Computer Algebra Recipes

138 CHAPTER 3. LINEAR ODE MODELS
> restart: with(plots): with(plottools):
Measuring y vertically downward, the origin y = 0 is chosen to be at the top
end of the bungee cord (length L) where it is attached to the bridge. If the cord
has a mass density ² per unit length, and g is the acceleration due to gravity,
the tension T in the cord is given by T = ²g(L ¡ y), which is entered.
> T:=epsilon*g*(L-y);
T := ²g(L ¡ y)
At the top end (y = 0), the cord has a tension T = ²gL, because the cord must
support its entire weight. At the bottom end, y = L andthetensioniszero.
The cord is now allowed to undergo a small transverse displacement dÃ(y; t)
at time t from the equilibrium position. The above expression for T will still
be valid. To understand this, consider the cord segment of arc length
ds = p (dy) 2 +(dÃ) 2 = p 1+(dÃ=dy) 2 dy
showninFigure3.9. Lettingμ be the angle with the vertical, and noting that
the forces still balance in the y-direction, then
(T cos μ) y ¡ (T cos μ) y+dy =(²ds) g; (3.4)
with cos μ = dy=ds. Assuming that dÃ=dy ¿ 1, one has ds ¼ dy and cos μ ¼ 1,
so the vertical force equation (3.4) reduces to
T (y) ¡ T (y + dy) =(²dy) g: (3.5)
Taylor expanding the left-hand side of (3.5) to the ¯rst power in dy, and dividing
both sides by dy, yields @T=@y = ¡²g. The expression T = ²g(L ¡ y) follows
on integrating and evaluating the constant at y =0.
In the Ã-direction, Newton's second law yields
(T sin μ) y+dy ¡ (T sin μ) y =(²ds) @2Ã @t2 (3.6)
with sin μ = dÃ=ds. For dÃ=dy ¿ 1, sin μ ¼ @Ã=@y, partial derivatives being
used because one also has time as an independent variable. So (3.6) becomes
μ
T (y) @Ã
μ
¡ T (y)
@y y+dy
@Ã
=(²dy)
@y y
@2Ã ; (3.7)
@t2 or, on Taylor expanding the left-hand side for small dy, dividing by dy, and
taking the limit dy ! 0,
μ
@
T (y)
@y
@Ã
= ²
@y
@2Ã : (3.8)
@t2 This equation of motion is now entered,
> eq:=diff(T*diff(psi(y,t),y),y)=epsilon*diff(psi(y,t),t,t);
μ
μ μ
2
2
@
@ @
eq := ¡²g Ã(y; t) + ²g(L ¡ y) Ã(y; t) = ² Ã(y; t)
@y @y2 @t2