In this introductory course on Ordinary Differential Equations, we first provide basic terminologies on the theory of differential equations and then proceed to methods of solving various types of ordinary differential equations. We handle first order differential equations and then second order linear differential equations. We also discuss some related concrete mathematical modeling problems, which can be handled by the methods introduced in this course. The lecture is self contained. However, if necessary, you may consult any introductory level text on ordinary differential equations. For example, "Elementary Differential Equations and Boundary Value Problems by W. E. Boyce and R. C. DiPrima from John Wiley & Sons" is a good source for further study on the subject. The course is mainly delivered through video lectures. At the end of each module, there will be a quiz consisting of several problems related to the lecture of the week.
9-1 Spring-mass system
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
64 個評分
Korea Advanced Institute of Science and Technology
64 個評分
從本節課中
APPLICATIONS OF SECOND ORDER EQUATIONS
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
In this last chapter of the course,
we handle two physical phenomena which involve
a linear second order constant of coefficients differential equations,
say the spring mass system and the motion of the pendulum.
First, let's consider the spring mass system.
Attach a mass m to a spring of length l,
which is suspended from a rigid support so that the spring is
stretched with elongation Δl and reaches its equilibrium state.
Then there are two forces acting on the mass: say
the gravitational force that is the weight w = mg.
m is the mass of the object and g is the gravitational acceleration which acting
downward and the spring's restoring force F_s acting upward.
Here, we adopt the convention that the downward direction is the positive.
Here's the un-stretched spring.
Then we put the mass m,
then we have an elongation given by the Δl.
Then by the Hooke's law,
when Δ is small compared to the original length l,
the restoring force F_s is proportional to the elongation.
In other words, F_s = -kΔl,
where the k, the positive constant k
is the so-called spring constant or the stiffness of the spring.
Since the mass is at its equilibrium state,
we should have mg - kΔl = 0.
In other words, mg is equal to the gravitational force
is the same as the restoring force k times Δl.
Now, let the mass be displaced further from
its equilibrium state as is shown in this figure by an amount x(t),
which is measured positive downward and then released.
And we are going to look at the motion of this mass.
Then, actually, there are four forces acting on the system: First,
the weight w = mg due to the gravitation and which is
acting downward and the spring force F_s = -k(Δl+x),
which is acting upward when the spring is extended.
In other words, the (Δl + x) is positive or
the spring force is acting downward when the spring is compressed.
In other words, (Δl + x) is negative.
Another force that we have is the damping force F_d due
to the viscosity of the fluid in which the mass is moving,
which is proportional to the mass of speed so that we have F_d = -cx'(t),
where the positive constant c
is the damping constant and the x' is the velocity of the object.
When the speed of the mass' of this is quite small.
Here the negative sign,
it means that the damping force is acting in
the direction opposite to the mass' moving direction.
Finally, there might be some external force f(t) acting on the mass.
Then by the Newton's second law of motion,
first we can think of the forces of the system is m times acceleration.
So m times x double prime must be equal to the total force acting on the system.
In other words, if we have the gravitational force mg and the spring force F_s and
the damping force F_d and any external force f. That is equal to mg - k(Δl +
x) - cx' + f.
Since we know that at the equilibrium state mg = kΔl.
So mg - kΔl cancelled out then simplifying the equation,
we get mx" + cx' + kx = f.
Which is a second order linear constant
coefficients possibly non-homogeneous differential equation.
We call this differential equation one.
As the differential equation for forced damped motion for general f,
if f is identically zero.
Then we call it as a free damped motion.
Divide the whole equation by m,
divide the whole equation by m. Then you are going to get x" + cx'/m + km/x and that
is equal to f/m,
to give this equation.
Let's set for the computational simplicity,
let's call c/m is equal to two times λ and the k/m,
let's say be equal to omega square and f/m
let's say it to be g. Then we get the second equation say x" + 2λx' + ω square x = g(t).
