) \approx \frac {R(y_0 + \Delta y) - R(y)} {\Delta y} $$

In your case, $y_0 = 0$ and $\Delta y$ is simply $y$, so $$ R'(0) \approx \frac {R(y) - R(0)} {y} \to R(y) \approx R(0) + R'(0)y$$

And what is $R(0)$, if $y = 0$ is the equilibrium position?

 

[DorelXD]

But I have a question. The derivate at point $y_0$ shouldn't be:

defined as the limit of $$ \frac {R(y_0 + \Delta y) - R(y_0)} {\Delta y} $$ when $\Delta y \to 0$ ?

You wrote:

$\frac {R(y_0 + \Delta y) - R(y_)} {\Delta y}$

Well, we found the tension when $y = 0$ :

R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 )

 

[voko]

But I have a question. The derivate at point $y_0$ shouldn't be:

defined as the limit of $$ \frac {R(y_0 + \Delta y) - R(y_0)} {\Delta y} $$ when $\Delta y \to 0$ ?

Yes you right, I made a mistake.

R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 )

You are forgetting the force of weight in the resultant. But, again, think of this physically: $y = 0$ is the equilibrium point. What is the net (resultant) force on a body in equilibrium?

 

[DorelXD]

Ahhhhh well, it is obviously 0 . I'm sorry for not answering so quick, but personally I'm not in a hurry, and I hadn't enough time to use the laptop or my smartphone :(

 

[voko]

Do you see then, that all that you need is to find $R'(0)$?

 

[DorelXD]

So, R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 ) + mg

And I need to derivate this function, right ?

 

[voko]

Yes, you need to differentiate $R(y)$.

 

[DorelXD]

But you said that I need to differentiate R(0), not R(y) .

 

[voko]

You cannot differentiate $R(0)$. It is a number, not a function. You differentiate the function with respect to its arguments, which gives you its derivative function, then you evaluate the derivative at 0.

 

[DorelXD]

But I've never differentiated something like that:

R(y) = k(y-c)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 ) + mg

It looks difficult...

 

[voko]

I am not sure you have to. If you have this problem while not having sufficient knowledge of calculus, it may be that you are supposed to approach it differently. I asked you earlier what you had learned in your class before, that might give you a clue.

 

[DorelXD]

There is indeed a different approach. I have just talked to somebody I know, and he gave me a hint. Actually he remindend me of some basic geomtery. Well, I will shortly post the beginning of the solution. You'll se that it's quite interesting, and that we both missed some things.

The solution I will post uses the aproximation of the trigonomteric functions when the angles are very small, and some basic euclidian geometry. It was fairly simple. I don't know how we both missed it...