Sheng-Fang Huang

2.4 Modeling: Free Oscillations (Mass–Spring System)

m

If we pull the body down a certain distance and then release it, it starts moving. We assume that it moves strictly vertically.

Physical Information Let y(t) be the displacement of the body at

time t. According to Newton’s second law

(1) Mass × Acceleration = my" = Force(牛頓 (nt) 為 Force單位 , 1 kg = 9.8nt)

where “Force” is the resultant of all the forces acting on the body.

Downward forces: positive. Upward forces: negative.

Physical InformationLet y = 0 in Fig. 32 (b). According to

Hooke’s law, the force F0 that cause the stretch s0 in the spring is,

(2) F0 = –ks0

k (> 0) is called the spring constant. The minus sign indicates that F0 points

upward. Stiff springs have large k.

Physical InformationF0 balances the weight W = mg of the body

(where g = 980 cm/sec2). Hence,

F0 + W = –ks0 + mg = 0.

These forces will not affect the motion. Spring and body are again at rest. This is called the static equilibrium of the system (Fig. 32b).

Physical InformationFrom the position y = 0 we pull the body

downward. This further stretches the spring by some amount y > 0.

By Hooke’s law this causes an (additional) upward force F1 in the spring,

F1 = –ky.

F1 is a restoring force which has the tendency to restore the system (y = 0).

Undamped System: ODE and SolutionEvery system has damping—otherwise it

would keep moving forever. If the effect of damping is negligible, then

F1 is the only force in (1) causing the motion. Hence (1) gives the model:

(3) my" + ky = 0. We obtain as a general solution

(4) y(t) = A cos ω0 t+B sin ω0 t,

The corresponding motion is called a harmonic oscillation.

Harmonic OscillationSince the trigonometric functions in (4)

have the period 2π/ω0.ω0/2π cycles per second ( called the frequency

of the oscillation) Another name for cycles/sec is hertz (Hz).

The sum in (4) can be combined into a phase-shifted cosine with amplitude and phase angle δ = arctan (B/A),

(4*) y(t) = C cos (ω0t – δ).

Fig. 33. Harmonic oscillations

•To find A:

y(0) = A

•To find B:

y’(0) = Bω0

B = y’(0) / ω0

Example 1 Undamped Motion.

If an iron ball of weight W = 98 nt stretches a spring 1.09 m, how many cycles per minute will this mass–spring system execute? What will its motion be if we pull down the weight an additional 16 cm (about 6 in.) and let it start with zero initial velocity?

Solution:

Fig. 34. Harmonic oscillation in Example 1

Damped System: ODE and SolutionsWe now add a damping force F2 = –cy'

where c > 0 is called damping constant Physically this can be done by connecting the

body to a dashpot (闊口盤 ). Assume this new force is proportional to the velocity y' = dy/dt.

To our model my" = –ky, so that we have my" = –ky – cy' , or

(5) my" + cy' + ky = 0.

Fig. 35. Damped system

continued

F1

F2

my’’

Damped System: ODE and SolutionsThe ODE (5) is homogeneous linear and

has constant coefficients. The characteristic equation is (divide (5) by m)

The roots of the equation is,

(6)

Discussion of the Three ClassIt is now most interesting that depending on the

amount of damping (much, medium, or little) there will be three types of motion corresponding to the three Cases I, II, II in Sec. 2.2:

Case I. OverdampingIf the damping constant c is so large that

c2 > 4mk, then λ1 and λ2 are distinct real roots. In this case the corresponding general solution of (5) is

(7) y(t) = c1e-(α-β)t + c2e-(α+β)t.

Damping takes out energy so quickly that the body does not oscillate. For t > 0 both exponents in (7) are negative

because α > 0, β > 0, and β2 = α2 – k/m < α2.

Case II. Critical DampingCritical damping is the border case

between nonoscillatory motions (Case I) and oscillations (Case III).

It occurs when c2 = 4mk, so that β = 0, λ1 = λ2 = –α. Then the corresponding general solution of (5) is

(8) y(t) = (c1 + c2t)e-αt.

Because e-αt is never zero, c1 + c2t can have at most one positive zero.

Case III. UnderdampingThe most interesting case which occurs when c2 < 4mk (c is small). Then in (6) is no longer real but pure imaginary, say,

(9) β = iω*

The roots of the characteristic equation are now complex conjugate,

λ1 = –α+ iω*, λ2 = –α– iω*

Damped OscillationsThe corresponding general solution is (10) y(t) = e-αt(A cos ω*t + B sin ω*t) = Ce-αt cos (ω*t – δ)

where C2 = A2 + B2 and tan δ = B/A, as in (4*).

This represents damped oscillations.Their curve lies between the dashed curves y =

Ce-αt and y = –Ce-αt, touching them when ω*t –δ is an integer multiple of π because these are the points at which cos (ω*t –δ) equals 1 or –1.

Damped OscillationsThe frequency is ω*/(2π) Hz (hertz,

cycles/sec). From (9) we see that the smaller c (> 0) is,

the larger is ω* and the more rapid the oscillations become.

If c approaches 0, then ω* approaches ω0=(k/m)1/2, giving the harmonic oscillation.

Fig. 38. Damped oscillation in Case III [see (10)]

Example 2 The Three Cases of Damped MotionHow does the motion in Example 1 change if

we change the damping constant c to one of the following three values, with y(0) = 0.16 and y'(0) = 0 as before?

(I) c = 100 kg/sec, (II) c = 60 kg/sec, (III) c = 10 kg/sec.

Fig. 39. The three solutions in Example 2