1
Homework 3 Solutions
1.84 CHAPTER 3
3-5. Differentiating the equation of motion for a simple harmonic oscillator,
0sinx A tω= (1)
we obtain
0 0cosx A t tω ω= (2)
But from (1)
0sinx
tA
ω = (3)
Therefore,
( )20cos 1t xω = − A (4)
and substitution into (2) yields
2 2
0
xt
A xω=
− (5)
Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval x at position x is given by
2 2 2 2
0 2
t x x
A x Aω π= =
− − x (6)
–A1–A2–A3 A3A2A1
∆t ⁄τ
x
This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for t at x = ±A. This occurs because at these points x = 0, and (2) is not valid.
3-6.
x1 x2
k
x
m1 m2
Suppose the coordinates of m and are and x and the length of the spring at equilibrium is . Then the equations of motion for m and are
1 2m 1x 2
1A 2m
( )1 1 1 2m x k x x= − − +�� A (1)
( )2 2 2 1m x k x x= − − +�� A (2)
2
90 CHAPTER 3
Substitution of (12) into (6) yields
( ) ( )
( ) ( )
( )
0 0
0 0
0
1 12
2
cos cos
i t i t i t i t
i t t i t ti t i t
Fe e e e
k
Fe e e e
k
Ft t t
k
ω ω ω ω
ω ωω ω
ω ω
− −+
− − − −
= − + −
= − + −
= − − (13)
Thus,
( )0 0cos cos ; F
t t t t tk
ω ω− = − − 0x x (14)
3-10. The amplitude of a damped oscillator is expressed by
( ) ( )1costx t Ae tβ ω δ−= + (1)
Since the amplitude decreases to 1 after n periods, we have e
1
21nT n
πβ β
ω= = (2)
Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped oscillations), 2 2
1 02ω ω β= − , we have
2
2 2 210 1 1 2 2
11
2 4n nω
ω ω ωπ π
= + = + (3)
Therefore,
1 2
12 2
0
11
4 nωω π
−
= + (4)
so that
12 2
2
11
8 nωω π
−
3-11. The total energy of a damped oscillator is
( ) ( ) ( )21 12 2
E t mx t kx t= +� 2 (1)
where
( ) ( )1costx t Ae tβ ω δ−= − (2)
( ) ( ) ( )1 1 1cos sintAe t tβx t β ω δ ω ω δ−= − − − −� (3)
2.
3
88 CHAPTER 3
2 sin2
4 cos2
ds dv a
dt dt
da
dt
φ φ
φ
= =
= − (8)
from which
2
24 cos2
dv a
dtφ
= −� (9)
Letting cos2
zφ
be the new variable, and substituting (7) and (9) into (1), we have
4maz mgz− =�� (10)
or,
04g
z za
+ =�� (11)
which is the standard equation for simple harmonic motion,
20 0z zω+ =�� (12)
If we identify
0g
ω =A
(13)
where we have used the fact that . 4a=A
Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673).
3-9. The equation of motion for 00 t t is
( ) ( )0mx k x x F kx F kx= − − + = − + +�� 0 (1)
while for , the equation is 0t t
( )0mx k x x kx kx= − − = − +�� 0 (2)
It is convenient to define
0x x= −
which transforms (1) and (2) into
m k F= − +�� ; 00 t t (3)
m k= −�� ; (4) 0t t
3.
OSCILLATIONS 89
The homogeneous solutions for both (3) and (4) are of familiar form ( ) i t i tt Ae Beω ω−= + , where
k mω = . A particular solution for (3) is F k= . Then the general solutions for (3) and (4) are
i t i tFAe Be
kω ω−
− = + + ; 00 t t (5)
i t i tCe Deω ω−+ = + ; t (6) 0t
To determine the constants, we use the initial conditions: ( ) 00x t x= = and x(t = 0) = 0. Thus,
( ) ( )0 0t t− − 0= = = =� (7)
The conditions give two equations for A and B:
( )
0
0
FA B
k
i A Bω
= + +
= −
(8)
Then
2F
A Bk
= = −
and, from (5), we have
( )0 1 cosF
x x tk
ω− = − = − ; 00 t t (9)
Since for any physical motion, x and must be continuous, the values of x� ( )0t t− = and
( )0t t− =� are the initial conditions for ( )t+ which are needed to determine C and D:
( ) ( )
( )
0 0
0 0
0 0
0 0
1 cos
sin
i t i t
i t i t
Ft t t Ce De
k
Ft t t i Ce De
k
ω ω
ω ω
ω
ω ω ω
−+
−+
= = − = +
= = = −� (10)
The equations in (10) can be rewritten as:
( )0 0
0 0
0
0
1 cos
sin
i t i t
i t i t
FCe De t
k
iFCe De t
k
ω ω
ω ω
ω
ω
−
−
+ = −
−− =
(11)
Then, by adding and subtracting one from the other, we obtain
( )
( )
0 0
0 0
12
12
i t i t
i t i t
FC e e
k
FD e e
k
ω ω
ω ω
−
−
= −
= −
(12)
4
OSCILLATIONS 89
The homogeneous solutions for both (3) and (4) are of familiar form ( ) i t i tt Ae Beω ω−= + , where
k mω = . A particular solution for (3) is F k= . Then the general solutions for (3) and (4) are
i t i tFAe Be
kω ω−
− = + + ; 00 t t (5)
i t i tCe Deω ω−+ = + ; t (6) 0t
To determine the constants, we use the initial conditions: ( ) 00x t x= = and x(t = 0) = 0. Thus,
( ) ( )0 0t t− − 0= = = =� (7)
The conditions give two equations for A and B:
( )
0
0
FA B
k
i A Bω
= + +
= −
(8)
Then
2F
A Bk
= = −
and, from (5), we have
( )0 1 cosF
x x tk
ω− = − = − ; 00 t t (9)
Since for any physical motion, x and must be continuous, the values of x� ( )0t t− = and
( )0t t− =� are the initial conditions for ( )t+ which are needed to determine C and D:
( ) ( )
( )
0 0
0 0
0 0
0 0
1 cos
sin
i t i t
i t i t
Ft t t Ce De
k
Ft t t i Ce De
k
ω ω
ω ω
ω
ω ω ω
−+
−+
= = − = +
= = = −� (10)
The equations in (10) can be rewritten as:
( )0 0
0 0
0
0
1 cos
sin
i t i t
i t i t
FCe De t
k
iFCe De t
k
ω ω
ω ω
ω
ω
−
−
+ = −
−− =
(11)
Then, by adding and subtracting one from the other, we obtain
( )
( )
0 0
0 0
12
12
i t i t
i t i t
FC e e
k
FD e e
k
ω ω
ω ω
−
−
= −
= −
(12)
90 CHAPTER 3
Substitution of (12) into (6) yields
( ) ( )
( ) ( )
( )
0 0
0 0
0
1 12
2
cos cos
i t i t i t i t
i t t i t ti t i t
Fe e e e
k
Fe e e e
k
Ft t t
k
ω ω ω ω
ω ωω ω
ω ω
− −+
− − − −
= − + −
= − + −
= − − (13)
Thus,
( )0 0cos cos ; F
t t t t tk
ω ω− = − − 0x x (14)
3-10. The amplitude of a damped oscillator is expressed by
( ) ( )1costx t Ae tβ ω δ−= + (1)
Since the amplitude decreases to 1 after n periods, we have e
1
21nT n
πβ β
ω= = (2)
Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped oscillations), 2 2
1 02ω ω β= − , we have
2
2 2 210 1 1 2 2
11
2 4n nω
ω ω ωπ π
= + = + (3)
Therefore,
1 2
12 2
0
11
4 nωω π
−
= + (4)
so that
12 2
2
11
8 nωω π
−
3-11. The total energy of a damped oscillator is
( ) ( ) ( )21 12 2
E t mx t kx t= +� 2 (1)
where
( ) ( )1costx t Ae tβ ω δ−= − (2)
( ) ( ) ( )1 1 1cos sintAe t tβx t β ω δ ω ω δ−= − − − −� (3)
5
NEWTONIAN MECHANICS—SINGLE PARTICLE 71
The equilibrium point (where 0ddU θ = ) that we wish to look at is clearly θ = 0. At that point,
we have ( )2 2 2d U d mg R bθ = − , which is stable for 2R b and unstable for R b . We can
use the results of Problem 2-46 to obtain stability for the case
2
2R b= , where we will find that the first non-trivial result is in fourth order and is negative. We therefore have an equilibrium at θ = 0 which is stable for 2R b and unstable for 2R b .
2-43. 3 2F kx kx α= − +
( )4
22
1 12 4
xU x F dx kx k
α= − = −∫
To sketch U(x), we note that for small x, U(x) behaves like the parabola 212
kx . For large x, the
behavior is determined by 4
2
14
xkα
−
U(x)
E0
E1
E2
E3 = 0
E4
x1 x2 x3
x4 x5 x
( )212
E mv U x= +
For E , the motion is unbounded; the particle may be anywhere. 0E=
For E (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may remain at rest where it is, but if perturbed slightly, it will move away from the equilibrium.
1E=
What is the value of ? We find the x values by setting 1E 0dUdx
= .
3 20 kx kx α= −
x = 0, ± α are the equilibrium points
( ) 2 21
1 1 12 4 4
U E k k k 2α α α± = = − = α
For E , the particle is either bounded and oscillates between 2E= 2x− and ; or the particle comes in from ± to ± and returns to ± .
2x
3x
4.
72 CHAPTER 2
For E , the particle is either at the stable equilibrium point x = 0, or beyond . 3 0= 4x x= ±
For E , the particle comes in from ± to 4 5x± and returns.
2-44.
m1
T
m1g
m2
T T
m2g
θ
From the figure, the forces acting on the masses give the equations of motion
11 1m m gx T= −�� (1)
22 2 2 cosm m g Tx θ= −�� (2)
where is related to by the relation 2x 1x
( )2
1 22 4
b xx
−d= − (3)
and ( )1cos 2d b xθ = − . At equilibrium, 1 2 0x x= =�� �� and T m1g= . This gives as the equilibrium values for the coordinates
110 2
1 2
4
4
m dx b
m m= −
− 2 (4)
220 2
1 24
m dx
m m=
− 2 (5)
We recognize that our expression is identical to Equation (2.105), and has the same requirement that
10x
2 1 2m m for the equilibrium to exist. When the system is in motion, the descriptive equations are obtained from the force laws:
( )2 1
112
( ) (4
m b xm gx
x−
2 )gx− =�� �� −
0
(6)
To examine stability, let us expand the coordinates about their equilibrium values and look at their behavior for small displacements. Let 1 1 1x x− and 2 2 2x x 0− . In the calculations, take terms in 1 and 2 , and their time derivatives, only up to first order. Equation (3) then becomes 2 1 2(m m 1)−� . When written in terms of these new coordinates, the equation of motion becomes
( )
( )
3 22 21 2
111 2 1 2
4
4
g m m
m m m m d
−= −
+�� (7)
6
NEWTONIAN MECHANICS—SINGLE PARTICLE 75
when 2=v c , we have 0.55 year10 3
ct = =
when v = 99% c, we have 99
6.67 years10 199
c= =t
2-51.
a) 2 02
0
( )mvdv dv b
m bv dt v tdt v m btv m
= − = − =+∫ ∫
Now let v(t) = v0/1000 , one finds 0
999138.7 hours
mt
v b= = .
t
v
b) 0
0
( ) lnt btv mm
x t vdtb m
+= =∫
We use the value of t found in question a) to find the corresponding distance
( ) ln(1000) 6.9 kmm
x tb
= =
2-52.
a) 2
02 2
4( ) 1
U xdU xF x
dx a a= − = − −
b)
x
U
When F = 0, there is equilibrium; further when U has a local minimum (i.e. 0dF dx ) it is stable, and when U has a local maximum (i.e. 0dF dx ) it is unstable.
5.
76 CHAPTER 2
So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0 is a stable equilibrium position.
c) Around the origin, 0 02 2
4 4U x UkF kx
a mω− − = =
ma
d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential,
2
0minmax 0 min
22
UmvU U v
m= = =
e) From energy conservation, we have
2 22 2
0 0min2 2
21
2 2U x Umv dx x
va dt m
+ = = = −mv
a
We note that, in the ideal case, because the initial velocity is the escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression,
022
20 00 0
2 2
8exp 1
ln ( )2 8 81 exp 1
xU
a tmam dx ma a x
tU U a xx U
ta ma
−+
= = =−
− +∫t x
t
x
2-53.
F is a conservative force when there exists a non-singular potential function U(x) satisfying F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations
yxFF
y x
∂∂=
∂ ∂
and so on.
a) In this case all relations above are satisfied, so F is indeed a conservative force.
2
1( , )2x
U bxayz bx c U ayzx cx f y z
x∂
= + + = − − − +∂
F = − (1)
where is a function of only y and z 1( , )f y z
2( , )yU
axz bz U ayzx byz f x zy
∂= + = − − +
∂F = − (2)