Introduction
We are going to consider a linear dynamical systems in which each mathematical model is a linear 2nd order differential equation with constant coefficients along with initial conditions specified at time t0 :
a2d2ydt2+ a1dydt+ a0y=gt, yt0= y0, y't0=y1
Spring-Mass System - Free Undamped Motion
Suppose a spring is suspended vertically from a rigid support and then a mass mis attached to its free end.
We wish to understand its motion when the spring is stretched(or compressed)
and then released. /
In order to keep the model as simple as possible, we ignore other external forces - free motion.
Hooke’s Law:
· The spring exerts a restoring force opposite to the direction of elongation.
· It is proportional to the amount of elongation, say s ; then we can state
F = ks
where k is a proportionality constant, called the spring constant.
After a mass m attached to a spring stretches the spring by an amount of s, it attains an equilibrium position. At the equilibrium position its weight W is balanced by the restoring force ks.
Recall Weight = mass ⋅ gravitational acceleration
= m⋅g
where m is measured in slugs (or kg) and g = 32 ft/s2( or 9.8 m/s2).
Then the equilibrium condition is
mg=ks
Now the mass is displaced by an amount of y from its equilibrium position and then released.
Newton’s 2nd Law:
Net Force = mass⋅acceleration
F= m⋅d2ydt2
Equating this with restoring force, we have
m⋅d2ydt2= -ky
Note that the restoring acts opposite to the direction of motion.
By dividing by m, we obtain
d2ydt2+kmy=0 or d2ydt2+ω2y=0 where ω2=km
This equation is called a simple harmonic motion or free undamped motion.
Note that this is a 2nd order homogeneous equation with constant coefficients.
We solve by using the method that was discussed in section 3.3:
Example A mass weighing 2 pounds stretches a spring 6 inches. Initially, the mass is released from a point 8 inches below the equilibrium position with an upward velocity of 43 ft/s.
Find the equation of simple harmonic motion.
Solution:
Graph of the solution:
Damped Spring-Mass Motion
Recall the solution of a harmonic motion involves sin and cos functions; oscillates forever with constant amplitude.
To make this model more realistic, we add damping.
Assume: The damping force is proportional to the speed of the mass and it acts as a restoring force.
Then, the damping force is
-βdydt
where β0 is called a damping constant.
With no other external forces, we get from Newton’s 2nd law
m⋅d2ydt2= -ky -βdydt
By dividing by m, we find a 2nd order DE, called free damped motion
d2ydt2+βm⋅dydt+kmy=0 or d2ydt2+2λ⋅dydt+ω2y=0
Again we can solve by using the method that was discussed in section 3.3:
Example A 16-pound weight is attached to a 5-foot-long spring. At equilibrium the spring measures 8.2 feet. If the weight is pushed up and released from rest at a point 2 feet above the equilibrium position, find the displacements y(t) if it is further known that the surrounding medium offers a resistance numerically equal to the instantaneous velocity.
Solution:
Graph of Solution: