# Rotating bar hanging on a spring (oscillation)

1. Mar 27, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'm trying to solve a basic problem about oscillations, but I struggle to get everything together when it comes to the differential equation...I hope someone can help me to understand it better :)

A thin bar of mass m and length l is pivotable about its hanging point A, and can oscillate with the help of a spring (with spring constant k) attached to its other end. Consider there is no deflection.
a) Put up the equation of movement and the resulting differential equation under the use of the displacement angle β.
b) What is the time period T of the system?
c) Give the amplitude A and the phase angle φ of the oscillation, if the bar is deflected at the angle β0 at time t = 0s and has an angle velocity ψ0.
A possible approach to the resolution of the oscillation equation is β(t) = A⋅cos(ω0⋅t + φ).

(see attached picture)

2. Relevant equations

equations related to oscillation, moment of inertia of a bar, maybe torque and angular momentum

3. The attempt at a solution

Okay this is probably very wrong, but I don't want to post the problem without at least trying something:

a) I've learned how to derive the differential equation for a simple harmonic motion when the spring is horizontal, but I wonder if that still holds when it is vertical, because of the force of gravity? Maybe it is the case because "there is no deflection"? If so:

m⋅a = - k⋅x
⇔ m⋅(d2x/dt) = - k⋅x(t)
⇔ m⋅(d2x/dt) + k⋅x(t) = 0
k = ω02⋅m
d2x/dt + ω02⋅x(t) = 0

But I believes that gives me for solution x(t) = A⋅cos(ω0⋅t + φ), right? Which is stated in the problem to the difference that it is for β(t)!

I also feel the moment of inertia of the bar should play a role, since L = I⋅ω = I⋅β'. Using the parallel axis theorem I get I = ⅓⋅m⋅l2 but I don't dare to go any further because I don't really understand what I am doing.. Could someone give me a clue or briefly describe me how to tackle such problems?

Thank you very much in advance.

Julien.

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2. Mar 27, 2016

### Nathanael

What does x represent? Aren't you looking for a differential equation for β?

Since β is the angle the rod about point A, you will want to consider the net torque about that point.

3. Mar 28, 2016

### haruspex

I assume "no deflection" means that the equilibrium position is with the rod horizontal. So you need to define whether x is the spring extension or the extension relative to the equilibrium position; then figure out the relationship to the deflection angle.

4. Mar 28, 2016

### JulienB

The net torque about point A is:

Στ = Fg⋅l/2 - Fs⋅l = ½⋅m⋅g⋅l + k⋅x⋅l

I am uncertain how I should define x: the most logical to me would be to say that x=0 describes the equilibrium position of the system, because I am searching an expression for β = sin-1(x/l).

But then when I consider the equilibrium position I run into a problem: x=0 ⇔ Fs = 0 but I know it is not true (mg + kx = 0 would be true with x defined as the equilibrium position of the spring alone).

I can see the problem is in my definition of the restoring force and of x, but I can't seem to find a coherent/consistent way around it.

Also, the restoring force is changing constantly so the equalities I can obtain for Στ= 0 or ΣF = 0 don't seem to hold when there is an oscillation.

Julien.

5. Mar 28, 2016

### haruspex

So let the equilibrium extension be x0 and x be the extension relative to that at some point.
What equation can you write relating x0 to mg etc?
What equation for net torque and angular acceleration when at relative extension x?

6. Mar 28, 2016

### JulienB

@haruspex Thank you for your answer. So when the bar is at point x0, we can say that ΣF = 0 ⇔ mg = -kx0
The net torque I believe is Στ = ½mgl + kxl, therefore the angular acceleration is α = Στ/I = (½mgl + kxl)/(⅓ml2) = (-3/2⋅kx0 + kx)/ml when substituting mg by -kx0.

Does that make sense before I go further?[/sub]

7. Mar 28, 2016

### haruspex

You are overlooking the support from the hinge. Use moments here.
I defined x as the relative extension, so the total extension would be x+x0. I think you'll find that works out better.

8. Mar 28, 2016

### JulienB

@haruspex Thank you again for your answer, it really helps me to make progress with this concept (hopefully).

Indeed I forgot to take in consideration the force of the support. I try again with the torque as starting point:

Στ = 0 ⇔ Fg⋅½⋅l - Fs⋅l = 0
⇔ ½⋅m⋅g⋅l + k⋅x⋅l = 0 (here I consider as I think you said x being the difference between the equilibrium position of the spring with charge and without charge)

Then I can define my angular acceleration:
α = Στ/I = [(3/2)⋅m⋅g + 3⋅k⋅x]/m⋅l

I also know that α = dω/dt = d2β/dt
⇒ d2β/dt + 3g/2l + 3k⋅x/ml = 0

Okay that already looks a little better, though I'm not sure what to do next. I could also say:
α = d2β/dt = -A⋅ω02⋅cos(ω0⋅t + φ)
⇔ d2β/dt + A⋅(k/m)⋅cos(√(k/m)⋅t + φ) = 0

What do you think? Am I going in the right direction?

Julien

Last edited: Mar 28, 2016
9. Mar 28, 2016

### JulienB

One of the guys studying with me just explained me that setting up x as the equilibrium position of the spring WITH the bar suspended on it makes possible to come up with an equation of motion without g!! Is that true, and if so is that "always" the case with such problems? I'm preparing an exam, that's why I'm curious about general methods to solve oscillation problems.

10. Mar 28, 2016

### JulienB

I attached a drawing I did from what I understood about the (non)role of the force of gravity in such oscillations. Can you tell me if that's in general the right way to consider the issue?

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11. Mar 28, 2016

### haruspex

Yes, that's what I was trying to lead you towards but you seem to have misunderstood something.
In post #6, your equilibrium eqation was nearly right, but you missed the factor 1/2 on the mg term. In post #8 you corrected that but reverted to x instead of x0. x0 is the extension in the equilibrium position.

In your angular acceleration equation in post #8, you have used x as though it is the total extension. Since it is only the extra extension, you need x0+x there. When you get that right, you should see g and x0 disappear.

12. Mar 28, 2016

### JulienB

@haruspex Aha that's interesting, I rewrite the whole development:

Στ = 0 ⇔ ½⋅m⋅g = -k⋅x0
⇒ α = [(3/2)⋅m⋅g + 3⋅k⋅(x+x0)]/ml2 = 3⋅k⋅x/m⋅l2

Then I can rewrite that equality in the form:

d2β/dt - 3⋅k⋅x/ml2 = 0

Is that the equation of movement relative to β we are asked to find? Lol I'm not even sure what the question is anymore!

I could substitute d2β/dt by -A⋅ω02⋅cos(ω0⋅t + φ), but I think that β(t) should be the solution to the differential equation, right? However substituting ω0 by √(k/m) then leads me to:

-A⋅cos(√(k/m)⋅t + φ) - 3⋅x/l2 = 0

Julien.

Last edited: Mar 28, 2016
13. Mar 29, 2016

### haruspex

14. Mar 29, 2016

### JulienB

@haruspex Thank you for your answer and sorry for those mistakes, I had it right on my sheet. So the differential equation is now:

d2β/dt2 - 3⋅k⋅x/(m⋅l) = 0

I can substitute k/m with ω02, and could I say that x/l ≈ sin β ≈ β or is that a crime? We do it often in class when the angle β is very small, which seems to be the case here.

The resulting equation would then be:

d2β/dt2 - 3⋅β⋅ω02 = 0

I fear the worse as I just realized the triangle isn't even rectangle...

15. Mar 29, 2016

### haruspex

That all looks fine, except that it is a bit misleading to use ω02 for k/l. Omega is generally used for a frequency, and nothing here has that frequency. Better to use it for the entire coefficient of the β term in the equation, so it would be 3k/l.

16. Mar 29, 2016

### JulienB

@haruspex thank you again for all your help, I really appreciate it. Yeah I realize the frequency is not always √(k/m) (is that only in case of simple harmonic motion?).

For question b), can I calculate the time period by using the formula T = 2π/ω0? When I input β(t) = A⋅cos(ω0⋅t + φ) in my equation of motion, I can solve for ω0 but I get a complex number (ω0 = i√(3k/m)). Is that normal?

Julien.

17. Mar 29, 2016

### haruspex

When I wrote post #13, advising you to express x in terms of beta, I realised you needed to get the sign right. But this morning I forgot about that, and indeed you have the sign wrong. You are measuring x as an extension, so it increases as the rod descends, but beta decreases that way.

18. Mar 29, 2016

### JulienB

@haruspex Aaah oh my god this is full of traps! In future situations I should probably define x with the same sign as β then.. Okay then I get a period T=2π√(m/3k). I guess (hope) that is correct.

For the last question, I imagine I have to solve β0 = A⋅cos(φ) and ψ0 = -A⋅φ2⋅cos(φ) for A and φ, right?

Julien.

19. Mar 29, 2016

### haruspex

Your first equation is right (β0) but there are at least two errors in the other. Please post your working.

With regard to getting the sign right, remember that the standard form of the SHM equation is $\ddot x+\omega^2x=0$. Thus, when x is positive the acceleration is negative, and vice versa. It is this that results in oscillation. If we change it to a minus sign then the system is unstable, accelerating away to infinity. The solution to the equation will be cosh, not cos.

20. Mar 30, 2016

### JulienB

@haruspex Okay I've redone it with a friend and came up with those results:

β0 = A⋅cos(φ)
ψ0 = -A⋅ω0⋅sin(φ)
I think ω0 is the same as ψ0, right? It's the angle velocity?

That takes me to:

A⋅sin(φ) = -1
φ = -tan-1(1/β0)
and
A = β0⋅cos-1(1/β0)

Hopefully that makes sense. Thanks for your help.

Julien.