Approximating SHM Homework: F_\theta=-mg\theta

[Karol]

Homework Statement

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.

Homework Equations

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

 

[Doc Al]

Homework Statement

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.

Homework Equations

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

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

 

[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

 

[Doc Al]

It says \alpha changes from 0 to 2\pi for a full oscillation.

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.

 

[Karol]

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

 

[haruspex]

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

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

 

[Karol]

Maybe, i don't remember the details now