Lecture One: Part II
Mass Spring Damper System: Deriving the motion
ESE112 Lecture 1
MSD Review
Created by Wikipedia user lzyvzl
Mass (m) is attached to Hook’s law spring
Recall Fs = −kx
Damping Force Fd = c dx
dt
ESE112 Lecture 1
Deriving Equations of Motion
From Newton
!
rearranging
f = ma
(1)
mẍ = −cẋ − kx
(2)
mẍ + cẋ + kx = 0
(3)
c
k
ẍ + ẋ + x
m
m
(4)
dividing by m
"
k
, be the natural frequency of the system and ζ =
if we let ω0 = m
the damping ratio, we get
ẍ + 2ζω0 ẋ + ω02 x = 0
√c ,
2 km
be
(5)
http://en.wikipedia.org/wiki/Damping
ESE112 Lecture 1
Solving the differential equation
What’s a differential equation?
“A differential equation is a mathematical equation for
an unknown function of one or several variables that
relates the values of the function itself and of its
derivatives of various orders.”
A simple instantiation of this is an equation
which relates the velocity, dx/dt, of an
object as a function of it’s position F(x) in
time.
ESE112 Lecture 1
Solution for a special case
ESE112 Lecture 1
Solution for a special case
Recall from calculus that if
y = eax
(1)
then
dy
= aex
dx
Now we can try to solve a simple differential equation
dx
=x
dt
(2)
(3)
ESE112 Lecture 1
Solution for a special case
Recall from calculus that if
y = eax
(1)
then
dy
= aex
dx
Now we can try to solve a simple differential equation
dx
=x
dt
(2)
(3)
let’s try the solution x = et
ESE112 Lecture 1
Applying it to a MSD
Recall
ẍ + 2ζω0 ẋ + ω02 x = 0
(1)
now assume that x = eγt then ẋ = γeγt and ẍ = γ 2 eγt substituting and dividing
by eγt we get
γ 2 + 2ζω0 γ + ω02 = 0
(2)
we can now solve for γ using quadratic formula
Quick Exercise:
Solve for gamma
now that we have solved for γ we can find x(t) (which we’ll leave as an
exercise for you to do on your own or in a future class)
ESE112 Lecture 1
Applying it to a MSD cont.
By choosing appropriate parameters we can “tune” the
behavior of the system
We can classify the system into three distinct categories.
1.Under-damped: ζ< 1. In this case the system oscillates with a frequency
equal to ωd = ωo √1-ζ2
2.Over-damped: ζ > 1. The system slowly returns to equilibrium
3.Critically Damped: ζ=1. The system returns to equilibrium
NOTE: Each of the three behaviors corresponds to the solution type of the
quadratic equation shown previously (complex, two real solutions, one real
solution)
ESE112 Lecture 1
Where does this all fit in
Background and history
MSD-like systems are found throughout
engineering
•Animal Running
•Circuit Design: RLC circuits
•Robotics (PD control)
•Automotive (Cruise control, car suspensions
etc)
Successful engineers take old ideas
from one field and use them to
innovate and create in another field
ESE112 Lecture 1
Where does this all fit in
Background and history
Linear Systems
During the first 3 (maybe all 4) years of
engineering you’ll see the world as a linear system
•Relatively easy to solve and understand
•Large body of theory
•Of course the world is rarely linear
•The trick is... can we make it look linear?
ESE112 Lecture 1
Let’s look at applet one
more time
• http://links.math.rpi.edu/applets/appindex/
springmass.html
ESE112 Lecture 1
Homework
Sealy (the bed company) is having a problem. The problem is
that people are so happy to get into their comfy Sealy beds
that they jump on them. Jumping so high and because the beds
are so springy they are flung off the bed onto the ground and
sue the company.
To combat this problem Sealy wants to design a new bed that
will keep the average adult (weighing between 55kg and
110kg) on their beds (but still provide the softest possible
landing). Can you help choose the damping and spring
constants for the bed assuming the bed is very thin (think
trampoline like), it is 30cm off the ground and the average
height (at it’s peak) of someone’s jump is 120 cm.
ESE112 Lecture 1