With the model just described, the motion of the mass continues indefinitely. Clearly, this doesn’t happen in the real world. In the real world, there is almost always some friction in the system, which causes the oscillations to die off slowly—an effect called damping.
Physical spring-mass systems almost always have some damping as a result of friction, air resistance, or a physical damper, called a dashpot (a pneumatic cylinder; see Figure 4).
Figure 4. A dashpot is a pneumatic cylinder that dampens the motion of an oscillating system.
Because damping is primarily a friction force, we assume it is proportional to the velocity of the mass and acts in the opposite direction. So the damping force is given by [latex]-bx\prime[/latex] for some constant [latex]b>0[/latex]. Again applying Newton’s second law, the differential equation becomes:
Just as before, we consider three cases, based on whether the characteristic equation has distinct real roots, a repeated real root, or complex conjugate roots.
Case 1: [latex]b^{2}>4mk[/latex]
In this case, we say the system is overdamped. The general solution has the form:
where both [latex]\lambda_1[/latex] and [latex]\lambda_2[/latex] are less than zero. Because the exponents are negative, the displacement decays to zero over time, usually quite quickly. Overdamped systems do not oscillate (no more than one change of direction), but simply move back toward the equilibrium position.
Figure 5. Behavior of an overdamped spring-mass system, with no change in direction (a) and only one change in direction (b).A [latex]16[/latex]-lb mass is attached to a [latex]10[/latex]-ft spring. When the mass comes to rest in the equilibrium position, the spring measures [latex]15[/latex] ft [latex]4[/latex] in. The system is immersed in a medium that imparts a damping force equal to [latex]\frac{5}{2}[/latex] times the instantaneous velocity of the mass. Find the equation of motion if the mass is pushed upward from the equilibrium position with an initial upward velocity of [latex]5[/latex] ft/sec. What is the position of the mass after [latex]10[/latex] sec? Its velocity?
The mass stretches the spring [latex]5[/latex] ft [latex]4[/latex] in., or [latex]\frac{16}{3}[/latex] ft. Thus, [latex]16=\left(\frac{16}3\right)k[/latex], so [latex]k=3[/latex]. We also have [latex]m=\frac{16}{32}=\frac12[/latex], so the differential equation is
so it is, effectively, at the equilibrium position. We have [latex]x^\prime(t)=10e^{-2t}-15e^{-3t}[/latex], so after [latex]10[/latex] sec the mass is moving at a velocity of
where [latex]\lambda[/latex] is less than zero. The motion of a critically damped system is very similar to that of an overdamped system. It does not oscillate.
However, with a critically damped system, if the damping is reduced even a little, oscillatory behavior results. From a practical perspective, physical systems are almost always either overdamped or underdamped (case 3, which we consider next). It is impossible to fine-tune the characteristics of a physical system so that [latex]b^{2}[/latex] and [latex]4mk[/latex] are exactly equal.
Figure 6. Behavior of a critically damped spring-mass system. The system graphed in part (a) has more damping than the system graphed in part (b).A [latex]1[/latex]-kg mass stretches a spring [latex]20[/latex] cm. The system is attached to a dashpot that imparts a damping force equal to [latex]14[/latex] times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from equilibrium with an upward velocity of [latex]3[/latex] m/sec.
We have [latex]mg=1(9.8)=0.2k[/latex], so [latex]k=49[/latex]. Then, the differential equation is
where [latex]\alpha[/latex] is less than zero. Underdamped systems do oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time.
Figure 7. Behavior of an underdamped spring-mass system.
Note that for all damped systems, [latex]\underset{t\to\infty}{\text{lim}}x(t)=0[/latex]. The system always approaches the equilibrium position over time.
A [latex]16[/latex]-lb weight stretches a spring [latex]3.2[/latex] ft. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point [latex]9[/latex] in. below equilibrium.
We have [latex]k=\frac{16}{3.2}=5[/latex] and [latex]m=\frac{16}{32}=\frac12[/latex], so the differential equation is
[latex]\frac12x^{\prime\prime}+x^\prime+5x=0[/latex], or [latex]x^{\prime\prime}+2x^\prime+10x=0[/latex].
Figure 8. (credit: modification of work by nSeika, Flickr)
For motocross riders, the suspension systems on their motorcycles are very important. The off-road courses on which they ride often include jumps, and losing control of the motorcycle when they land could cost them the race.
This suspension system can be modeled as a damped spring-mass system. We define our frame of reference with respect to the frame of the motorcycle. Assume the end of the shock absorber attached to the motorcycle frame is fixed. Then, the “mass” in our spring-mass system is the motorcycle wheel. We measure the position of the wheel with respect to the motorcycle frame. This may seem counterintuitive, since, in many cases, it is actually the motorcycle frame that moves, but this frame of reference preserves the development of the differential equation that was done earlier. As with earlier development, we define the downward direction to be positive.
When the motorcycle is lifted by its frame, the wheel hangs freely and the spring is uncompressed. This is the spring’s natural position. When the motorcycle is placed on the ground and the rider mounts the motorcycle, the spring compresses and the system is in the equilibrium position (Figure 9).
Figure 9. We can use a spring-mass system to model a motorcycle suspension.
This system can be modeled using the same differential equation we used before:
[latex]mx^{\prime\prime}+bx^\prime+kx=0[/latex].
A motocross motorcycle weighs [latex]204[/latex] lb, and we assume a rider weight of [latex]180[/latex] lb. When the rider mounts the motorcycle, the suspension compresses [latex]4[/latex] in., then comes to rest at equilibrium. The suspension system provides damping equal to [latex]240[/latex] times the instantaneous vertical velocity of the motorcycle (and rider).
Set up the differential equation that models the behavior of the motorcycle suspension system.
We are interested in what happens when the motorcycle lands after taking a jump. Let time [latex]t=0[/latex] denote the time when the motorcycle first contacts the ground. If the motorcycle hits the ground with a velocity of [latex]10[/latex] ft/sec downward, find the equation of motion of the motorcycle after the jump.
Graph the equation of motion over the first second after the motorcycle hits the ground.
We have defined equilibrium to be the point where [latex]mg=ks[/latex], so we have
The differential equation found in part a. has the general solution:
[latex]x(t)=c_1e^{-8t}+c_2e^{-12t}[/latex].
Now, to determine our initial conditions, we consider the position and velocity of the motorcycle wheel when the wheel first contacts the ground. Since the motorcycle was in the air prior to contacting the ground, the wheel was hanging freely and the spring was uncompressed. Therefore the wheel is 4 in. ([latex]\frac{1}{3}[/latex] ft) below the equilibrium position (with respect to the motorcycle frame), and we have [latex]x(0)=\frac{1}{3}[/latex]. According to the problem statement, the motorcycle has a velocity of 10 ft/sec downward when the motorcycle contacts the ground, so [latex]x^\prime(0)=10[/latex]. Applying these initial conditions, we get [latex]c_1=\frac{7}{2}[/latex] and [latex]c_2=-(\frac{19}{6})[/latex], so the equation of motion is
[latex]x(t)=\frac72e^{-8t}-\frac{19}6e^{-12t}[/latex].
The graph is shown in Figure 10. Figure 10. Graph of the equation of motion over a time of one second.
