Approximation of SHM

1. Apr 29, 2014

Karol

1. The problem statement, all variables and given/known data
The restoring force of a pendulum is $F_\theta=-mg\sin\theta$
and is approximated to $F_\theta=-mg\theta$.
The period is $T=2\pi\sqrt{\frac{L}{g}}$, but can be expressed as the infinite series:
$$T=2\pi\sqrt{\frac{L}{g}}\left( 1+\frac{1^2}{2^2}\sin^2\frac{\theta}{2}+\frac{1^2}{2^2}\frac{3^2}{4^2}\sin^4\frac{\theta}{2}+...\right)$$
What is this approximation and of what? i don't think it's a Maclaurin series.

2. Relevant equations
Maclaurin series of sin(x):
$$\sin(x)\cong 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$$

2. Apr 29, 2014

Staff: Mentor

When solving for the period of a large amplitude pendulum you get a nasty elliptical integral. You can express that integral in terms of a series, the first term being the familiar period for small amplitudes.

Try this for more details: LARGE-ANGLE MOTION OF A SIMPLE PENDULUM

3. May 5, 2014

Karol

Change of variables

In the link you gave me:
http://api.viglink.com/api/click?fo...M&txt=LARGE-ANGLE MOTION OF A SIMPLE PENDULUM
there is equation 7:
$$\sin(\varphi)=\frac{\sin(\vartheta/2)}{\sin(\alpha/2)}$$
It says $\alpha$ changes from 0 to 2$\pi$ for a full oscillation.
First, when $\alpha=0$ then the denominator=0
Secondly, $\alpha$ changes from $-\vartheta$ to $+\vartheta$
so $\sin(\varphi)$ changes from $-\varphi$ $+\varphi$

4. May 5, 2014

Staff: Mentor

I believe it is $\varphi$ that ranges from 0 to 2$\pi$ for a full oscillation, not $\alpha$. $\alpha$ is the initial angle of the pendulum, when released from rest.

5. May 6, 2014

Karol

Right, $\varphi$ ranges from 0 to 2$\pi$
$\vartheta$ is the initial angle

6. May 6, 2014

haruspex

No, as Doc Al posted, $\alpha$ is the initial angle. $\vartheta$ varies between $-\alpha$ and $+\alpha$.

7. May 7, 2014

Karol

Maybe, i don't remember the details now