University Physics I - Classical Mechanics, 2019

272 CHAPTER 11. SIMPLE HARMONIC MOTION
11.6 Advanced Topics
11.6.1 Mass on a spring damped by friction with a surface
Consider the system depicted in Figure 11.2 in the presence of friction between the block and the
surface. Let the coefficient of kinetic friction be μ k and the coefficient of static friction be μ s . As
usual, we will assume that μ s ≥ μ k .
As the mass oscillates, it will experience a kinetic friction force of magnitude F k = μ k mg, inthe
direction opposite the direction of motion; that is to say, a force that changes direction every half
period. As shown in section 11.2.2, this force does not change the frequency of the motion, but it
displaces the equilibrium position in the direction of the force, which is to say, closer to the starting
point for each half-swing. As a result of that, the amplitude for each half-swing is reduced from
the previous one.
Let the original equilibrium position (in the absence of friction) be x 0 = 0. Suppose we displace
the mass a distance A to the right (call this position, the starting point for the first half-swing,
x 1 = A), and let go. In the presence of friction, the equilibrium position for this first half-swing
becomes the point x ′ 0 = F k /k = μ k mg, so the real amplitude of this first half-oscillation will be
A 1 = x 1 − x ′ 0 = A − x′ 0 , and the resulting motion will be
x(t) =x ′ 0 + A 1 cos(ωt) (first half-period, 0 ≤ t ≤ π/ω) (11.32)
The mass then stops, momentarily, at t = π/ω, atthepositionx 2 = x ′ 0 − A 1, and turns around
for the second half-swing. However, now the external force has reversed direction, so the new
equilibrium position is at −x ′ 0 , and the new amplitude is A 2 = −x ′ 0 − x 2 = A 1 − 2x ′ 0 = A − 3x′ 0 .
The motion for this next half-period is then
x(t) =−x ′ 0 + A 2 cos(ωt) (second half-period, π/ω ≤ t ≤ 2π/ω) (11.33)
Continuing the process, we see that A 1 = A−x ′ 0 , A 2 = A−3x ′ 0 , A 3 = A−5x ′ 0 ...A n = A−(2n−1)x ′ 0 .
Of course, this can’t go on forever, since we require the amplitude to be a positive quantity; so the
motion will consist of only n half-periods, where n is an integer such that A − (2n − 1)x ′ 0 > 0 but
A − (2n +1)x ′ 0 < 0. (That is to say, n is equal to the integer part of (A/x′ 0 +1)/2.)
The figure shows an example of how this would go, for the following choice of parameters: period
T =1s,μ k =0.1, and A =0.18 m. Note that, since x ′ 0 depends only on the ratio m/k =1/ω2 ,there
is no need to specify m and k separately. We get ω =2π/T =2π rad/s, x ′ 0 = μ kg/ω 2 =0.0248 m,
and (A/x ′ 0 +1)/2 =4.13, which means that the motion will go on for 4 half-periods before stopping.