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{\Large\bf Examples of Second Order ODE's}

\vspace{.3in}

{\bf Mechanical Systems/Vibrations}

\vspace{.2in}

Newton's second law asserts that the sum of the external forces
acting on an object at any instant in time is equal to the
product of the object's mass and acceleration.  This has far
reaching implications for ODE's.

Consider a mass attached to a spring that is allowed to move in
a straight line in the horizontal direction. Let x(t), v(t),
and a(t) denote the position, velocity, and acceleration,
respectively, of the particle. We assume that x=0 is the
equilibrium position of the spring. Hooke's law asserts that
the force exerted on the mass by the spring is F(t)=-kx(t),
where k is the spring constant. Newton's second law states: 
{\em Mass times Acceleration = Sum of External Forces}.
Using the fact that a(t)=x''(t) we obtain the equation
$$
mx''(t)=-kx 
$$
This is a linear second-order ode. It is accompanied by
the initial position x(0) and the initial velocity x'(0).

Notice that according to Hooke's law the force exerted by the
spring depends linearly on the position x(t). In many applications
the force depends nonlinearly on x(t). One example is Duffing's
model. The differential equation is 
$$
mx''(t)=-k(x+ex^3)
$$
Here e is a positive constant. This is an example of a
nonlinear second-order ode.



Finally, suppose that there is damping in the spring-mass system.
Damping might be provided by a dashpot that exerts a continuous
force that is proportional to the velocity (F(t)=-cv(t), where
c is a constant). The total force is a sum of force due to the
spring and the damping. The differential equation is  
$$
mx''(t)= -cx'(t)-kx(t)
$$

\vspace{.3in}

{\bf Electrical Circuits}

\vspace{.2in}

An RLC circuit consists of a resistor, an inductor, and capacitor
in series with a voltage source. Let us assume that the resistance
is R, the inductance is L, the capacitance is C, and the electromotive
force is E(t). Let I(t) denote the current in the circuit. The
current satisfies the differential equation:
$$
LI''(t) + RI'(t) + \frac{1}{C}I(t) = E'(t)
$$
This is an example of a linear second-order ode. To completely
solve this problem the initial current I(0) and its derivative
I'(0) must be specified.

\vspace{.3in}

{\bf One-Dimensional Free Fall Motion}

\vspace{.2in}

Suppose that a particle initially at height $s_{0}$ is thrown
straight upward with velocity $v_{0}$.  Let s(t) denote the
particle's height, v(t) the particle's velocity, and a(t) the
objects acceleration at time t. If air resistance is neglected,
then by Newton's second law we have ma(t)=-mg. Using the
fact that a(t)=s''(t) and eliminating the mass, we obtain the
equation
$$
s''=-g
$$
The initial position s(0) and initial velocity s'(0)=v(0) must
be specified. This is linear second-order ode.

Now suppose that air resistance is proportional to velocity.
$$
ms'' = -mg - cs'(t)
$$
This is another example of a linear second-order ode.

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