ok, lets start with some Physics.
Hooke's Law suppose that a flexible spring is suspended vertically from a rigid support and the mass "m" is attached to its free end. The amount of stretch, or elongation, or the spring will of course depend on the mass; masses with different weights stretch the spring by differing amounts. By Hooke's Law, the spring itself exerts a restoring force "F" opposite to the direction of elongation and proportional to the amount of elongation "s". Simply stated, F= ks, where "k" is a constant of proportionality called the spring constant. The spring is essentially characterized by the number "k". For example, if a mass weighing 10 pounds stretches a sprig 1/2, then 10 =k(1/2) implies k= 20 lb/ft. Necessarily then, a mass weighing, say, 8 pounds stretches the same spring only 2/5 foot.
Newton's Second Law after a mass "m' is attached to a spring, it stretches the spring by the amount "s" and attain a position of equilibrium at which its weight "W" is balanced by the restoring force "ks". Recall that weight is define by W=mg, where mass is measured in slugs, Kilograms, or grams and g=32 ft/s2, 9.8 m/s2, or 980 cm/s2, respectively. As indicated in the figure (1b), the condition of equilibrium is mg=ks or mg-ks=0. If the mass is displaced by an amount o "x" from its equilibrium position, the restoring force of the spring is then k(x+s). Assuming that there are not retarding forces acting on the system and assuming that the mass vibrates free of other external forces -- free motion-- we can equate Newton's second law with the net, or resultant, force of the restoring force and the weight:
md2x/dt2 = -k(x+s) + mg = -ks + mg - ks
-ks + mg = 0
The negative sign the example above indicate that the restoring force of the spring acts opposite to the direction of motion. Furthermore, we can adopt the convention that displacements measured below the equilibrium position are positive.
Differential Equation of free Undamped Motion by dividing the above equation by mass "m" we obtain the second-order differential equation d2x/dt2 + (k/m)x = 0 or
d2x/dt2 + w2x = 0,
where w2= k/m. the equation above is said to describe simple harmonic motion or free undamped motion.
Two equation working worth working. The two differential equations
y'' + k2y = 0
and
y'' - k2y = 0,
k real, are important in applied mathematics. For y'' + k2y = 0, the auxiliary equation m2 + k2 = 0 has imaginary roots mi= ki and m2= -ki, where i=(-1)^(1/2). The general solution of DE is seen to be
y = c1cos kx + c2 sin kx or y = c1cos wt + c2 sin wt
The period of the free vibrations described by the formula above is T = 2p/w, and the frequency if f=1/T=w/2p. For example, for x(t)= 2cos 3t - 4sin 3t the period is 2p/3 and the frequency is 3/2p. The former number means that the graph of x(t) repeats every 2p/3 units; the later number means that there are three cycles of the graph every 2p units or, equivalently, that the mass undergoes 3/2p over complete vibrations per unit time.
Example:
A mass weighing 2 pounds stretches a spring 6in. At t=0 the mass is released from a point 8in below the equilibrium point with an upward velocity of 4/3 ft/s. determine the equation of free motion.
solution
first we have to convert in (inches) into ft (feet). 6 in = 1/2 ft; 8 in = 2/3 ft. in addition we must convert the units of weight given in pounds into units of mass. from W=mg, m=W/g we have m=2/32=1/16 slug. also, from Hooke's law, 2=k(1/2) implies the spring constant is k= 4lb/ft. hence from the equation md2x/dt2=-kx
(1/16)d2x/dt2=-4x
(1/16)d2x/dt2 + 4x=0
or
d2x/dt2 + 64x=0
The initial displacement and the initial velocity are x(0)=2/3, x'(0)=-4/3, where the negative sign in the last condition is a consequence of the fact that the mass is given an initial velocity in the negative, or upward, direction.
Now from w2=64 or w=8, so the general solution of the differential equation is y = c1cos 8t + c2 sin 8t.
Applying the initial conditions to x(t) and x'(t) gives c1=2/3 and c2=-1/6.
Thus the equation of the motion is x(t) = (2/3)cos 8t - (1/16)sin 8t.
Alternative form of x(t) When c1¹0 and c2¹0, the actual amplitude A of the free vibrations is not obvious from the inspection of the equation x(t) = c1cos wt + c2 sin wt. For example, although the mass of the above problem is initially displaced 2/3 ft beyond the equilibrium position, the amplitude of the vibrations is a number larger that 2/3. Hence it is often convenient to convert a solution of form y = c1cos wt + c2 sin wt to the simpler form x(t)=Asin(wt+f), where A=(c12+c22)^(1/2), and f is a phase define by sinf=c1/A, cosf=c2/A, and tanf=c1/c2
Alternative form of solution.
In the view of the foregoing discussion, we can write solution y = (2/3)cos 8t - (1/16)sin 8t, in a alternative form x(t)=Asin(8t+f). computation of the amplitudes is straightforward, A=((2/3)2+(-1/6)2)^(1/2)=(17/36)^(1/2)»0.69ft, but some care should be exercised when computing the phase angle defined by tanf= c1/ c2. With c1=2/3 and c2=-1/6 we find tanf=-4, and a calculator the give tan-1=-1.326 radians. we must take f to be the second quadrant angle f=p+(-1.326)=1.816 radians. thus we have x(t)=(17/36)^(1/2)sin(8t+1.816).
the period of the function is T=2p/8=p/4
