Mathematical Tools for Physics

4—Differential Equations 87

For small values oft, the first terms in the power series expansion of this result are

x(t) =

v 0

ω′

[1−γt+γ^2 t^2 / 2 −...][ω′t−ω′^3 t^3 /6 +...] =v 0 t−v 0 γt^2 +...

The first term is what you should expect, as the initial velocity isvx=v 0. The negative sign in the next term
says that it doesn’t move as far as it would without the damping, but analyze it further. Does it have the right
size as well as the right sign? It is−v 0 γt^2 =−v 0 (b/ 2 m)t^2. But that’s an acceleration:axt^2 / 2. It says that the
acceleration just after the motion starts isax=−bv 0 /m. Is that what you should expect? As the motion starts,
the mass hasn’t gone very far so the spring doesn’t yet exert much force. The viscous friction is however−bvx.
Set that equal tomaxand you see that−v 0 γt^2 has precisely the right value.
What is the energy for this damped oscillator? The kinetic energy ismv^2 / 2 and the potential energy for
the spring iskx^2 / 2. Is the sum constant? No.

d

dt

1

2

(

mv^2 +kx^2

)

=mv

dv

dt

+kx

dx

dt

=vx

(

max+kx

)

=−bvx^2 =Fx,frictvx

“Force times velocity” is a common expression for power, and this says that the total energy is decreasing according
to this formula. The energy decreases exponentially on average.

4.2 Forced Oscillations
What happens if the equation is inhomogeneous? That is, what if there is a term that doesn’t involvexor its
derivatives at all. In this harmonic oscillator example, apply an extra external force. Maybe it’s a constant; maybe
it’s an oscillating force; it can be anything you want not involvingx.

m

d^2 x

dt^2

=−kx−b

dx

dt

+Fext(t) (11)

The key result that you need for this class of equations is very simple to state and not too difficult to implement.
It is a procedure for attacking any linear inhomogeneous differential equation and consists of three steps.

1. Temporarily throw out the inhomogeneous term [here Fext(t)] and completely solve the resulting
   homogeneous equation. In the current case that’s what I just got through doing when I solved the
   equationmd^2 x

/

dt^2 +bdx

/

dt+kx= 0. [xhom(t)]