Let us now focus on damped motion. Let us rewrite the equation 𝑚𝑥00+𝑐𝑥0+𝑘𝑥
as 𝑥00+ 2𝑝𝑥0+𝜔2 0𝑥 =0, where 𝜔0= r 𝑘 𝑚, 𝑝 = 𝑐 2𝑚. The characteristic equation is
𝑟2+2𝑝𝑟+𝜔2 0=0.
Using the quadratic formula we get that the roots are 𝑟 =−𝑝±
q
𝑝2−𝜔2
0.
The form of the solution depends on whether we get complex or real roots. We get real roots if and only if the following number is nonnegative:
𝑝2−𝜔2 0 = 𝑐 2𝑚 2 − 𝑘 𝑚 = 𝑐2−4𝑘𝑚 4𝑚2 . The sign of𝑝2−𝜔2
0is the same as the sign of𝑐2−4𝑘𝑚. Thus we get real roots if and only if
𝑐2−4𝑘𝑚 is nonnegative, or in other words if𝑐2 ≥ 4𝑘𝑚.
Overdamping 0 25 50 75 100 0 25 50 75 100 0.0 0.5 1.0 1.5 0.0 0.5 1.0 1.5
Figure 2.3: Overdamped motion for several differ- ent initial conditions.
When 𝑐2 − 4𝑘𝑚 > 0, the system is over-
damped. In this case, there are two distinct real roots 𝑟1 and 𝑟2. Both roots are nega-
tive: As
q
𝑝2−𝜔2
0is always less than𝑝, then
−𝑝±q𝑝2−𝜔2
0is negative in either case.
The solution is
𝑥(𝑡) =𝐶1𝑒𝑟1𝑡+𝐶2𝑒𝑟2𝑡.
Since𝑟1, 𝑟2are negative,𝑥(𝑡) →0 as𝑡 → ∞.
Thus the mass will tend towards the rest position as time goes to infinity. For a few sample plots for different initial conditions,
see .
No oscillation happens. In fact, the
graph crosses the 𝑥-axis at most once. To see why, we try to solve 0 = 𝐶1𝑒𝑟1𝑡 +𝐶2𝑒𝑟2𝑡.
Therefore, 𝐶1𝑒𝑟1𝑡 =−𝐶2𝑒𝑟2𝑡 and using laws of exponents we obtain
−𝐶1
𝐶2 =𝑒
This equation has at most one solution𝑡 ≥0. For some initial conditions the graph never crosses the𝑥-axis, as is evident from the sample graphs.
Example 2.4.2: Suppose the mass is released from rest. That is 𝑥(0) = 𝑥0 and 𝑥0(0) = 0. Then
𝑥(𝑡) = 𝑥0
𝑟1−𝑟2 𝑟1𝑒
𝑟2𝑡 −𝑟
2𝑒𝑟1𝑡.
It is not hard to see that this satisfies the initial conditions. Critical damping
When 𝑐2− 4𝑘𝑚 = 0, the system is critically damped. In this case, there is one root of multiplicity 2 and this root is−𝑝. Our solution is
𝑥(𝑡) =𝐶1𝑒−𝑝𝑡+𝐶
2𝑡𝑒−𝑝𝑡.
The behavior of a critically damped system is very similar to an overdamped system. After all a critically damped system is in some sense a limit of overdamped systems. Since these equations are really only an approximation to the real world, in reality we are never critically damped, it is a place we can only reach in theory. We are always a little bit underdamped or a little bit overdamped. It is better not to dwell on critical damping. Underdamping 0 5 10 15 20 25 30 0 5 10 15 20 25 30 -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0
Figure 2.4: Underdamped motion with the envelope curves shown.
When 𝑐2 −4𝑘𝑚 < 0, the system is under-
damped. In this case, the roots are complex. 𝑟 =−𝑝± q 𝑝2−𝜔2 0 =−𝑝± √ −1 q 𝜔2 0−𝑝2 =−𝑝±𝑖𝜔1, where𝜔1= q 𝜔2 0−𝑝2. Our solution is 𝑥(𝑡)=𝑒−𝑝𝑡 𝐴cos(𝜔1𝑡) +𝐵sin(𝜔1𝑡), or 𝑥(𝑡)=𝐶𝑒−𝑝𝑡 cos(𝜔1𝑡−𝛾).
An example plot is given in . Note that we still have that𝑥(𝑡) →0 as𝑡 → ∞. The figure also shows theenvelope curves𝐶𝑒−𝑝𝑡and−𝐶𝑒−𝑝𝑡. The solution is the oscillating line between the two envelope curves. The envelope curves give the maximum amplitude of the oscillation at any given point in time. For example, if you are bungee jumping, you are really interested in computing the envelope curve as not to hit the concrete with your head.
The phase shift 𝛾shifts the oscillation left or right, but within the envelope curves (the envelope curves do not change if𝛾changes).
Notice that the angularpseudo-frequency becomes smaller when the damping𝑐 (and hence𝑝) becomes larger. This makes sense. When we change the damping just a little bit, we do not expect the behavior of the solution to change dramatically. If we keep making𝑐 larger, then at some point the solution should start looking like the solution for critical damping or overdamping, where no oscillation happens. So if 𝑐2 approaches 4𝑘𝑚, we want𝜔1to approach 0.
On the other hand, when 𝑐gets smaller,𝜔1approaches𝜔0(𝜔1is always smaller than
𝜔0), and the solution looks more and more like the steady periodic motion of the undamped
case. The envelope curves become flatter and flatter as𝑐 (and hence𝑝) goes to 0.
2.4.4
Exercises
Exercise2.4.2: Consider a mass and spring system with a mass𝑚=2, spring constant𝑘 =3, and damping constant𝑐 =1.
Set up and find the general solution of the system. a)
Is the system underdamped, overdamped or critically damped? b)
If the system is not critically damped, find a𝑐 that makes the system critically damped. c)
Exercise2.4.3: Do for𝑚 =3,𝑘 =12, and𝑐 =12.
Exercise2.4.4: Using the mks units (meters-kilograms-seconds), suppose you have a spring with spring constant 4N/m. You want to use it to weigh items. Assume no friction. You place the mass
on the spring and put it in motion.
You count and find that the frequency is 0.8 Hz (cycles per second). What is the mass? a)
Find a formula for the mass𝑚given the frequency𝜔in Hz. b)
Exercise 2.4.5: Suppose we add possible friction to . Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz.
Find𝑘 (spring constant) and𝑐 (damping constant). a)
Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency.
b)
For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
c)
∗We do not call𝜔
Exercise2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find𝑐). You have a spring with spring constant𝑘 =5N/m. You take the
spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction?
Exercise2.4.101: A mass of2kilograms is on a spring with spring constant𝑘 newtons per meter