Determine the period of small oscillations

[Jorgen1224]

Homework Statement

Two balls of mass m are attached to ends of two, weigthless metal rods (lengths l₁ and l₂). They are connected by another metal bar.
Determine period of small oscillations of the system

Homework Equations

E_k=mv²/2
v=dx/dt
Conversation of energy
2πsqrt(M/k)

The Attempt at a Solution

That is actually the problem. I know the solution, but i don't really understand it and i wanted to ask if anyone could explain it to me or help me understand. I'm especially curious about the 1st and 3rd lines.

1st equation i don't understand at all, where did it come from? This root makes me think that pythagora's theorem was used, but how did it even turn into this fraction at the end?

I understand the beginning of 3rd equation, i suppose that's pythagoras theorem there, but why written in derivatives? It's basically v_x and v_y And how did this square root turn into a different one, but times dx/dt? Apparentaly this root is equal 1.

 

[gneill]

You know that $\cos(\theta) = \sqrt{1- \sin(\theta)^2}$. Consider the small angle approximation for your scenario.

 

[haruspex]

i don't really understand it

There is a small angle approximation in the diagram itself. In reality, the horizontal displacements will not be equal. The exact relationship is $l_1l_2(1-\cos(\theta_1-\theta_2))=h(l_2\sin(\theta_2)-l_1\sin(\theta_1))$, where h is the horizontal distance between the string supports. But the LHS is order θ², whereas the RHS is order θ, so to a first order approximation $l_2\sin(\theta_2)=l_1\sin(\theta_1)$.
Further approximations in the algebra are just more of the same.

Handling approximations correctly can be a bit tricky. You might keep the first two order terms (a constant and θ term) and discard the rest (θ² and up) only to find that the terms you kept all cancel later and you should have kept the quadratic term. Generally speaking, though, sin(θ) can be safely turned into θ (discarding θ³ and up) and cos(θ) into 1-½θ².

At the end of eqn 3, there is a factor like √(1+x²). This can be expanded by the standard method as 1+½x²-x⁴/8... At this stage, it can be seen that the 1 is not going to cancel with anything, so we can safely discard the x² term onwards.

 

[Jorgen1224]

Okay, now i think i do understand 3rd equation, but I'm still stuck at the first one. In this equationl₁l₂(1-cos(θ₁-θ₂))=h(l₂sin(θ₂)-l₁sin(θ₁)) if we make an approximation l_2\sin(\theta_2)=l_1\sin(\theta_1) doesn't that make the right side equal to zero?

And should i use here l₁l₂(1-cos(θ₁-θ₂)) formula for cos(θ₁-θ₂)=cosθ₁cosθ₂+sinθ₁sinθ₂? If so, what then? I don't see it disappearing anyhow.

I'm sorry, this is my first time ever with such approximations and I'm getting lost in math here a little bit too.

 

[Jorgen1224]

And where did that relationship come from? i mean l₁l₂(1-cos(θ₁-θ₂))=h(l₂sin(θ₂)-l₁sin(θ₁))

 

[haruspex]

And where did that relationship come from? i mean l₁l₂(1-cos(θ₁-θ₂))=h(l₂sin(θ₂)-l₁sin(θ₁))

That's just geometry. To get it, express the length of rod connecting the masses in terms of the other rod lengths, the distance between the supports, and the angles the hanging rods make to the vertical. Equate it to that same length when the angles are zero.
But I only went through that exercise because it seemed to me that the assumption in the diagram that the horizontal displacements ("x") are near enough equal was not quite obvious.
So the choice for you is to accept the diagram as given or develop the equation I posted and make the approximation from there.

 

[haruspex]

In this equationl1l2(1-cos(θ1-θ2))=h(l2sin(θ2)-l1sin(θ1)) if we make an approximation l_2\sin(\theta_2)=l_1\sin(\theta_1) doesn't that make the right side equal to zero?

Yes, that's the point. As we make the two angles smaller and smaller, the left hand side gets smaller like θ², whereas the sin θ terms on the right only get smaller like θ. So the difference between $l_2\sin(\theta_2)$ and $l_1\sin(\theta_1)$ must get smaller like θ².

 

[SammyS]

That's just geometry. To get it, express the length of rod connecting the masses in terms of the other rod lengths, the distance between the supports, and the angles the hanging rods make to the vertical.

As I read the problem, I interpret it as: the motion is entirely in the vertical plane,

 

[haruspex]

As I read the problem, I interpret it as: the motion is entirely in the vertical plane,

In the vertical plane containing the three rods, yes. What did I write that suggested otherwise?

 

[SammyS]

In the vertical plane containing the three rods, yes. What did I write that suggested otherwise?

The following from Post #6 .

That's just geometry. To get it, express the length of rod connecting the masses in terms of the other rod lengths, the distance between the supports, and the angles the hanging rods make to the vertical. Equate it to that same length when the angles are zero.

Admittedly, I did not make an attempt to replicate your result with that or any other assumption.

A second look: My considering that you might think that the motion was perpendicular to the vertical plane (passing through the rod hangers) is rather silly. (DUH ! ) No way would $x$ be even close to the same for the two balls.

 

[Jorgen1224]

And I'm still extremely confused. I do understand now the beginning of the first equation, but I don't see where did x²/2l_1,2 come from. How was this calculated?

 

[haruspex]

And I'm still extremely confused. I do understand now the beginning of the first equation, but I don't see where did x²/2l_1,2 come from. How was this calculated?

It is the first two terms in the Taylor series for $(1-\frac{x^2}{l^2})^{\frac 12}$.
See https://en.m.wikipedia.org/wiki/Taylor_series