Mathematical Physics Vol 1

5.10 Examples

299

mathematical pendulum is thus moving along the arc ( ϕ ∗ 1 , ϕ ∗ 2 ) which contains the point ϕ = 0. From (5.276) and (5.273) it follows that

v 2 0 2 Rg

h Rg

cos ϕ ∗ = −

= cos ϕ 0 −

and as v 2 0 / 2 Rg > 0, it also follows that cos ϕ 0 > cos ϕ ∗ , and thus | ϕ 0 | < | ϕ ∗ 0 | , so the initial position ϕ 0 is in the interval ( ϕ ∗ 1 , ϕ ∗ 2 ) . Such movement of the mathematical pendulum is called oscillatory movement , as the point oscillates between positions that correspond to zeros ϕ ∗ 1 and ϕ ∗ 2 . 2. The function f ( ϕ ) has a second order zero, i.e. from the condition f ( ϕ )= 0, that is cos ϕ ∗ = − h Rg , sin ϕ ∗ = q 1 − ( h / Rg ) 2 = 0 ⇒ h Rg = 1 . (5.279) The analysis shows that the point reaches a position in which it stops and remains there, because the velocity and the acceleration are equal to zero. 3. In the case when | h / Rg | > 1, i.e. there are no real zeroes, and the point does not stop ( v > 0), there is progressive movement. Our task is to determine the final equation of the movement, i.e. find ϕ ( t ) . The analysis shows that only the first case is interesting, because that is when the point is performing oscillatory movement (moves to a point where it stops, then changes direction and continues to move...). Observe the equation (5.274) R 2 ˙ ϕ 2 = 2 Rg cos ϕ + h Rg . (5.280) where | h / Rg | < 1. Let us introduce the substitution h Rg = − cos γ (5.281) where 0 < γ < π is the angular amplitude . From equations (5.280) and (5.281) (separable differential equation) we obtain s R g d ϕ d t = p 2 ( cos ϕ − cos γ ) . (5.282) This equation can be expressed in the form 20 . s R g d ( ϕ / 2 ) d t = r sin 2 γ 2 − sin 2 ϕ 2 , (5.283) that is, using a substitution (introducing a new variable u )

ϕ 2

γ 2

sin

= u sin