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sourlemon
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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) [tex]\frac{d^{2}\phi}{dt^{2}} + \frac{g}{l}sin\theta=0[/tex]
[tex]\theta(0) = \alpha[/tex]
[tex]\theta'(0) = 0[/tex]
where g is the acceleration due to gravity and [tex]\theta(t)[/tex] 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 [tex]\alpha, 0 < \alpha < \pi[/tex]. 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
[tex]\left(\frac{d\theta}{dt}\right)^{2} = \frac{2g}{l}(cos\theta - cos\alpha)[/tex]
and hence
[tex]dt = - \sqrt{\frac{l}{2g}}\frac{d\theta}{\sqrt{cos\theta - cos\alpha}}[/tex]
(b) Use the trigonometric identity [tex]cos x = 1 - 2sin^{2}(\frac{x}{2})[/tex] to express dt by
[tex]dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}[/tex]
(c)Make the change of variables [tex]sin(\theta/2) = sin (\alpha/2)sin\phi [/tex] to rewrite dt in the form
[tex]dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}[/tex]
where [tex]k = sin(\alpha/2)[/tex].
(d) The period [tex]P(\alpha)[/tex] 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 [tex]\alpha to -\alpha back to \alpha[/tex]. Show that the period of oscillation is given by
[tex](19) P(\alpha) = 4\sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}[/tex]
where [tex]k = sin(\alpha/2)[/tex]. [Hint: The period is just four times the time it takes the pendulum to go from [tex]\theta = 0 to \theta = \alpha[/tex]]
The integral in (19) is called an elliptic integral of the first kind and is denoted by [tex]F(k, \pi/2[/tex]. As you might expect, the period of the simple pendulum depends on the length "l" of the rod and the initial displacement [tex]\alpha[/tex]. In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from [tex]\pi/8[/tex] to [tex]15\pi/16[/tex] (for fixed "l"). What happens as [tex]\alpha[/tex] approaches [tex]\pi [/tex]
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
[tex]f(y) = \frac{d}{dy}F(y).[/tex]
Then the quantity
(8) [tex]E(t) = \frac{1}{2}y't^{2} - F(y(t))[/tex]
is constant; i.e.,
(9) [tex]\frac{d}{dt}E(t) = 0[/tex].
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 [tex]sin(\theta/2) = sin (\alpha/2)sin\phi [/tex] to rewrite dt in the from
[tex]dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}[/tex]
to
[tex]dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}[/tex]
where [tex]k = sin(\alpha/2)[/tex].
I'm not sure yet how to start d, but I've start c. I think my problem is changing [tex]d\theta[/tex] to [tex]d\phi[/tex].
My teacher said to take the derivative of both. So
[tex]sin(\theta/2) = sin (\alpha/2)sin\phi [/tex]
becomes
[tex]1/2cos(\theta/2) = sin(\alpha/2)cos\phi [/tex]
Since I'm changing [tex]d\theta[/tex] to [tex]d\phi[/tex], I took the assumption that [tex]sin(\alpha/2) [/tex]is a constant.
Plugging that in the equation I get
[tex]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}}[/tex]
In part b, [tex]cos x = 1 - 2sin^{2}(\frac{x}{2})[/tex], so I figure I can try to do that too. So
[tex]cos(\theta/2) = 1 - 2sin^{2}(\frac{\theta}{4})[/tex]
But how can I change [tex]1 - 2sin^{2}(\frac{\theta}{4})[/tex] to something with [tex]\alpha[/tex] or [tex]\phi[/tex]? I can't figure out what triognometric properties I can use.
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