Download 2 <t< 0 sin(πt/2)

Quiz # 2
Answer
ALL Questions
1. Find the Fourier series for the periodic extension of
f (t) =
½
0,
−2 < t < 0
sin(πt/2), 0 < t < 2
2. Find an expression for the central force that will provide radial quadratic spiral
orbits given by
r = bθ2 .
Here, r and θ are polar coordinates and b is a constant.
3. A particle P of unit mass is moving under a central force to the origin O, show that
the equation of its path must satisfy
d2 u
F (u)
+u= 2 2,
2
dθ
hu
where r = 1/u and θ are polar coordinates, F (u) is the strength of the force, and h
is the angular momentum.
Find the law of force to the pole under which a particle can describe the curve
r3 = a3 {1 + cos(3θ)} ,
where a is a positive constant.
If F (u) = µu2 + λu3 , where µ and λ are positive constants, and P is projected with
speed V at right angles to OP from a point where OP = r0 and θ = 0, find the
equation of the orbit on the assumption that λ < V 2 r02 < 2µr0 + λ.
An “apse” is defined as a point on an orbit at a maximum or minimum distance from
the center of force. Show that at an apse, du/dθ = 0, and that between consecutive
apses, the radius vector turns through an angle π/(1 − λV −2 r0−2 )1/2 .
4. (a) A particle of unit mass is moving under a central force µ/r2 in an elliptic orbit
of semi-major axisqa and eccentricity e. Show that its angular momentum about the
center of force is µa(1 − e2 ).
(b) The period of Neptune is 164.8 years. Show that it is about thirty times as far
from the sun as the earth is.
5. A body of mass m is projected with speed u in a medium that exerts a resistance
force of magnitude (i ) mk|v|, or (ii) mK|v|2 , where k and K are positive constants
and v is the velocity of the body. Gravity can be ignored. Determine the subsequent
motion in each case, i.e., calculate the position coordinate x as a function of time.
Verify that the motion is bounded in case (i), but not in case (ii).
2
Solutions
1. The Fourier series is given by
∞
X
1
f (t) = a0 +
(an cos(nt) + bn sin(nt)) ,
2
n=1
where
µ
¶
∙
µ
1Z 2
1Z 2
πt
1
πt
a0 =
dt f (t) =
dt sin
= − cos
2 0
2 0
2
π
2
¶¸2
=
0
2
π
µ
¶
µ
¶
µ
¶
µ
¶
µ
¶
µ
¶
1Z 2
nπt
1Z 2
nπt
πt
an =
dt f (t) cos
=
dt cos
sin
2 0
2
2 0
2
2
1Z 2
1Z 2
nπt
nπt
πt
bn =
dt f (t) sin
=
dt sin
sin
2 0
2
2 0
2
2
But,
a1 =
1Z 2
dt sin(πt) = 0
4 0
and for n > 1, we have
an
(
Ã
!
Ã
1Z 2
(n + 1)πt
(n − 1)πt
=
dt sin
− sin
4 0
2
2
"
Ã
!
Ã
!)
1
1
1
(n + 1)πt
(n − 1)πt
=
cos
−
cos
2π n + 1
2
n−1
2
2 1
, for n even
= − 2
πn −1
= 0 , otherwise .
!#0
2
(1)
We also have for n > 1
bn
µ
¶
µ
¶
nπt
πt
1Z 2
=
dt sin
sin
2 0
2
2
(
Ã
!
Ã
!)
1Z 2
(n − 1)πt
(n + 1)πt
=
dt cos
− cos
4 0
2
2
= 0
and b1 = 1/2.
3
(2)
2. Since u = 1/bθ2 , we have
d2 u
6
= θ−4 = 6bu2
2
dθ
b
This gives
d2 u
F (u)
+
u
=
u(1
+
6bu)
=
.
dθ2
mh2 u2
Thus, the magnitude of the attractive force is
Ã
6b
mh2
F = 3 1+ 2
r
r
!
.
3. (b) From
u=
1
1
a (1 + cos 3θ)1/3
we have
du
1
sin 3θ
=
dθ
a (1 + cos 3θ)4/3
and
d2 u
1 3 cos 3θ − cos2 3θ + 4
=
dθ2
a
(1 + cos 3θ)7/3
So, we have
d2 u
+ u = 5a3 u4
2
dθ
But, this last equation is equal to F/mh2 u2 . Therefore,
F = (5a3 mh2 ) u6
(b) Here
4
F = µu2 + λu3 ,
h = V r0 .
So,
d2 u
µ + λu
+
u
=
dθ2
V 2 r02
which we rewrite as
Ã
!
d2 u
λ
µ
+ 1− 2 2 u= 2 2
2
dθ
V r0
V r0
The coefficient of u on the left-hand side
γ2 ≡ 1 −
λ
V 2 r02
is positive since we have λ < V 2 r02 . So, we have as solution
µ
u=
V
2 r2 γ 2
0
+ A cos γθ ,
where A is a constant. The total energy is conserved. The initial kinetic energy is
1
T = V2 .
2
The potential energy is
U =−
λ
µ
− 2 .
r
2r
So that the total (initial) energy is
´
µ
λ
1 ³
1
E = V2− −
= 2 V 2 r02 − 2µr0 − λ
2
r0 2r0
2r0
which is negative since V 2 r02 < 2µr0 + λ. The coefficient A” is obtained by substituting the equation of the orbit into the energy equation
Ã
du
dθ
!2
+ u2 =
This is the same as
5
2
(E − U) .
V 2 r02
Ã
du
dθ
!2
+ γ 2 u2 −
2µ
2E
u= 2 2 .
2
2
V r0
V r0
After some algebra, we obtain
1
A=
γ
s
µ2
2E
+ 2 2 .
4 2
4
V r0 γ
V r0
4. (a) From our lecture notes, we have for the total energy
E=−
µm
2a
and the eccentricity is given in terms of the energy by
e=
s
1+
2Eh2
mµ2
and E = −µm/2a follows since l = a(1 − e2 ) and l = h2 /µ. Substituting the value
of E from the first equation into the second, we obtain
h=
q
µa(1 − e2 ) .
(b) According to Kepler’s third law, we have
τN2 = C a3N
for Neptune, and
τE2 = C a3E
for earth. where C is a constant, τ is the periodic time and a is the semi-major axis.
Therefore, from these two equations, we obtain
µ
aN
τN
=
aE
τE
¶2/3
6
= (164.8)2/3 ≈ 30.6
5. (i) Suppose that the body starts at the origin and moves along the positive x axis.
Then, the equation of motion is
m
dv
= −mkv ,
dt
where v = ẋ. Separating the variables, rewrite this equation as
Z
Z
dv
= −k dt .
v
This leads to
ln v = −kt + C ,
where C is a constant. But, v = u at t = 0. Therefore, C = ln u. So, the velocity at
time t is given by
v = u e−kt
Integrating this equation and using the initial conditions, we obtain
x=
u ³ −kt ´
1e
k
This motion is bounded e−kt → 0 as t → ∞bso that x → u/k.
(ii) The equation of motion is
m
dv
= −mKv 2
dt
assuming the body starts at the origin and moves along the positive x axis. Separate
the variables and integrate to obtain
Z
Z
dv
=
−K
dt ,
v2
which gives
1
− = −Kt + C
v
The constant C is obtained from the initial conditions and we have
v=
7
u
Ktu + 1
Integrating this equation with respect to time and applying the initial conditions,
we obtain
x=
1
ln(Kut + 1) ,
K
which tends to infinity as t → ∞ and so the motion is unbounded.
The acceleration in polar coordinates is
a = ar r̂ + aθ θ̂
1d 2
(r ω)θ̂
r dt
= (r̈ − ω 2 r)r̂ + 2ṙω θ̂ .
= (r̈ − ω 2 r)r̂ +
(3)
Applying
d
r̂ = ω θ̂ ,
dt
d
θ̂ = −ωr̂ ,
dt
to obtain
d
d
a =
(ar r̂ + aθ θ̂)
dt
dt
= (ȧr r̂ + ȧθ θ̂) + (ar ω θ̂ − aθ ωr̂) .
Setting the coefficient of θ̂ in Eq. (4) equal to zero, we obtain
ȧθ + ar ω = 2r̈ω + (r̈ − ω 2 r)ω = 0 .
This equation gives
1
r̈ = ω 2 r .
3
Multiply this equation by ṙ and integrate over time. We have
ṙ2 =
´
1³ 2
r − a2 .
3
Rewrite as
dr
1
√
= √ dt .
2
2
r −a
3
Integrating, we obtain
8
(4)
cosh
−1
µ ¶
x
1
=√ t
a
3
or
Ã
t
x = a cosh √
3
9
!
.