Simple Pendulum - Linear Second-Order Equation

[sourlemon]

Homework Statement

In section 4.7, we discussed the simple pendulum consisting of a mass m suspended by a rod of length l having negligible mass and derived the nonlinear initial value problem.

(17) \frac{d^{2}\phi}{dt^{2}} + \frac{g}{l}sin\theta=0

\theta(0) = \alpha

\theta'(0) = 0

where g is the acceleration due to gravity and \theta(t) is the angle of the rod makes with the vertical at time t. Here it is assumed that the mass is released with zero velocity at an initial angle \alpha, 0 < \alpha < \pi. We would like to determine the equation of motion for the pendulum and its period of oscillation.

(a) Use equation (17) and the energy integral lemma discussed in Section 4.7 to show that

\left(\frac{d\theta}{dt}\right)^{2} = \frac{2g}{l}(cos\theta - cos\alpha)

and hence

dt = - \sqrt{\frac{l}{2g}}\frac{d\theta}{\sqrt{cos\theta - cos\alpha}}

(b) Use the trigonometric identity cos x = 1 - 2sin^{2}(\frac{x}{2}) to express dt by

dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}

(c)Make the change of variables sin(\theta/2) = sin (\alpha/2)sin\phi to rewrite dt in the form

dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

where k = sin(\alpha/2).

(d) The period P(\alpha) of the pendulum is defined to be the time required for the pendulum to swing from one extreme to the other and back - that is, from \alpha to -\alpha back to \alpha. Show that the period of oscillation is given by

(19) P(\alpha) = 4\sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

where k = sin(\alpha/2). [Hint: The period is just four times the time it takes the pendulum to go from \theta = 0 to \theta = \alpha]

The integral in (19) is called an elliptic integral of the first kind and is denoted by F(k, \pi/2. As you might expect, the period of the simple pendulum depends on the length "l" of the rod and the initial displacement \alpha. In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from \pi/8 to 15\pi/16 (for fixed "l"). What happens as \alpha approaches \pi

Homework Equations

Lemma 3.
Let y(t) be a solution to the differential equation

(7) y" = f(y)

where f(y) is a continuous function that does not depend on y' or the independent variable t. Let F(y) be an indefinite integral of f(y), that is

f(y) = \frac{d}{dy}F(y).

Then the quantity

(8) E(t) = \frac{1}{2}y't^{2} - F(y(t))

is constant; i.e.,

(9) \frac{d}{dt}E(t) = 0.

The Attempt at a Solution

Lolz writing this problem alone takes forever. There's actually a part e, but I figure that out, so I didn't post it. As for part a and b, I got that too. I'm only having problem with part c and d.

(c)Make the change of variables sin(\theta/2) = sin (\alpha/2)sin\phi to rewrite dt in the from

dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}

to

dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

where k = sin(\alpha/2).

I'm not sure yet how to start d, but I've start c. I think my problem is changing d\theta to d\phi.

My teacher said to take the derivative of both. So

sin(\theta/2) = sin (\alpha/2)sin\phi

becomes

1/2cos(\theta/2) = sin(\alpha/2)cos\phi

Since I'm changing d\theta to d\phi, I took the assumption that sin(\alpha/2)is a constant.

Plugging that in the equation I get

dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{1}{2}\frac{cos(\theta/2)}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\alpha/2)sin^{2}\phi}}

In part b, cos x = 1 - 2sin^{2}(\frac{x}{2}), so I figure I can try to do that too. So

cos(\theta/2) = 1 - 2sin^{2}(\frac{\theta}{4})

But how can I change 1 - 2sin^{2}(\frac{\theta}{4}) to something with \alpha or \phi? I can't figure out what triognometric properties I can use.

 

[siddharth]

My teacher said to take the derivative of both. So

sin(\theta/2) = sin (\alpha/2)sin\phi

1/2cos(\theta/2) = sin(\alpha/2)cos\phi

That's not right. What happened to the derivative? You should get,

<br /> \frac{1}{2} \cos{\frac{\theta}{2}} \left(\frac{d \theta}{d \phi}\right) = \sin {\frac{\alpha}{2}} \cos \phi

You need to substitute for d\theta and \sin\frac{\theta}{2}} in the equation.

 

[sourlemon]

Lolz, thank you for not running away after seeing my post. Lolz I feel so dumb. What a careless mistake. Thank you so much siddharth!

I have c solved :D now I must tackle d :(

(d) The period P(\alpha) of the pendulum is defined to be the time required for the pendulum to swing from one extreme to the other and back - that is, from \alpha to -\alpha back to \alpha. Show that the period of oscillation is given by

(19) P(\alpha) = 4\sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

where k = sin(\alpha/2). [Hint: The period is just four times the time it takes the pendulum to go from \theta = 0 to \theta = \alpha]

The integral in (19) is called an elliptic integral of the first kind and is denoted by F(k, \pi/2. As you might expect, the period of the simple pendulum depends on the length "l" of the rod and the initial displacement \alpha. In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from \pi/8 to 15\pi/16 (for fixed "l"). What happens as \alpha approaches \pi?

From part c, I have the equation

dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

where k = sin(\alpha/2).

So integrating both sides to have the period, I would have

P(\alpha) = \sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}

Since \sqrt{\frac{l}{g}} is constant, I can move that out of the integral.

I am little confuse on how you determine that the integration is from 0 to \frac{\pi}{2}. What I figure is depending on where the pendulum is release, the most it can make is from 0 to \pi. But for a period, the pendulum needs to swing from \alpha to -\alpha back to \alpha. So would it make sense if I integrate it from 0 to \pi and multiply it by 2?...I'm a bit stuck here. I can see that if you release the pendulum from 0, it's the same as it is release from \pi, so with that, can I conclude that from 0 to \frac{\pi}{2} is the same as \frac{\pi}{2} to \pi. That's why you can say that the integral is from 0 to \frac{\pi}{2}, multiply by 4?

What happens as \alpha approaches \pi?

If the \alpha from 0 to \frac{\pi}{2}, then the period decrease. But if \alpha is from \frac{\pi}{2} to \pi, then the period increase. Is that right?

 

[sourlemon]

*sigh* turned out I missed a lot of things. I mistyped in the first post, it's suppose to be d\theta not d\phi.

Let's start from the beginning.

(17) \frac{d^{2}\theta}{dt^{2}} + \frac{g}{l}sin\theta=0

Rearranging the equation,
\frac{d^{2}\theta}{dt^{2}} = - \frac{g}{l}sin\theta = \theta&#039;&#039; = f(\theta)

Following the lemma energy stated above,

f(\theta) = \frac{d}{d\theta}F(\theta) = \frac{d}{d\theta}(\frac{g}{l}cos\theta + C)

E(t)= \frac{1}{2}\theta&#039;(t)^{2} - F(\theta(t))

Plugging the known data \theta(0) = \alpha and \theta&#039;(0) = 0

E(t) = \frac{1}{2}(0)^{2} - \alpha = - \alpha

E(t)= \frac{1}{2}\theta&#039;(t)^{2} - F(\theta(t)) = \frac{1}{2}\theta&#039;(t)^{2} - \frac{g}{l}cos\theta = - \alpha

\frac{1}{2}\theta&#039;(t)^{2} = \frac{g}{l}cos\theta - \alpha

\theta&#039;(t)^{2} = 2\left(\frac{g}{l}cos\theta - \alpha\right) =

\theta&#039;(t)^{2} = 2\frac{g}{l}\left(cos\theta - \frac{g}{l}\alpha\right)


Did I miss something? I want to end up with the equation below.

\left(\frac{d\theta}{dt}\right)^{2} = \frac{2g}{l}(cos\theta - cos\alpha)