Deriving the Simple Pendulum Solution: Second Order and Cosine

[Taylor_1989]

Homework Statement

Hi guys I am having a problem deriving this solution for a simply pendulum. Could someone please help me.

My issue is taking the second order and getting into just cos. I have attempted a solution which is shown below.

Homework Equations

The Attempt at a Solution

$$\theta^{''}+\frac{g}{l}\theta=0$$

$$ p^2+\frac{g}{l}\theta=0$$

$$\omega^2=\frac{g}{l}$$

So general solution: $$\theta (t) = Ae^{i\sqrt{\omega t }}+Be^{-i\sqrt{\omega t }}$$[/B]using the fact that: $$e^{(\pm) i \omega t}=cos{\omega t}\pm sin{i\omega t }$$

I get the general equation:

$$\theta (t)= (A+B)cos(\omega t)+ i(A-B)sin(\omega t)$$

I am now confused on what to do next. I was thinking that beacuse I am only looking for real values the i component would be equal to 0. But I do no think this is the case, I am just really can't figure why.

 

[DrClaude]

So general solution: $$\theta (t) = Ae^{i\sqrt{\omega t }}+Be^{-i\sqrt{\omega t }}$$

There should be no square root there. The next step should be to convert this equation to a single exponential with a phase shift.

 

[Ray Vickson]

Homework Statement

Hi guys I am having a problem deriving this solution for a simply pendulum. Could someone please help me.

View attachment 112484

My issue is taking the second order and getting into just cos. I have attempted a solution which is shown below.

I get the general equation:

$$\theta (t)= (A+B)cos(\omega t)+ i(A-B)sin(\omega t)$$

I am now confused on what to do next. I was thinking that beacuse I am only looking for real values the i component would be equal to 0. But I do no think this is the case, I am just really can't figure why.

As Dr. Claude has said, you should be looking at $e^{\pm i \omega t}$, without the square root. Anyway, $A$ and $B$ are just some constants; they can be complex, so you could end up with real $C_1 = A+B$ and $C_2 = i(A-B)$. Can you see how to write $C_1 \cos (\omega t) + C_2 \sin (\omega t)$ as $C_3 \cos (\omega t + \phi)$ of $C_4 \sin (\omega t + \lambda)?$ (Hint: it is a standard trick, used everywhere.)

 

[pasmith]

Homework Statement

Hi guys I am having a problem deriving this solution for a simply pendulum. Could someone please help me.

View attachment 112484

My issue is taking the second order and getting into just cos. I have attempted a solution which is shown below.

Homework Equations

The Attempt at a Solution

$$\theta^{''}+\frac{g}{l}\theta=0$$

$$ p^2+\frac{g}{l}\theta=0$$

$$\omega^2=\frac{g}{l}$$

So general solution: $$\theta (t) = Ae^{i\sqrt{\omega t }}+Be^{-i\sqrt{\omega t }}$$[/B]using the fact that: $$e^{(\pm) i \omega t}=cos{\omega t}\pm sin{i\omega t }$$

I get the general equation:

$$\theta (t)= (A+B)cos(\omega t)+ i(A-B)sin(\omega t)$$

I am now confused on what to do next. I was thinking that beacuse I am only looking for real values the i component would be equal to 0. But I do no think this is the case, I am just really can't figure why.

A and B are not real; they are complex. But as e^{i\omega t} and e^{-i\omega t} are complex conjugates and we want \theta to be real, we must have A and B as complex conjugates, A = \frac 12 (C - iD), B = \frac12 (C + iD). Thus
<br /> A + B = C \in \mathbb{R}, \\<br /> i(A - B) = D \in \mathbb{R}.

 

[Taylor_1989]

Ok thank you for the respones. I think I have figured it.

$$Mcosx+Nsinx=Rcos(x-\alpha)$$
$M= cos (\alpha)$ & $N= sin(\alpha)$

$$\sqrt{M^2+N^2}=R$$

I have notice that $$sinx+cosx $$
is similar to $$cos(x-\frac{\pi}{4})$$

Am i on the right lines I am now just trying to figure the relation. I have a feeling that

$$M=N=\frac{\pi}{4}$$

 

[haruspex]

.
$$Mcosx+Nsinx=Rcos(x-\alpha)$$
...
I have a feeling that
$$M=N=\frac{\pi}{4}$$

wrong feeling. cos(a-b)=cos(a)cos(b)+sin(a)sin(b). Apply that to the first equation above.

 

[Taylor_1989]

@haruspex so If i apply to the above I get the following:

$M=\cos \alpha$ and $N=\sin \alpha$ But I am not seeing the connection. Surley R is just $$
\sqrt{M^2+N^2}=R
$$
Mod edit: Fixed broken tex - To OP: don't put a space between \ and alpha

 

[haruspex]

$M=\cos \alpha$ & $N=\sin \alpha$

If that were true then R=1. How did the R disappear when following my suggestion?

Surley R is just

√(M²+N²)=R​

That is correct, but it does not tell you what M and N are individually.

 

[Taylor_1989]

@haruspex
Not matter which I got about it, I keep get $$\frac{N}{M}=tan\alpha $$. I have tried doing via subsituion and still come back with the same result:

$$M=(M^2+N^2)cos^2\alpha$$
expanding this out still gives the same result as the above. I am really not sure how to get M and N independent of each other, other than say $M=Rcos\alpha$ & $N=Rsin\alpha$

Mod edit: Don't use & inside a tex expression -- it causes the expression to not render. Also, use \ in front of alpha, theta, etc., not /.

 

[Taylor_1989]

@pasimth

A and B are not real; they are complex. But as e^{i\omega t} and e^{-i\omega t} are complex conjugates and we want \theta to be real, we must have A and B as complex conjugates, A = \frac 12 (C - iD), B = \frac12 (C + iD). Thus
<br /> A + B = C \in \mathbb{R}, \\<br /> i(A - B) = D \in \mathbb{R}.

how did u form the above equations

 

[Ray Vickson]

@pasimthhow did u form the above equations

You formed them yourself in post #1!

 

[haruspex]

$M=R\cos\alpha$ & $N=R\sin\alpha$

Exactly.
By the way, the latex comes out a bit better if you also put \ in front of trig functions and log functions. Don't leave a space after \.

 

[haruspex]

@pasimthhow did u form the above equations

Look at your last equation in post #1. The left hand side must be real, the cos and sin functions return real numbers, and the equation has to be true for all t. The only way this can be is if the coefficients of the trig functions are both real. So A+B is real and i(A-B) is real. It's easy to show that A and B must be complex conjugates.

And you meant pasmith.

 

[Taylor_1989]

Ok I have deiced to forget everything and start from the begin, because something is just sinking in here, so I came up with what hopefully will be correct method:

So my train of thought is as follows
So I have two solution ( by the way I have striped the coefficient for now to just simplify), which are shown below:

$$ \theta(t)$$