Simple Pendulum- Real formula

1. Nov 11, 2009

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

The motion of a simple pendulum is only approximately simple harmonic. How small should the amplitude should be for the approximation to hold good?. Obtain the general expression for the time period of a simple pendulum. How much does the actual time period differ from the approximate time period when the amplitude is 15 degree?

2. Relevant equations

3. The attempt at a solution
On obtaining the period of oscillation using SHM, we approximate sin(a)=a in radians. I was thinking of keeping it as sin(a). Any help with the first question? The amplitude one.

2. Nov 11, 2009

gabbagabbahey

The approximation $\sin\theta\approx\theta$ is derived via Taylor series expansion. To determine how accurate it is, you need to look at the Taylor series remainder.

3. Nov 11, 2009

$$\ T=2pi \sqrt{L/g}(1\theta^2/16+11\theta^4/3072+173\theta^6/737280+22931\theta^8/1321205760+...)$$

That was the equation I got for time period. But I cant approximate the smallest value of amplitude.

Last edited: Nov 11, 2009
4. Nov 11, 2009

gabbagabbahey

Surely you mean

$$T=2\pi\sqrt{\frac{L}{g}}\left(1+\frac{1}{16}\theta_0^2+\frac{11}{3072}\theta_0^4+\frac{173}{737280}\theta_0^6+\ldots\right)$$

where $\theta_0$ is the initial amplitude (angular displacement) of the pendulum....right?

Read the question again. You aren't asked to approximate the smallest value of amplitude.

5. Nov 12, 2009

"How small should the amplitude should be for the approximation to hold good?"
It is given in the question. I dont know how to do it.

6. Nov 12, 2009

gabbagabbahey

The approximation they are referring to is

$$T=2\pi\sqrt{\frac{L}{g}}\left(1+\frac{1}{16}\theta _0^2+\frac{11}{3072}\theta_0^4+\frac{173}{737280}\theta_0^6+\ldots\right)\approx2\pi\sqrt{\frac{L}{g}}$$

...how large can you make $\theta_0$ before this is no longer a good approximation?