Download Newton`s Second Law Spring/Mass Systems: Free Undamped

Newton’s Second Law
Spring/Mass Systems: Free Undamped Motion
Hooke's Law stated that the restoring force F of a spring opposite to the direction of elongation
and proportional to its total elongation. The equation given by F = ks where F, the restoring
force, s, amount of elongation and k, spring constant. For example, if a mass weighing 14
pounds stretches a spring ½ foot, then 14 = k(1/2) and k = 28 lbs/ft.
Before proceed to Newton’s Second Law, we define the weight, W = mg where mass is
measured in slugs, grams, or kilograms. For example, g = 32 ft. /s2 or 9.8 m / s2 or 980 cm / s2.
The condition for equilibrium is mg = ks or mg - ks = 0. If the mass is displaced by an amount
x from its equilibrium position, the restoring force of the spring is then k(x + s). Assume that
there are no retarding forces acting on the system and assuming that the mass vibrates free of
other forces (free motion), we can write Newton’s Second Law with resultant force and the
weight given by
m
d 2x
 k ( s  x)  mg
dt 2
= kx  mg  ks  kx (1)
Newton's 2nd Law of Motion
The negative sign in (1) show that the restoring force acts opposite to the direction of motion.
Next,
m
d 2x
 kx
dt 2
Divide by m both side equations
d 2 x k

x
dt 2
m
Let  2 
k
m
Then
d 2x
d 2x
2
and



x
2x  0
2
2
dt
dt
(2)
The last equation is called Simple Harmonic Motion or Free Undamped Motion.
In order to solve equation (2), we need two initial condition, x(0)  x0 , the amount of initial
displacement, x '(0)  x1 , the initial velocity of the mass.
Method of Solution
d 2x
To solve 2   2 x  0 , we need to find the auxiliary equation associated to second order
dt
homogenous equation.
So, we have m 2   2  0 so m   i .
The solution given by,
xt   A cost   B sin t 
The period of free vibrations for equation (3) is given by
F
1 

T 2
(3)
T
.
Equation (3) is called Equation of Motion
By using two initial conditions, we should easy to determine A and B.
2

and the frequency,
Ex: Free Undamped Motion:
A mass weighing 4 lbs stretches a spring 8 inches. At t = 0, the mass is released from a point
10 inches below the equilibrium position with an upward velocity of 3/4 ft/sec. Determine the
equation of free motion. Determine the period of free vibrations and its frequency.
Solution:
8 inches = 8/12 ft. = 2/3 ft. = s
W
4lbs
1
W = mg implies m  
 slug
2
g 32 ft / sec
8
From Hooke's Law, F = ks, we have
2
4  k 
3
So k = 6 lb/ft.
Since m
1 d 2x
d 2x
we
have
 6 x and


kx
8 dt 2
dt 2
d 2x
 48 x  0
dt 2
10 5
3
 ft and x '(0) 
(the mass has an initial velocity in the negative
12 6
4
or upward direction)
x(0) = 10 inches =
From
d 2x
 48 x  0 ,  2  48 so   48
dt 2
The general solution to the DE is


xt   A cos 48t  B sin
Note that
x t    48 A sin



48t


48t  48 B cos 48t

Since x (0) 
5
, then
6
5
 A cos0   B sin 0  implies
6
5
A
6
3
Since x '(0) 
,
4

3
5

48 sin 0  48 B cos0
4
6
3
 48 B
4
3
3
3
B


16
4 48
4 3 16
So

Thus, the equation of motion is
5
3
xt   cos 48t 
sin 48t .
6
16




The period of free vibrations, T 
The frequency is F 
1

T
1
2

  


2 3


2
48
2 3



2 3