Springs and small oscillations

In summary, the conversation discussed a problem involving a rod of length l and mass m, pivoted at one end and held by springs at its midpoint and far end. The springs had a spring constant k and their pull was perpendicular to the rod at equilibrium. The goal was to find the frequency of small oscillations around the equilibrium position. The conversation went on to discuss how to approach the problem, including calculating torque and using terms of order θ^2 and higher. The importance of paying attention to statements about the direction of the forces was also emphasized. The conversation concluded with a recommendation to keep more terms than needed until a single equation is derived, and to be alert for the possibility of a wrong prediction.
  • #1
madafo3435
55
15
[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.
 

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  • #2
madafo3435 said:
approaching for small θ
How about dropping terms of order ##\theta^2## and higher ?
(in this case actually ##\theta^3## and higher :smile: , I think)
 
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  • #3
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:
madafo3435 said:
pulling in opposite directions
 
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  • #4
haruspex said:
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 ...
 
  • #5
BvU said:
How about dropping terms of order ##\theta^2## and higher ?
(in this case actually ##\theta^3## and higher :smile: , I think)
I thought about that, but ... this order if it can be despised?
 
  • #6
madafo3435 said:
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.
 
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  • #7
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
 
  • #8
haruspex said:
##\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
 
  • #9
madafo3435 said:
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.
 
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  • #10
haruspex said:
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
 
  • #11
madafo3435 said:
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##.
 
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  • #12
haruspex said:
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?
 
  • #13
madafo3435 said:
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.
 
  • #14
haruspex said:
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?
 
  • #15
madafo3435 said:
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.
 
  • #16
haruspex said:
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)##
 
  • #17
No. $$a+\mathcal O(x) \qquad{\sf if} \qquad a \ne 0$$ and
$$bx+\mathcal O(x^2) \qquad{\sf if} \qquad a= 0$$
 
  • #18
BvU said:
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
 
  • #19
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
 
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  • #20
BvU said:
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
 

Related to Springs and small oscillations

1. What is a spring?

A spring is a flexible object, usually made of metal, that can be stretched or compressed and has the ability to return to its original shape. It stores potential energy when it is stretched or compressed and releases it when it returns to its original shape.

2. What is the relationship between a spring's force and its displacement?

The force exerted by a spring is directly proportional to its displacement from its equilibrium position. This relationship is described by Hooke's Law, which states that the force (F) equals the spring constant (k) multiplied by the displacement (x): F = -kx.

3. What is a small oscillation?

A small oscillation is a type of periodic motion where the object moves back and forth around a stable equilibrium point. In the case of a spring, this would be the point where the spring is neither stretched nor compressed. The amplitude of the oscillation is small, meaning the displacement is much smaller than the length of the spring.

4. What factors affect the frequency of a spring's oscillation?

The frequency of a spring's oscillation is affected by its mass, spring constant, and the amplitude of the oscillation. A heavier mass or a stiffer spring will result in a lower frequency, while a larger amplitude will result in a higher frequency.

5. How is the period of a spring's oscillation calculated?

The period of a spring's oscillation is calculated by dividing the time it takes for one complete oscillation (T) by the number of oscillations (n): T = t/n. It can also be calculated by taking the inverse of the frequency (f): T = 1/f. The period is measured in seconds and is a measure of how long it takes for the spring to complete one full cycle of oscillation.

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