Physics 5300, Theoretical Mechanics Spring 2015
Assignment 1
Given: Tue, Jan 13,
Due Tue Jan 20
The problems numbers below are from Classical Mechanics, John R. Taylor, University
Science Books (2005).
Problem 1
Solution:
Taylor 5.5
We start with the form
I:
x(t) = C1 eiωt + C2 e−iωt
(1)
We have
eiωt = cos(ωt) + i sin(ωt)
(2)
e−iωt = cos(ωt) − i sin(ωt)
(3)
Thus we get
x(t) = C1 [cos(ωt) + i sin(ωt)] + C2 [cos(ωt) − i sin(ωt)]
= (C1 + C2 ) cos(ωt) + i(C1 − C2 ) sin(ωt)
≡ B1 cos(ωt) + B2 sin(ωt)
(4)
Thus we have
II :
B1 = (C1 + C2 ),
B2 = i(C1 − C2 )
(5)
Now consider the expression
III :
x(t) = A cos(ωt − δ)
(6)
This gives
x(t) = A cos(ωt) cos δ + A sin(ωt) sin δ
(7)
Comparing to II, we have
A cos δ = B1 ,
A sin δ = B2
(8)
Thus we have
A2 (cos2 δ + sin2 δ) = A2 = B12 + B22
(9)
q
A = B12 + B22
(10)
which gives
1
and
tan δ =
B2
,
B1
δ = tan−1
B2
B1
(11)
Now consider the expression
IV :
x(t) = Re[Ceiωt ]
(12)
We write
C = |C|eiφ
(13)
x(t) = Re[|C|ei(ωt+φ) ] = |C|Re[ei(ωt+φ) ] = |C| cos(ωt + φ)
(14)
Then we have
Comparing to III, we have
|C| = A,
Now start with
φ = −δ
(15)
1
x(t) = Re[Ceiωt ] = [Ceiωt + C ∗ e−iωt ]
2
(16)
1
1
C1 = C, C2 = C ∗
2
2
(17)
x(t) = B1 cos(ωt) + B2 sin(ωt)
(18)
ẋ(t) = −B1 ω sin(ωt) + B2 ω cos(ωt)
(19)
x(0) = x0 = B1 ,
(20)
This is of the form I with
Problem 2
Taylor 5.7
Solution:
(a) We have
Thus
ẋ(0) = v0 = B2 ω
Thus
B1 = x0 ,
B2 =
v0
ω
(21)
(b) We have
r
ω=
k
=
m
r
50
= 10
.5
B1 = x0 = 3
B2 =
v0
50
=
=5
ω
10
2
(22)
(23)
(24)
(c) We have
x(t) = B1 cos(ωt) + B2 sin(ωt) = 3 cos(ωt) + 5 sin(ωt)
(25)
To get x(t) = 0 we need
tan(ωt) = −
3 cos(ωt) + 5 sin(ωt) = 0,
3
5
(26)
We have
3
= −.54, −.54 + π = 2.6, . . .
5
Thus the earliest positive time when x(t) = 0 is
− tan−1
t=
2.6
2.6
=
= .26 s
ω
10
(27)
(28)
Similarly, we have
ẋ(t) = −B1 ω sin(ωt) + B2 ω cos(ωt) = −30 sin(ωt) + 50 cos(ωt)
(29)
We get ẋ(t) = 0 when
tan(ωt) =
5
3
(30)
This gives
ωt = 1.03
and therefore
t=
Problem 3
Solution:
1.03
= .103
ω
(31)
(32)
Taylor 5.22
(a) The equation for oscillations is
ẍ + 2ω0 ẋ + ω02 x = 0
(33)
x(t) = C1 e−ω0 t + C2 te−ω0 t
(34)
x(0) = 0
(35)
C1 e−ω0 t + C2 te−ω0 t C1 = C1 = 0
(36)
ẋ = −ω0 C1 e−ω0 t − C2 ω0 te−ω0 t + C2 e−ω0 t
(37)
The solution is
We have
which gives
The velocity is
3
Setting this to v0 at t = 0 we get
C2 = v 0
(38)
x(t) = v0 te−ω0 t
(39)
x(0) = C1 = x0
(40)
Thus we have
(b) Now we have
ẋ(0) = −ω0 C1 + C2 = 0,
C2 = ω0 C1
(41)
Thus
At t =
2π
ω0 ,
x(t) = x0 e−ω0 t + ω0 x0 te−ω0 t
(42)
x = x0 e−2π + 2πx0 e−2π = x0 (1 + 2π)e−2π
(43)
we have
Problem 4
Solution:
Taylor 5.33
We have
x(t) = A cos(ωt − δ) + e−βt [B1 cos(ω1 t) + B2 sin(ω1 t)]
(44)
Thus
ẋ(t) = −Aω sin(ωt−δ)−βe−βt [B1 cos(ω1 t)+B2 sin(ω1 t)]+e−βt [−B1 ω1 sin(ω1 t)+B2 ω1 cos(ω1 t)]
(45)
Thus we have
x0 = A cos(δ) + B1
(46)
v0 = −Aω sin(δ) − βB1 + B2 ω1
(47)
B1 = x0 − A cos δ
(48)
Thus
B2 =
Problem 5
Solution:
1
[v0 + Aω sin δ + β(x0 − A cos δ)]
ω1
(49)
f02
(ω02 − ω 2 )2 + 4β 2 ω 2
(50)
Taylor 5.41
We have
A2 =
4
The peak of A2 occurs at ω = ω0 , which gives
A2 =
f02
f02
≈
4β 2 ω 2
4β 2 ω02
(51)
Thus A2 has half this maximum value when
(ω02
f02
f02
f02
≈
=
2
2
− ω 2 )2 + 4β 2 ω 2
(ω0 − ω 2 )2 + 4β 2 ω0
8β 2 ω02
(52)
This gives
(ω02 − ω 2 )2 + 4β 2 ω02 = 8β 2 ω02
(53)
(ω02 − ω 2 )2 = 4β 2 ω02
(54)
ω02
2
− ω = ±2βω0
(55)
We can write this as
(ω + ω0 )(ω − ω0 ) ≈ 2ω0 (ω − ω0 ) = ±2βω0
(56)
ω − ω0 = ±β
(57)
which gives
Thus ω = ω0 ± β, and the full width at half-maximum is
Problem 6
Solution:
Thus
(ω0 + β) − (ω0 − β) = 2β
(58)
1
cos A cos B = [cos(A + B) + cos(A − B)]
2
(59)
1
cos(nωt) cos(mωt) = [cos((n + m)ωt) + cos((n − m)ωt)]
2
(60)
Taylor 5.47
(i) We have
We get
Z
τ
2
Z
dt cos(nωt) cos(mωt) =
− τ2
τ
2
− τ2
1
dt [cos((n + m)ωt) + cos((n − m)ωt)]
2
If n = 0, m = 0, then we get
Z τ
Z τ
2
2
1
dt [cos((n + m)ωt) + cos((n − m)ωt)] =
dt = τ
2
−τ
−τ
2
2
5
(61)
(62)
If n = m 6= 0, then we have
Z
Z τ
2
dt cos(nωt) cos(mωt) =
− τ2
τ
2
− τ2
1
dt [cos((2n)ωt) + 1]
2
τ
1
1
1
[
sin((2n)ωt)]|−2 τ + τ
2
2 (2nω)
2
= τ
=
(63)
If n 6= m, then n − m 6= 0, and n + m 6= 0. (The latter statement is true because n, m
are positive numbers.) Thus we get
Z τ
Z τ
2
2
1
dt cos(nωt) cos(mωt) =
dt [cos((n + m)ωt) + cos((n − m)ωt)]
τ
τ
2
−
−
2
2
τ
1
1
1
=
[
sin((n + m)ωt) +
sin((n − m)ωt)]|−2 τ
2
2 (m + n)ω
(m − n)ω
= 0
(64)
where in the last step we have used the fact that
τ
ω =π
2
(65)
(ii) We have
Thus
1
sin A sin B = [− cos(A + B) + cos(A − B)]
2
(66)
1
sin(nωt) sin(mωt) = [− cos((n + m)ωt) + cos((n − m)ωt)]
2
(67)
We get
Z
τ
2
Z
dt sin(nωt) sin(mωt) =
− τ2
τ
2
− τ2
1
dt [− cos((n + m)ωt) + cos((n − m)ωt)]
2
(68)
If either n = 0 or m = 0, then the integrand vanishes. Thus we must have n > 0, m > 0.
If n 6= m then we get
Z τ
Z τ
2
2
1
dt sin(nωt) sin(mωt) =
dt [− cos((n + m)ωt) + cos((n − m)ωt)]
τ
τ
2
−
−
2
2
τ
1
1
1
=
[−
sin((n + m)ωt) +
sin((n − m)ωt)]|−2 τ
2
2 (m + n)ω
(m − n)ω
= 0
(69)
6
where in the last step we have used the fact that
ω
τ
=π
2
(70)
If n = m, then we get
Z
τ
2
Z
dt sin(nωt) sin(nωt) =
− τ2
τ
2
− τ2
=
=
1
dt [− cos(2nωt) + 1]
2
τ
1
1
1
[−
sin((2n)ωt)]|−2 τ + τ
2
2 (2n)ω
2
τ
2
(71)
(iii) We have
Thus
1
cos A sin B = [sin(A + B) − sin(A − B)]
2
(72)
1
cos(nωt) sin(mωt) = [sin((n + m)ωt) − sin((n − m)ωt)]
2
(73)
We get
Z
τ
2
Z
dt cos(nωt) sin(mωt) =
− τ2
τ
2
− τ2
1
dt [sin((n + m)ωt) − sin((n − m)ωt)]
2
(74)
If either m = 0, then the integrand vanishes. Thus we must have m > 0. If n 6= m then
we get
Z
τ
2
Z
dt cos(nωt) sin(mωt) =
− τ2
τ
2
− τ2
1
dt [sin((n + m)ωt) − sin((n − m)ωt)]
2
τ
1
1
1
[−
cos((n + m)ωt) +
cos((n − m)ωt)]|−2 τ
2
2 (m + n)ω
(m − n)ω
= 0
=
(75)
where in the last step we have used the fact that for any number p
τ
τ
cos(pω ) = cos(pω(− ))
2
2
7
(76)
If m = n, then we have
Z
τ
2
Z
dt cos(nωt) sin(nωt) =
− τ2
τ
2
− τ2
1
dt [sin((2n)ωt) − 0]
2
τ
1
1
[−
cos((2n)ωt)]|−2 τ
2
2 (2n)ω
= 0
=
(77)
8