Equation of Motion for a Simple Pendulum

Printable View

October 30th 2009, 10:11 AM
Aryth

Equation of Motion for a Simple Pendulum

I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta$

Where l is length of the pendulum.

Doing some research, I keep finding that the solution is:

$\theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)$

Where $\theta_0$ is the initial condition at time t=0

Now, here's my attempt at a solution:

Since it is linear and homogeneous, we can assume a solution:

$\theta = e^{kt}$

Where k is an arbitrary constant.

This leads to the indicial equation:

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

The solution to which is:

$k = \pm i\sqrt{\frac{g}{l}}$

This leads to two solutions:

$\theta_1 = e^{i\sqrt{\frac{g}{l}}t}$
$\theta_2 = e^{-i\sqrt{\frac{g}{l}}t}$

A linear combination of these leads to the general homogeneous solution:

$\theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}$

Expanding the exponentials into their trigonometric form, we get:

$\theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right]$ $+ B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]$

$\theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)$

From here I'm not sure how to proceed...
October 30th 2009, 11:07 AM
Mush

Quote:

Originally Posted by Aryth

I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta$

Where l is length of the pendulum.

Doing some research, I keep finding that the solution is:

$\theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)$

Where $\theta_0$ is the initial condition at time t=0

Now, here's my attempt at a solution:

Since it is linear and homogeneous, we can assume a solution:

$\theta = e^{kt}$

Where k is an arbitrary constant.

This leads to the indicial equation:

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

The solution to which is:

$k = \pm i\sqrt{\frac{g}{l}}$

This leads to two solutions:

$\theta_1 = e^{i\sqrt{\frac{g}{l}}t}$
$\theta_2 = e^{-i\sqrt{\frac{g}{l}}t}$

A linear combination of these leads to the general homogeneous solution:

$\theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}$

Expanding the exponentials into their trigonometric form, we get:

$\theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right]$ $+ B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]$

$\theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)$

From here I'm not sure how to proceed...

The equation for the simple pendulum SHOULD BE... $\ddot{\theta} = -\frac{g}{l} \sin(\theta)$ . If you use the process I gave you with THAT equation, then you get $\theta = \theta_0 \cos(\sqrt{\frac{g}{l} t})$ . You won't get the same solution if you use the equation you started with, which is the small angle approximation of the proper equation of motion...

EDIT: Perhaps I'm wrong about that actually..
October 30th 2009, 11:50 AM
Aryth

Quote:

Originally Posted by Mush

The equation for the simple pendulum SHOULD BE... $\ddot{\theta} = -\frac{g}{l} \sin(\theta)$ . If you use the process I gave you with THAT equation, then you get $\theta = \theta_0 \cos(\sqrt{\frac{g}{l} t})$ . You won't get the same solution if you use the equation you started with, which is the small angle approximation of the proper equation of motion...

EDIT: Perhaps I'm wrong about that actually..

You're somewhat right... I actually meant that the solution was:

$\theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t + \phi\right)$

How do you proceed to get that solution?
October 30th 2009, 01:02 PM
Mush

Quote:

Originally Posted by Aryth

You're somewhat right... I actually meant that the solution was:

$\theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t + \phi\right)$

How do you proceed to get that solution?

Are you familiar with Laplace Tranforms? That's the method I would use, as it makes the problem so much easier.

So we start from the equation with the small angle approximation $\ddot{\theta} = - \frac{g}{l} \theta$

If we assume that the pendulum is going to be in free vibration when released from some angle $\theta_0$ , then the initial conditions of a pendulum as as follows: $\theta(0) = \theta_0$ , $\dot{\theta}(0) = 0$ .

Then perform the Laplace Transform of both sides of the equation, noting that:

$\mathcal{L} ( \ddot{\theta}) = s^2 \Theta(s) - s \theta(0) - \dot{\theta}(0)$

$\mathcal{L}( k \theta )= k \Theta(s)$ for some constant $k$ (in this case $k = -\frac{g}{l}$

Once you've done that, insert your initial conditions into the new transformed equation, and rearrange to make $\Theta(s)$ the subject. Then use the following rule to perform the inverse Laplace Transform to get $\theta(t)$ :

$\mathcal{L}^{-1} \bigg(K \frac{s}{s^2 + \omega^2}\bigg) = K \cos(\omega t)$ , for some constant $K$ . (In order to get your equation into the form you need to use this inverse laplace transform rule, it might be helpful to know that $\frac{g}{l} = \bigg(\sqrt{\frac{g}{l}}\bigg)^2$ )
October 31st 2009, 02:20 AM
HallsofIvy

Quote:

Originally Posted by Aryth

I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta$

Where l is length of the pendulum.

Doing some research, I keep finding that the solution is:

$\theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)$

Where $\theta_0$ is the initial condition at time t=0

Now, here's my attempt at a solution:

Since it is linear and homogeneous, we can assume a solution:

$\theta = e^{kt}$

Where k is an arbitrary constant.

This leads to the indicial equation:

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

The solution to which is:

$k = \pm i\sqrt{\frac{g}{l}}$

This leads to two solutions:

$\theta_1 = e^{i\sqrt{\frac{g}{l}}t}$
$\theta_2 = e^{-i\sqrt{\frac{g}{l}}t}$

A linear combination of these leads to the general homogeneous solution:

$\theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}$

Expanding the exponentials into their trigonometric form, we get:

$\theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right]$ $+ B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]$

$\theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)$

From here I'm not sure how to proceed...

Take C= A+ B and D= i(A- B) so your answer becomes $\theta_h= C cos(\left(\sqrt{\frac{g}{l}}}t\right)+ D sin\left(\sqrt{\frac{g}{l}}t\right)$

Now there is a trig identity that says that cos(x+ y)= cos(x)cos(y)- sin(x)sin(y). That would be of this form if we could take [tex]x= \sqrt{\frac{g}{l}}}t\right[/itex] and y such that cos(y)= C, sin(y)= -D.

Unfortunately, we must have $cos^2(y)+ sin^2(y)= 1$ and it may not happen that $C^2+ D^2= 1$ . To fix that, multiply and divide by $\sqrt{C^2+ D^2}$ . Then we have $\sqrt{C^2+ D^2}\left(\frac{C}{\sqrt{C^2+ D^2}}cos\left(\sqrt{\frac{g}{l}}}t\right)+ \frac{D}{\sqrt{C^2+ D^2}}sin\left(\sqrt{\frac{g}{l}}}t\right)\right)$ .

Now, we do have $\left(\frac{C}{\sqrt{C^2+ D^2}}\right)^2+ \left(\frac{D}{\sqrt{C^2+ D^2}}\right)^2= \frac{C^2+ D^2}{C^2+ D^2}= 1$ .

We have $y(t)= \sqrt{C^2+ D^2}cos\left(\sqrt{\frac{g}{l}}}t\right+ \phi)$ where $\phi= arccos\left(\frac{C}{\sqrt{C^2+ D^2}}\right)$