Exercise 2.7.16

Answers

1.

Use Theorem 2.34. The auxiliary polynomial is

t^{2} + \frac{g}{l}

. Hence the basis of the solution space is

{e^{it \sqrt{\frac{g}{l}}}, e^{- it \sqrt{\frac{g}{l}}}}

{\cos t \sqrt{\frac{g}{l}}, \sin t \sqrt{\frac{g}{l}}}

by Exercise 2.7.8. So the solution should be of the form

𝜃 (t) = C_{1} \cos t \sqrt{\frac{g}{l}} + C_{2} \sin t \sqrt{\frac{g}{l}}

for some constants $C_{1}$ and $C_{2}$ .

2.

Assume that

𝜃 (t) = C_{1} \cos t \sqrt{\frac{g}{l}} + C_{2} \sin t \sqrt{\frac{g}{l}}

for some constants $C_{1}$ and $C_{2}$ by the previous argument. Consider the two initial conditions

𝜃 (0) = C_{1} \sqrt{\frac{g}{l}} = 𝜃_{0}

and

𝜃^{'} (0) = C_{2} \sqrt{\frac{g}{j}} = 0 .

Thus we get

C_{1} = 𝜃_{0} \sqrt{\frac{l}{g}}

and

C_{2} = 0 .

So we get the unique solution

𝜃 (t) = 𝜃_{0} \sqrt{\frac{l}{g}} \cos t \sqrt{\frac{g}{l}} .

3.

The period of

\cos t \sqrt{\frac{g}{l}}

2 π \sqrt{\frac{l}{g}}

. Since the solution is unique by the previous argument, the pendulum also has the same period.

jephianlin

2011-06-27 00:00