Length/Time Period squared Relationship of a Simple Pendulum

[Thinker8921]

I know the L/T^2 relationship of a simple pendulum gives a constant- g/(2pi)^2.
Could anyone please show me how it is derived?
Thanks in advance.

 

[Andrew Mason]

I know the L/T^2 relationship of a simple pendulum gives a constant- g/(2pi)^2.
Could anyone please show me how it is derived?
Thanks in advance.

The force on the pendulum bob is:

F = -mgsin\theta = ma = mL\ddot\theta

For small angles, sin\theta \approx \theta so you have the differential equation:

\ddot\theta = -\frac{g}{L}\theta the solution of which is:

\theta = A_0\sin(\omega t + \phi) where \omega^2 = g/L with \omega = 2\pi/T

AM

 

[Thinker8921]

Thankyou for the reply. I think I should have been a little clearer. I am in high school and this depth is a little new so I don't understand the whole explanation.
-The theta sign with the 2 dots*, I am thinking it is angular acceleration? That way, mLa will be torque of the pendulum bob.
- I get the differential, however not the next step with the A zero sign and phi sign. What do they represent. Is it (Started calculus a week ago).
Please could you clear my doubts.
Thanks again.
*How do I insert the actual symbols in this?

 

[Andrew Mason]

Thankyou for the reply. I think I should have been a little clearer. I am in high school and this depth is a little new so I don't understand the whole explanation.

Welcome to PF, by the way.

-The theta sign with the 2 dots*, I am thinking it is angular acceleration? That way, mLa will be torque of the pendulum bob.

That's right.

- I get the differential, however not the next step with the A zero sign and phi sign. What do they represent. Is it (Started calculus a week ago).

These are parameters for the initial condition. The A₀ (I could have said \theta_0) is the maximum value of \theta, which occurs when the sin term = 1. The \phi is simply to adjust for the phase - ie. when the amplitude maximum occurs in relation to t. For example if amplitude was maximum at t=0, you would set \phi = \pi/2 so that the sin term gave a value of 1 (which is the maximum value for sin).

The idea here is that if you take the second derivative of \theta_0\sin (\omega t + \phi) you get -\omega^2\theta_0\sin (\omega t + \phi), which is simply -\omega^2 x the original function. Generally if the second derivative is -\omega^2 multiplied by the original function, the function must be a some sort of sine wave with frequency \omega. You will learn how to solve this kind of equation when you study differential equations.

*How do I insert the actual symbols in this?

You have to use Latex. See https://www.physicsforums.com/showthread.php?t=8997" for help on Latex.

AM

 

[Thinker8921]

Thanks, this has made it clearer to me.