Differential Equations – Section 5.1 – Homework: 1, 3, 5, 11, 21, 23, 29
Hooke’s Law -
F  ks , k = spring constant, s = amount of elongation, F =restoring force
Newton’s Second Law – Free Undamped Motion or Simple Harmonic Motion
d 2x
  2 x  0 where  2  k / m ,
2
dt
Solution:
x(t )  c1 cos  t  c2 sin  t
x(0)  x0 =the amount of initial displacement
x(0)  x1 = the initial velocity of the mass
The period of free vibrations is T  2 /  and the frequency is f  1 / T   / 2 .
A maximum of x(t) is a positive displacement corresponding to the mass’s attaining a maximum distance
below the equilibrium position, whereas a minimum of x9t) is a negative displacement corresponding to
the mass’s attaining a maximum height above the equilibrium position. We refer to either case as an
extreme displacement of the mass. The particular solution is the equation of motion.
#2
#6
x(t )  c1 cos  t  c2 sin  t  A sin  t    where
c
c
and  is the phase angle defined by sin   1 and cos   2 .
A
A
If you need to find the amplitude of motion, recall from trig:
A  c1  c2
2
#8
2
Free Damped Motion – We will assume that the damping force is given by a constant multiple of
be modeled by
Case I:
dx
. The motions will
dt

k
d 2x
dx
 2
  2 x  0 where 2  ,  2  .
2
m
m
dt
dt
2   2  0
The system is said to be overdamped. The solution is
x(t )  e t  c1e

2  2 t
 c2 e 
This equation represents a smooth an nonoscillatory motion.
Case II:     0 The system is said to be critically damped. The solution is
Any slight decrease in the damping force would result in oscillatory motion.
2
Case III:
x(t )  e
#22
2
2   2  0
 t
c cos
1
The system is said to be underdamped. The solution is
  2 t  c2 sin  2  2 t
2
.
x(t )  e t c1  c2 t  .
2  2 t
.

