Transcript Document

Sheng-Fang Huang
2.4 Modeling: Free Oscillations
(Mass–Spring System)
If we pull the body down a certain distance and then release it, it
starts moving. We assume that it moves strictly vertically.
m
Physical Information

Let y(t) be the displacement of the body at time t.
According to Newton’s second law
Mass × Acceleration = my" = Force
(牛頓(nt)為Force單位, 1 kg = 9.8nt)
(1)
where “Force” is the resultant of all the forces acting
on the body.


Downward forces: positive.
Upward forces: negative.
Physical Information
 Let 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 Information
 F0 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 Information
 From 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 Solution
 Every 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 Oscillation
 Since 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 phaseshifted 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 Solutions
 We 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
F2
F1
my’’
continued
Damped System: ODE and Solutions
 The 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 Class
 It 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. Overdamping
 If 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
Damping
 Critical 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.interesting
Underdamping
 The most
case which occurs when c < 4mk
2
(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 Oscillations
 The 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 Oscillations
 The 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 Motion
 How 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