# Double Pendulum and matrices

1. May 8, 2013

### CAF123

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

3. The attempt at a solution

I would have write my work out here, but I have not managed to display matrices next to each other yet.

My problem is putting it into matrix form. The form I have is shown in the below attachement. The matrix I get for A, (at the bottom of the attachment) does not allow me to correctly compute the normal modes later, so I have made an error somewhere (besides the numbers look messy and there does not seem to be any motivation for taking g/2a outside the matrix given my work).

The a',b,c,d... etc at the top of the attachment just allowed me to determine the matrix elements more easily.

Many thanks.

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2. May 8, 2013

### TSny

Check to see if the off-diagonal terms of your matrix M should have the 2 in the denominators.

3. May 8, 2013

### CAF123

Hi TSny,
I checked it again and I still end up with the terms $ma^2/2$ on the non diagonal.

My method was to write out matrices M and K but with all their elements labelled a,b,c,d.... Then I multiplied by $\dot{\theta},\,\theta$ and their corresponding transposes through matrix multiplication. What I got was $$\frac{1}{2} (\dot{\theta_1}^2 a' + \dot{\theta_2} \dot{\theta_1}[c+b] + \dot{\theta_2}^2) + \frac{1}{2} ( \theta_1^2 e + \theta_2 \theta_1 [g + f] + \theta_2^2)$$

Then I compared this to the form in the question and this gave me the elements a',b,c.. as listed at the very top of the attachment.

EDIT: I think I see the error.

4. May 8, 2013

### CAF123

I wrote the general solution as $$\underline{\theta} = \underline{\theta}_1 + \underline{\theta}_2$$ I have the initial condition that the system is at rest so I take $\underline{\theta}(0) = 0$. Similarly, I am given the condtion that the upper pendulum experiences a disturbance such that $\dot{\theta}_1 (0) = 1$.

However, these are two initial conditions for four parameters ($\beta^{(1)}, \beta^{(2)}, \phi^{(1)}, \phi^{(2)}$) But given that the system is at rest, I thought I could also say that individually, $\theta_1(0) = 0$ and likewise for $\theta_2 (0)$. This gives me $$\phi^{(1)} = \phi^{(2)} = \frac{\pi}{2} (+ n\pi)$$ which then gives $$\langle -1/\sqrt{3}, 1 \rangle\beta^{(1)} = \frac{1}{\frac{\sqrt{3}}{2}\sqrt{3 + \sqrt{3}}}$$ but I am not sure how to find $\beta^{(2)}$

Last edited: May 8, 2013
5. May 8, 2013

### TSny

Can you show more detail on getting the last equation. Looks like you have a vector on the left but a number on the right. (?)

6. May 8, 2013

### CAF123

My expression was : $$\theta = \beta^{(1)} \frac{\sqrt{3}}{2} <-1/\sqrt{3}, 1> \cos (w_1 t -\phi^{(1)}) + \beta^{(2)} \frac{\sqrt{3}}{2} < 1\sqrt{3}, 1> \cos (w_2 t - \phi^{(2)})$$
where the first term is $\theta_1$ and the latter $\theta_2$.

Setting $\theta_1$ equal to 0 gives me $\phi^{(1)}$. Likewise for $\theta_2$ which gives me $\phi^{(2)}$. Subbing these condtions into the initial condition $\theta (0) = 0,$ just gives me 0=0 (since cos{pi/2} = 0). So then I differentiated to obtain $\dot{\theta}_1$ to get $$1 = \beta^{(1)} \frac{\sqrt{3}}{2} <-1/\sqrt{3}, 1> \sin (\phi^{(1)})$$ Put in $\phi^{(1)}$ and I get the result posted before.

Perhaps what they mean when they say $\dot{\theta}_1(0) = 1$ is that this is equal to the vector <1,1>? I am always going to have a vector on the RHS so could this be the case?

7. May 8, 2013

### TSny

The impulse is such that at t = 0 you have $\dot{\theta}_1(0) = 1$ and $\dot{\theta}_2(0) = 0$ as well as $\theta_1(0) = 0$ and $\theta_2(0) = 0$

The phase constants are $\pi/2$ so you can write $$\theta = \beta^{(1)} \frac{\sqrt{3}}{2} <-1/\sqrt{3}, 1> \sin (w_1 t ) + \beta^{(2)} \frac{\sqrt{3}}{2} < 1\sqrt{3}, 1> \sin (w_2 t)$$
The initial conditions $\theta_1(0) = 0$ and $\theta_2(0) = 0$ are then satisfied and you only need to find $\beta^{(1)}$ and $\beta^{(2)}$ such that $\dot{\theta}_1(0) = 1$ and $\dot{\theta}_2(0) = 0$. That is, $\dot{\theta}(0) = <1,0>$

You don't really need to include the normalizing constants $\frac{\sqrt{3}}{2}$ in $\theta$ since they can be absorbed in the $\beta$'s. But it's ok if you want to include them.

8. May 8, 2013

### CAF123

Thanks TSny, but could you explain why we must have $\dot{\theta}_2 (0) = 0$?

9. May 8, 2013

### TSny

I think that's just the interpretation of giving an impulse to the first mass. The impulse sets the first mass immediately into motion while the second mass is still at rest immediately after the impulse.

10. May 8, 2013

### CAF123

So instantaneously particle 2 is stationary after particle 1 is given the impulse. It makes sense - thanks again.