spring mass damper solution

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General Solution to an Unforced Spring-Mass-Damper

System

Andrew Cox

February 11, 2013



Chapter 1

Background

m

k

c

x

Figure 1.1: General Unforced Spring-Mass-Damper System

Consider a generic freely moving spring-mass-damper system described by the followingdifferential equation:

x + 2ρωn x + ω2

nx = 0; (1.1)

where x describes the displacement from the unstretched length of the spring, ρ is thedamping ratio, and ωn is the natural frequency. The initial conditions are x(0) = a, x(0) = b.

This equation is often written slightly differently:

mx + cx + kx = 0 (1.2)where m is the mass, c is the viscous damping strength, and k is the spring constant as

shown in Figure 1.1. The relationships between the two are as follows:

ωn =

k

m

1



ρ = c

2√

km

This type of problem shows up in all sorts of applications, so it is useful to have a generalsolution. The easiest way to find a solution is to use Laplace transforms.

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Chapter 2

Solution

First we take the Laplace of both sides of the equation:

L x + 2ρωn x + ω2

nx =

L[0]

X (s)

s2 + 2ρωns + ω2

n

= 0 (2.1)

Technically the Laplace transform above is incorrect; we must include the initial condi-tions when taking L[x] and L[x].

L[x] = s2X (s) − sx(0)− x(0)

L[x] = sX (s)− x(0)

However, including the initial conditions at this stage only makes the algebra more tedious,so we’re going to say they’re all equal to zero for now and plug them in later when the

algebra is kinder.The case where X (s) = 0 is a trivial solution, which isn’t interesting, so we’ll onlyconsider the other term. Using the quadratic equation, we solve for s:

s = −ρωn ± ωn

ρ2 − 1 (2.2)

The general solution is then:

x(t) = C 1 exp−ρωn + ωn

ρ2 − 1

t

+ C 2 exp−ρωn − ωn

ρ2 − 1

t

(2.3)

However, we don’t know whether or not

ρ2 − 1 is real or not, so we must evaluate this

solution in three pieces.

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2.1 ρ > 1

When ρ > 1, the system is ‘over-damped’. We first apply the initial position condition,x(0) = a and find that:

a = C 1 + C 2 (2.4)Next, we take the derivative and apply the second initial condition, x(0) = b:

b = C 1(−ρωn + ωn

ρ2 − 1) + C 2(−ρωn − ωn

ρ2 − 1) (2.5)

By solving equations 2.4 and 2.5, we find that

C 1 = a + b + a(ρωn − ωn

ρ2 − 1)

2ωn

ρ2 − 1(2.6)

C 2 = −b − a(ρωn − ωn ρ2 − 1)

2ωn

ρ2 − 1(2.7)

Plugging these constants into equation 2.3 yields the equation for the displacement of the mass over time when the system is over-damped.

Figure 2.1: Overdamped System Response (x( 0) = 1m, x( 0) = 0.25ms

)

As shown in Figure 2.1, the spring returns to it’s equilibrium position of 0 meters. Notethat there are no oscillations: the response doesn’t ever cross the x-axis.

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2.2 ρ = 1

When ρ = 1, the system is said to be ‘critically damped’. Equation 2.2 now simplifies to

s = −ωn (2.8)

with a multiplicity of 2. The general solution is now written slightly differently becauseof the identical roots:

x(t) = C 1 exp(−ωnt) + C 2t exp(−ωnt) (2.9)

Plugging in the initial conditions like we did for the last case yields the following:

C 1 = a (2.10)

C 2 = b + ωna (2.11)

Plugging all of these equations together yields the solution for the critically dampedsystem:

x(t) = a exp(−ωnt) + (b + ωna)t exp(−ωnt) (2.12)

Figure 2.2: Critically Damped System Response (x( 0) = 1m, x( 0) = 0.25ms )

Figure 2.2 shows the critically damped response. The spring returns to its equilibriumposition more quickly than the over damped case.

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2.3 ρ < 1

The last case is the case of the under-damped system. Now we know that the term insidethe square root is negative and we can rewrite it accordingly:

s = −ρωn ± ωn

1− ρ2

i (2.13)We can also rewrite the general solution now that we know the roots are complex:

x(t) = exp(−ρωnt)

C 1 exp(iωn

1− ρ2) + C 2 exp(−iωn

1 − ρ2)

(2.14)

= exp(−ρωnt)

A cos

ωn

1− ρ2t

+ B sin

ωn

1 − ρ2t

(2.15)

By plugging in the initial conditions (be careful to use the power and chain rules correctly)we find that

A = a (2.16)

B = b + ρωna

ωn

1 − ρ2(2.17)

By combining these equations, we find that the solution for the under-damped system is

x(t) = exp(−ρωnt)

a cos

ωn

1 − ρ2t

+ b + ρωna

ωn 1− ρ2

sin

ωn

1 − ρ2t

(2.18)

Alternatively, it is often useful to write this response in terms of a single sinusoid with aphase angle:

x(t) = A exp(−ρωnt)sin(ωnt + φ) (2.19)

where

φ = arctan

ωna

b + ρωna

(2.20)

A = a

sin(φ) (2.21)

Figure 2.3 shows the underdamped response to the initial conditions. The mass willoscillate several times. The weaker the damping, the more oscillations will occur.

6



Figure 2.3: Underdamped System Response (x( 0) = 1m, x( 0) = 0.25ms )

2.4 MATLAB Code

1 % General Spring−Mass−Damper System Plots (Unforced)

2 %

3 % Author: Andrew Cox

4 % Version: January 11, 2013

5

6 clear;

7 close all;

8 clc;

9

10 w n = 2;

11 a = 1; %Initial position

12 b = 0.25; %Initial velocity

13 t = linspace(0,10, 2ˆ14);

14

15 %First, overdamped case

16 rho = 1.5;

17 w d = w n*sqrt(rhoˆ2 − 1);

18

19 C1 = a + (b + a*

(rho*

w n −

w d))/(2*

w d);

20 C2 = (−b − a*(rho*w n − w d))/(2*w d);

21

22 x = C1*exp((−rho*w n + w d)*t ) + C 2*exp((−rho*w n − w d)*t);

23

24 h1 = figure(1);

25 plot(t, x);

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26 xlabel('Time [s]');

27 ylabel('Displacement [m]');

28 title('Overdamped System Response');

29 grid on;

30

31 %Next, critically damped case

32 C 1 = a ;33 C2 = b + w n*a;

34

35 x = ( C 1 + C 2*t).*exp(−w n*t);

36

37 h2 = figure(2);

38 plot(t, x);



41 title('Critically−Damped System Response');

42 grid on;

43

44 %Finally, underdamped case

45 rho = 0.5;

46 phi = atan(w n*a/(b + rho*w n*a));

47 A = a/sin(phi);

48

49 x = A*exp(−rho*w n*t).*sin(w n*t + phi);

50

51 h3 = figure(3);

52 plot(t, x);



55 title('Underdamped System Response');

56 grid on;

57

58 %Save the plots to file

59 saveas(h1, 'Overdamped System', 'png');

60 saveas(h2, 'Critically Damped System', 'png');

61 saveas(h3, 'Underdamped System', 'png');

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spring mass damper solution

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