4—Differential Equations 70Does 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:
x(t)≈v 0 t−v 0 γt^2 =v 0 t−v 0
b
2 m
t^2 =v 0 t+
1
2
−bv 0
m
t^2
The last term says that the acceleration starts asax=−bv 0 /m, as required.
In Eq. (4.8) I assumed that the two roots of the quadratic, the twoα’s, are different. What if
they aren’t? Then you have just one value ofαto use in defining the solutioneαtin Eq. (4.9). You
now have just one arbitrary constant with which to match two initial conditions. You’re stuck. See
problem4.11to understand how to handle this case (critical damping). It’s really a special case of what
I’ve already done.
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.
If Fx=max=−kx+Fx,frict, then
dE
dt
=
d
dt
1
2
(
mv^2 +kx^2
)
=mv
dv
dt
+kx
dx
dt
=vx
(
max+kx
)
=Fx,frictvx (4.11)
“Force times velocity” is a common expression for power, and this says that the total energy is decreasing
according to this formula. For the wet friction used here, this isdE/dt=−bv^2 x, and 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 involve
xor 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) (4.12)
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 [hereFext(t)] and completely solve the
resulting homogeneous equation. In the current case that’s what you just saw when Iworked out the solution to the differential equationmd^2 x
/
dt^2 +bdx
/
dt+kx= 0.
[xhom(t)]
- Find anyonesolution to the full inhomogeneous equation. Note that for step one you
have to have all the arbitrary constants present; for step two you do not. [xinh(t)]
3. Add the results of steps one and two. [xhom(t) +xinh(t)]
I’ve already done step one. To carry out the next step I’ll start with a particular case of theforcing function. IfFext(t)is simple enough, you should be able toguessthe answer to step two. If it’s
a constant, then a constant will work forx. If it’s a sine or cosine, then you can guess that a sine or
cosine or a combination of the two should work. If it’s an exponential, then guess an exponential —
remember that the derivative of an exponential is an exponential. If it’s the sum of two terms, such