8.4 Oscillating 1D Systems: A Second Order ODE
241
we choose the body to be at rest, but moved away from its equilibrium position:
x(0) = X 0 , x
(0) = 0 .
The exact solution of (8.42) with these initial conditions is x(t) = X 0 cos ωt. This
can easily be verified by substituting into (8.42) and checking the initial conditions.
The solution tells us that such a spring-mass system oscillates back and forth as
described by a cosine curve.
The differential equation (8.42) appears in numerous other contexts. A classical
example is a simple pendulum that oscillates back and forth. Physics books derive,
from Newton’s second law of motion, that
mLθ
+ mg sin θ = 0,
where m is the mass of the body at the end of a pendulum with length L, g is the
acceleration of gravity, and θ is the angle the pendulum makes with the vertical.
Considering small angles θ , sin θ ≈ θ , and we get (8.42) with x = θ , ω =
√
g/L,
x(0) = Θ, and x (0) = 0, if Θ is the initial angle and the pendulum is at rest at
t = 0.
8.4.2 Numerical Solution
We have not looked at numerical methods for handling second-order derivatives,
and such methods are an option, but we know how to solve first-order differential
equations and even systems of first-order equations. With a little, yet very common,
trick we can rewrite (8.42) as a first-order system of two differential equations.
We introduce u = x and v = x = u as two new unknown functions. The
two corresponding equations arise from the definition v = u and the original
equation (8.42):
u
= v,
(8.43)
v
= −ω
2 u .
(8.44)
(Notice that we can use u = v to remove the second-order derivative from
Newton’s second law.)
We can now apply the Forward Euler method to (8.43)–(8.44), exactly as we did
in Sect. 8.3.2:
u n+1 − u n
Δt
= v
n ,
(8.45)
v n+1 − v n
Δt
= −ω
2 u
n ,
(8.46)
241
we choose the body to be at rest, but moved away from its equilibrium position:
x(0) = X 0 , x
(0) = 0 .
The exact solution of (8.42) with these initial conditions is x(t) = X 0 cos ωt. This
can easily be verified by substituting into (8.42) and checking the initial conditions.
The solution tells us that such a spring-mass system oscillates back and forth as
described by a cosine curve.
The differential equation (8.42) appears in numerous other contexts. A classical
example is a simple pendulum that oscillates back and forth. Physics books derive,
from Newton’s second law of motion, that
mLθ
+ mg sin θ = 0,
where m is the mass of the body at the end of a pendulum with length L, g is the
acceleration of gravity, and θ is the angle the pendulum makes with the vertical.
Considering small angles θ , sin θ ≈ θ , and we get (8.42) with x = θ , ω =
√
g/L,
x(0) = Θ, and x (0) = 0, if Θ is the initial angle and the pendulum is at rest at
t = 0.
8.4.2 Numerical Solution
We have not looked at numerical methods for handling second-order derivatives,
and such methods are an option, but we know how to solve first-order differential
equations and even systems of first-order equations. With a little, yet very common,
trick we can rewrite (8.42) as a first-order system of two differential equations.
We introduce u = x and v = x = u as two new unknown functions. The
two corresponding equations arise from the definition v = u and the original
equation (8.42):
u
= v,
(8.43)
v
= −ω
2 u .
(8.44)
(Notice that we can use u = v to remove the second-order derivative from
Newton’s second law.)
We can now apply the Forward Euler method to (8.43)–(8.44), exactly as we did
in Sect. 8.3.2:
u n+1 − u n
Δt
= v
n ,
(8.45)
v n+1 − v n
Δt
= −ω
2 u
n ,
(8.46)