4—Differential Equations 68If there’s friction (and there’salwaysfriction), the force has another term. Now how do you describe
friction mathematically? The common model for dry friction is that the magnitude of the force is
independent of the magnitude of the mass’s velocity and opposite to the direction of the velocity. If
you try to write that down in a compact mathematical form you get something like
F~friction=−μkFN~v
|~v|
(4.2)
This is hard to work with. It can be done, but I’m going to do something different. (See problem4.31
however.) Wet friction is easier to handle mathematically because when you lubricate a surface, the
friction becomes velocity dependent in a way that is, for low speeds, proportional to the velocity.
F~friction=−b~v (4.3)
Neither of these two representations is a completely accurate description of the way friction works.
That’s far more complex than either of these simple models, but these approximations are good enough
for many purposes and I’ll settle for them.
Assume “wet friction” and the differential equation for the motion ofmis
m
d^2 x
dt^2
=−kx−b
dx
dt
(4.4)
This is a second order, linear, homogeneous differential equation, which simply means that the highest
derivative present is the second, the sum of two solutions is a solution, and a constant multiple of a
solution is a solution. That the coefficients are constants makes this an easy equation to solve.
All you have to do is to recall that the derivative of an exponential is an exponential.det/dt=et.
Substitute this exponential forx(t), and of course it can’t work as a solution; it doesn’t even make
sense dimensionally. What iseto the power of a day? You need something in the exponent to make it
dimensionless,eαt. Also, the functionxis supposed to give you a position, with dimensions of length.
Use another constant:x(t) =Aeαt. Plugthisinto the differential equation (4.4) to find
mAα^2 eαt+bAαeαt+kAeαt=Aeαt
[
mα^2 +bα+k
]
= 0
The product of factors is zero, and the only way that a product of two numbers can be zero is if one of
the numbers is zero. The exponential never vanishes, and for a non-trivial solutionA 6 = 0, so all that’s
left is the polynomial inα.
mα^2 +bα+k= 0, with solutions α=
−b±
√
b^2 − 4 km
2 m
(4.5)
The position function is then
x(t) =Aeα^1 t+Beα^2 t (4.6)
whereAandBare arbitrary constants andα 1 andα 2 are the two roots.
Isn’t this supposed to be oscillating? It is a harmonic oscillator after all, but the exponentials
don’t look very oscillatory. If you have a mass on the end of a spring and the entire system is immersed