# Period of a simple pendulum

1. Jan 24, 2006

### Hootenanny

Staff Emeritus
I am attempting to derrive the equation for the time period of a simple pendulum, the notes we have been given show some hints to where we should be aiming to get. I have managed to get to
$$\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0$$
However, the next line should be
$$\theta = \theta_{max} \sin \sqrt{\frac{g}{L}} \theta$$
I've tried intergrating, taking $$\theta$$ out as a factor and multiplying by $$dt^2$$ but cannot get the correct solution. Any help would be much appreciated.

2. Jan 24, 2006

### Integral

Staff Emeritus
Last edited by a moderator: Apr 21, 2017
3. Jan 24, 2006

### dextercioby

The general solution to your ODE is

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

where $\omega=\sqrt{\frac{g}{L}}$ and the exact form of the coefficients depends on the 2 initial conditions.

Daniel.

4. Jan 24, 2006

### Hootenanny

Staff Emeritus
I don't unstand why the general solution should contain trig functions are where the $$t$$ goes.

Last edited: Jan 24, 2006
5. Jan 24, 2006

### Hootenanny

Staff Emeritus
If anyone has the time could they show me a step by step method?

6. Jan 24, 2006

### krab

The differential equation indicates you want that the sum of a function and its second derivative to be zero (aside from the g/L factor that can be normalized out). The trig functions sin and cos have exactly this property. You seem to be wanting to derive the solution simply by manipulating the equation. It's not that simple. Welcome to the wonderful world of differential equations.

7. Jan 24, 2006

### Staff: Mentor

By studying the differential equation, you will realize that the solution must be a function whose 2nd derivative is proportional (within a constant) to the function itself. Trig functions satisfy that requirement. (This might help a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c3)

8. Jan 24, 2006

### Hootenanny

Staff Emeritus
So where do I go from $$\theta (t)=A\sin \omega t +B\cos \omega t$$? Do I just sub it in for theta?

9. Jan 24, 2006

### krab

Right! If it works, then how could it be wrong? You will find directly that it works provided $\omega=\sqrt{\frac{g}{L}}$.

10. Jan 25, 2006

### Hootenanny

Staff Emeritus
Do I have to prove that $$\omega=\sqrt{\frac{g}{L}}$$ or can I just state it?

11. Jan 25, 2006

### Hootenanny

Staff Emeritus
Do I just leave the A and B in or do I have to sub something in for them?

12. Jan 25, 2006

### krab

Did you actually try the solution? Did you plug it into the differential equation? It will be pretty obvious what happens and why omega MUST satisfy $\omega=\sqrt{\frac{g}{L}}$.

Re A and B, these are just the amplitude. You can equally well write the solution as
$$\theta (t)=A\sin (\omega t +\phi)$$
Where A is theta_max and phi is another constant. You can determine the constants by the initial conditions: What is theta at t=0? What is the speed of the bob (Ld theta/dt) at t=0?

13. Jan 25, 2006

### Hootenanny

Staff Emeritus
Yeah I did, but messed it up and couldn't get it to work. I'm going to try again now.

14. Jan 25, 2006

### Hootenanny

Staff Emeritus
After manipulating I got $$\frac{g}{L}\theta - \omega^2 A\sin (\omega t + \phi ) = 0$$ I think its correct.

Last edited: Jan 25, 2006
15. Jan 26, 2006

### Hootenanny

Staff Emeritus
Ive got $$\frac{g}{L}\theta = \omega^2 A\sin (\omega t + \phi )$$ but i dont know how to now prove that $\omega=\sqrt{\frac{g}{L}}$ Help Please.

16. Jan 26, 2006

### krab

What is theta equal to? Look at my last equation in Post #12. I don't know how you fail to see the cancellation.

17. Jan 26, 2006

### Hootenanny

Staff Emeritus
O dear, it's been staring me in the face! I can't beleive I missed it! Sorry. Thank's for all you help.

18. Jan 26, 2006

### Hootenanny

Staff Emeritus
Now I have $$\theta = \theta_{max} \sin(\sqrt{\frac{g}{L}} t - \alpha )$$ How Can I remove the $\alpha$ ?

Last edited: Jan 26, 2006
19. Jan 26, 2006

### Hootenanny

Staff Emeritus