What Is the Frequency of Small Oscillations for a Pivoted Rod with Two Springs?

[madafo3435]

[Moved from technical forums, so no template]
Summary:: A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations around the equilibrium position.

(I attach my drawing below)

The Attempt at a Solution

first calculated the torque:

τ=-F(L/2)sen(θ+90°)-F'Lsen(θ+90°)+mg(L/2)sen(θ)

τ=-F(L/2)cos(θ)-F'Lcos(θ)+mg(L/2)sen(θ)
​

approaching for small θ:

τ=-F(L/2)(1-θ^2/2)-F'L(1-θ^2/2)+mg(L/2)θ
​

replacement:

F=K(L/2)θ

F'=KLθ​

then i use:

τ=Iθ''​

I have problems here, I cannot continue, I appreciate any help.

 

[BvU]

approaching for small θ

How about dropping terms of order $\theta^2$ and higher ?
(in this case actually $\theta^3$ and higher , I think)

 

[haruspex]

As @BvU implies, you do not need the cosine factors. For small oscillations the lever arm is just L or L/2.
You also need to pay attention to this statement:

pulling in opposite directions

 

[madafo3435]

As @BvU implies, you do not need the cosine factors. For small oscillations the lever arm is just L or L/2.
You also need to pay attention to this statement:

II was thinking about that, but I don't understand why the cosine factor is negligible ...

 

[madafo3435]

How about dropping terms of order $\theta^2$ and higher ?
(in this case actually $\theta^3$ and higher , I think)

I thought about that, but ... this order if it can be despised?

 

[haruspex]

II was thinking about that, but I don't understand why the cosine factor is negligible ...

$\theta^2$ can be ignored provided some $\theta$ term remains, i.e. don't all cancel. The equation will then be a good approximation for all sufficiently small $\theta$.
Sometimes you only later find all lower order terms cancel and have to go back and keep more terms.
A trap I have seen an official answer fall into was to discard some $x^2$ terms early and not others that snuck in later. When all the $x$ terms cancelled, the resulting equation was invalid, but this wasn't noticed.

 

[madafo3435]

Thanks for that, I would like to tell you, even so, I would like to be more sure when to ignore these terms in a future problem, so could you please recommend some reading that can help me to be more persuasive with these issues ... thank you very much

 

[madafo3435]

$\theta^2$ can be ignored provided some $\theta$ term remains, i.e. don't all cancel. The equation will then be a good approximation for all sufficiently small $\theta$.
Sometimes you only later find all lower order terms cancel and have to go back and keep more terms.
A trap I have seen an official answer fall into was to discard some $x^2$ terms early and not others that snuck in later. When all the $x$ terms cancelled, the resulting equation was invalid, but this wasn't noticed.

Thanks for that, I would like to tell you, even so, I would like to be more sure when to ignore these terms in a future problem, so could you please recommend some reading that can help me to be more persuasive with these issues ... thank you very much

 

[haruspex]

Thanks for that, I would like to tell you, even so, I would like to be more sure when to ignore these terms in a future problem, so could you please recommend some reading that can help me to be more persuasive with these issues ... thank you very much

The safe way is to keep more terms than you think you will need until you have it all in one equation. At that point, you can usually figure out fairly easily which terms vanish first as the variable tends to zero. But in general it is a problem in limits, and you have probably seen limits problems that are quite subtle, so I can't give you a standard procedure.

More usually, in the real world, it is not hard once it is down to a single equation. The skill is in seeing in advance which terms can be discarded while building towards that equation. This saves time, but you do need to be alert to the possibility of a wrong prediction. One indicator may be that you end up with the 'small' variable ending up with a nonzero constant value, or vanishing from the equation entirely.

 

[madafo3435]

The safe way is to keep more terms than you think you will need until you have it all in one equation. At that point, you can usually figure out fairly easily which terms vanish first as the variable tends to zero. But in general it is a problem in limits, and you have probably seen limits problems that are quite subtle, so I can't give you a standard procedure.

More usually, in the real world, it is not hard once it is down to a single equation. The skill is in seeing in advance which terms can be discarded while building towards that equation. This saves time, but you do need to be alert to the possibility of a wrong prediction. One indicator may be that you end up with the 'small' variable ending up with a nonzero constant value, or vanishing from the equation entirely.

I understand your point, my way to rectify before replacing one expression with another, is to see if both expressions are equivalent in the desired limit, only when I verify this replacement, and one can realize that in this case cos (theta) ~ 1 when theta tends to 0

 

[haruspex]

my way to rectify before replacing one expression with another, is to see if both expressions are equivalent in the desired limit

No, it's not quite that simple.
Suppose we have $e^y=\cos(x)$, and we are interested in y as a function of x when x is small.
On each side, the function equates to 1 in the limit, but replacing each with 1 is not useful. Keeping the first two terms each side we get $2y=-x^2$.

 

[madafo3435]

No, it's not quite that simple.
Suppose we have $e^y=\cos(x)$, and we are interested in y as a function of x when x is small.
On each side, the function equates to 1 in the limit, but replacing each with 1 is not useful. Keeping the first two terms each side we get $2y=-x^2$.

You are right, and what if I have two equivalent functions and that they are infinitesimal in x, could we replace these functions for values close to x no? ... On the other hand, what you tell me confuses me a little, because in the problem of the beginning , I would be replacing 1-theta ^ 2/2 by 1, do you understand my confusion?

 

[haruspex]

in the problem of the beginning , I would be replacing 1-theta ^ 2/2 by 1

Right, but in this case there is a $\theta ^ 1$ term elsewhere in the equation which is not going to get cancelled, so the $\theta ^ 2$ terms can be discarded. In my example in post #11 all terms lower than $x^2$ disappeared.

 

[madafo3435]

Right, but in this case there is a $\theta ^ 1$ term elsewhere in the equation which is not going to get cancelled, so the $\theta ^ 2$ terms can be discarded. In my example in post #11 all terms lower than $x^2$ disappeared.

So what you want to tell me is that for very small values of theta, any integer power of this value greater than 1 will be much closer to 0. Is this correct?

 

[haruspex]

So what you want to tell me is that for very small values of theta, any integer power of this value greater than 1 will be much closer to 0. Is this correct?

Each increase in power means the term will shrink faster as the variable approaches zero.
Consider the expression $a+bx+cx^2$. If a is nonzero then the least order approximation is a. If a is zero but b is nonzero then the least order approximation is bx, etc.

 

[madafo3435]

Each increase in power means the term will shrink faster as the variable approaches zero.
Consider the expression $a+bx+cx^2$. If a is nonzero then the least order approximation is a. If a is zero but b is nonzero then the least order approximation is bx, etc.

can i see it like this?
$a + bx + cx ^ 2 = a + o(bx + cx^2)$ and $bx + cx ^ 2 = bx + o(x)$

 

[BvU]

No. $$a+\mathcal O(x) \qquad{\sf if} \qquad a \ne 0$$ and
$$bx+\mathcal O(x^2) \qquad{\sf if} \qquad a= 0$$

 

[madafo3435]

No. $$a+\mathcal O(x) \qquad{\sf if} \qquad a \ne 0$$ and
$$bx+\mathcal O(x^2) \qquad{\sf if} \qquad a= 0$$

You are right, and I understand with intuition what they have explained to me, but I need a rigorous explanation of why I can replace $1-\frac{\theta ^ 2}{2}$ with $\theta$, it is intuitive for me, but it bothers me a little not to have a more rigorous understanding of what that i am accepting

 

[BvU]

You do not replace $1-\frac{\theta ^ 2}{2}$ by $\theta$. You replace it by 1. Or rather:
You write $\ \theta\,(1-\frac{\theta ^ 2}{2})\$ as $\quad \theta + \mathcal O(\theta^3)\$ and ignore terms of higher order than $\theta$.

Every time you see 'small ... ' in a problem statement, you are supposed to work out the lowest order (with non-zero coefficient) only. cf the mathematical pendulum

 

[madafo3435]

You do not replace $1-\frac{\theta ^ 2}{2}$ by $\theta$. You replace it by 1. Or rather:
You write $\ \theta\,(1-\frac{\theta ^ 2}{2})\$ as $\quad \theta + \mathcal O(\theta^3)\$ and ignore terms of higher order than $\theta$.

Every time you see 'small ... ' in a problem statement, you are supposed to work out the lowest order (with non-zero coefficient) only. cf the mathematical pendulum

yes, I was wrong, my intention was to replace by. At the moment I will face the exercises as indicated, and I will make the effort to understand the insignificant error between these jumps and orders