3-14.
For the case of overdamped oscillations, x(t) and x ( t) are expressed by
x( t) = e− β t ⎡⎣ A 1eω 2t + A 2e−ω 2t⎤⎦
(
) (
(1)
)
x ( t) e− β t ⎡⎣ −β A 1eω 2t + + A 2e−ω 2t + A 1ω 2eω 2t − A 2ω 2e−ω 2t ⎤⎦
(2)
where ω 2 = β 2 − ω 02 . Hyperbolic functions are defined as
ey + e− y
cosh y =
,
2
ey − e− y
sinh y =
2
(3)
or,
ey = cosh y + sinh y ⎤
⎥
−y
e = cosh y − sinh y ⎥⎦
(4)
Using (4) to rewrite (1) and (2), we have
x( t) = ( cosh β t− sinh β t) ⎡⎣( A 1 + A 2 ) cosh ω 2t+ ( A 1 − A 2 ) sinh ω 2t⎤⎦
and
x ( t) = ( cosh β t− sinh β t) ⎡⎣( A 1ω 2 − A 1β ) ( cosh ω 2t+ sinh ω 2t)
− ( A 2β + A 2ω 2 ) ( cosh ω 2t− sinh ω 2t) ⎤⎦
(5)
3-24. As requested, we use Equations (3.40), (3.57), and (3.60) with the given values to
evaluate the complementary and particular solutions to the driven oscillator. The amplitude of
the complementary function is constant as we vary ω, but the amplitude of the particular
solution becomes larger as ω goes through the resonance near 0.96 rad ⋅ s−1 , and decreases as ω
is increased further. The plot closest to resonance here has ω ω 1 = 1.1, which shows the least
distortion due to transients. These figures are shown in figure (a). In figure (b), the ω ω 1 = 6
plot from figure (a) is reproduced along with a new plot with A p = 20 m ⋅ s−2 .
ω/ω1 = 1/9
2
1
1
0
0
0
–1
ω/ω1 = 1.1
ω/ω1 = 1/3
–2
–1
0
10
20
30
0
10
ω/ω1 = 3
20
0
30
10
ω/ω1 = 6
0.5
0.5
0
0
–0.5
–0.5
20
xc
xp
x
Legend:
–1
0
10
20
30
–1
0
10
20
t (s)
30
t (s)
30
t (s)
(a)
0.5
0.5
0
0
–0.5
–0.5
Ap = 1
–1
0
5
10
15
20
25
Ap = 20
30
(b)
–1
0
5
10
15
20
25
30
The equation describing the car’s motion is
3-40.
m
d2y
= − k( y − a sin ω t)
dt2
where y is the vertical displacement of the car from its equilibrium position on a flat road, a is
the amplitude of sine-curve road, and
k = elastic coefficient =
ω=
2π v0
λ
dm × g 100 × 9.8
=
= 98000 N /m
dy
0.01
= 174 rad/s with v0 and λ being the car’s speed and wavelength of sine-curve road.
The solution of the motion equation can be cast in the form
y( t) = B cos(ω 0t+ β ) +
aω 02
sin ω t with
ω 02 − ω 2
ω0 =
k
= 9.9 rad/s
m
We see that the oscillation with angular frequency ω has amplitude
A=
aω 02
= −0.16 m m
ω 02 − ω 2
The minus sign just implies that the spring is compressed.
3-43.
a)
Potential energy is the elastic energy:
U (r)=
1
k(r− a)2 ,
2
where m is moving in a central force field. Then the effective potential is (see for example,
Chapter 2 and Equation (8.14)):
U eff(r)= U (r)+
l2
1
l2
2
=
−
+
k
(
r
a
)
2m r2 2
2m r2
where l= m vr = m ω r2 is the angular momentum of m and is a conserved quantity in this
problem. The solid line below is U eff(r); at low values of r, the dashed line represents
1
l2
. At large values of r,
k(r− a)2 , and the solid line is dominated by
2m r2
2
1
U eff(r)≅ U (r)= k(r− a)2 .
2
Potential energy
U (r)=
U (r ) =
l2
2mr 2
1
k ( r − a )2
2
U eff (r )
r
b)
In equilibrium circular motion of radius r0 , we have
k( r0 − a) = m ω 02r0 ⇒ ω 0 =
c)
k( r0 − a)
m r0
For given (and fixed) angular momentum l, V(r) is minimal at r0, because V ′(r)r= r = 0 , so
0
we make a Taylor expansion of V(r) about r0 ;
3m ω 02 (r− r0 )2 K (r− r0 )2
1
V (r)= V (r0 )+ (r− r0 )V ′(r0 )+ (r− r0 )2V ′′(r0 )+ ...≈
=
2
2
2
where K = 3m ω 02 , so the frequency of oscillation is
ω=
K
= 3ω 0 =
m
3k(r0 − a)
m r0