1. No category

PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
Reading: Taylor, Chapter 6, all and 7.1-7.3
Homework: Due in class March 23. All problems count equally towards your grade.
(Numbers refer to problems in Classical Mechanics, J. R. Talyor, 2005 Edition). This
counts as two problem sets.
(1) Consider an oscillator described by a coordinate x obeying the equation
dx
d2 x
+4
+ x = cos (t)
2
dt
dt
The general solution is the sum of a particular solution xp and the general solution of the associated homogeneous equation, xh . To construct the particular
solution we note that the oscillator frequency Ω = 1 and the drive frequency ω = 1.
The damping coefficient τ1 = 2. This implies the phase angle is π/2 while the
q
2
amplitude factor R = 1/ (ω 2 − Ω2 )2 + 4 ωτ 2 = 1/4 so
cos t − π2
xp =
4
For the homogeneous solution we note that we are in the overdamped case with
the two decay constants given as
r
√
1
1
1
2 =2± 3
= ±
−
Ω
τ±
τ
τ2
Thus the general solution is
π
2
sin t −
A+ − τt+
A− − τt−
v(t) = −
e
−
e
−
τ+
τ−
4
π
2
− τt
x(t) = A+ e
+
+ A− e
− τt
−
cos t −
+
4
implying
(a) Find x(t > 0) and v(t > 0) assuming that both x and v = dx/dt equal 0 at
t = 0.
x(t = 0) = 0 and v(t = 0) = 0 means
A+ + A− = 0
−
1
A+ A− 1
−
+ =0
τ+
τ−
4
2
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
The first equation implies A− = −A+ ; using this in the second gives
1
A+ = √
8 3
so
t
cos t − π2
1
−τ
− τt
e + −e − +
x(t) = √
4
8 3
(b) Find x(t > 0 and v(t > 0) assuming that at t = 0, x = 0 but v = 0.5
We still have A+ = −A− but now
A+ A− 1
1
−
+ =
−
τ+
τ−
4
2
so
1
A+ = − √
8 3
and so
t
cos t − π2
1
−τ
− τt
e + −e − +
x(t) = − √
4
8 3
(c) Sketch (on one graph) the phase plane portraits showing the evolution of
these two solutions for t > 0.
0.4
0.2
0.0
-0.2
-0.4
-0.4
-0.2
0.0
0.2
0.4
Figure 1. Phase plane portraits: Solid line: solution to (a); dashed line,
solution to (b). Notice how the solution converges to the circle that is the
particular solution.
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
3
(2) Consider a general harmonically driven oscillator obeying the equation
d2 x 2m dx
+
+ kx = Fd cos (Ωd t)
dt2
τ dt
We saw in class that the steady state solution (at long times, after the transients
have died away) is of the form
m
x(t) = RFd cos (Ωd t − φ)
with R and φ determined by m, τ and k.
Over one quarter period, x changes from −RFd to 0. Compare the energy put
in to the oscillator over this time by the driving force to the energy taken out of
the oscillator by friction.
The velocity is
v(t) = −RFd Ωd sin (Ωd t − φ)
A time at which x = −RF is t = (π + φ)/ΩD . One quarter period later t = t =
( 3π
2 + φ)/ΩD .
Energy put into oscillator is
Z
Ein =
dtF v = −RFD2 ΩD
Z
3π +φ
2
2ΩD
π+φ
ΩD
dt cos (ΩD t) sin (ΩD t − φ)
Using cosAsinB = (Sin(A + B) − Sin(A − B))/2 we get
Ein
RFD2 ΩD
=−
2
Z
3π +φ
2
ΩD
π+φ
Ωd
dt (sin (2ΩD t − φ) − sin (φ))
or
Ein =
RFD2
RFD2
(Cos (3π + φ) − Cos(2π + φ)) +
sinφ
4
4
or
RFD2
RFD2
Cos (φ) +
sinφ
2
4
Energy taken out of oscillator is
Ein = −
Z
Eout =
2m 2 2 2
dtF v =
R FD ΩD
τ
Z
3π +φ
2
2ΩD
π+φ
ΩD
dt sin2 (ΩD t − φ) =
mR2 FD2 ΩD
2τ
(3) 5.43 (variant): answer this question but instead of using the data in the book
assume that the car sinks by 2 cm if one 80kG passenger climbs in and that the
axle assembly has total mass 10kG.
Here we need to estimate the spring constant for the wheel assembly and then note
that the spring constant, plus the mass of the wheel assembly, implies an oscillation
frequency for the car.
4
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
(a) Car without passenger has mass M and is supported by the 2 axle assemblies.
Each axle assembly has 2 springs, each of spring constant K. Thus the extension
of the springs H is given by
4KH = gM
We add a person, mass 80kG and H → H + 0.02m. Thus approximating g =
10m/sec2
4K × 0.02m = g(80kG)
so
80 × 10
= 104 N/m2
4 × 0.02
(b) Because each axle has 2 wheels the total spring constant for the axle is 2K =
2 × 104 N/m2 . The mass of the axle assembly is 10kG so the oscillation frequency
ω0 is
r
r
2K
2 × 104
=
≈ 45/sec
ω0 =
maxle
10
K=
This implies that the period 2π/ω0 ≈ 0.13 sec. If the car is driven over a road with
bumps spaced 0.8m apart then then the velocity at which these bumps will resonate
with the suspension of the car is 0.8/0.13 ≈ 6m/sec
(4) 6.1 and 6.16 The infinitesimal path length is
p
dl = dx2 + dy 2 + dz 2
Expressing x, y, z in polar coordinates and taking the differential gives
dz = −Rsinθdθ
dx = Rcosθcosφdθ − Rsinθsinφdφ
dy = Rcosθsinφdθ + Rsinθcosφdφ
so
dx2 + dy 2 + dz 2 = R2
dθ2 + sin2 θdφ2 = R2
1 + sin2 θ
so
s
dl = R
1+
dφ
dθ
s
2
dθ
and
Z
θstop
L=R
1+
θstart
dφ(θ)
dθ
2
dθ
dφ
dθ
2 !
dθ2
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
5
Because the “Lagrangian” (argument of the integral) has no explicit dependence
on θ we may construct the conserved “energy” as
2
s
dφ
dθ
1
dφ(θ) 2
= −r
H=r
2 − 1 +
2
dθ
dφ(θ)
1 + dφ(θ)
1
+
dθ
dθ
This implies that dφ
dθ is constant, which is the equation for a great circle path, in
other words a path traced out by drawing the circle that is centered at the origin of
the sphere and connects the two points.
(5) 6.22
Noting that ds2 = dx2 + dy 2 we have
s
1−
dx =
dy
ds
2
ds
so that
Z
s
Z
A=
ydx =
1−
y
dy
ds
2
ds
Again the absence of explicit dependence on s gives us a conserved “energy” which
is
2
s
2
y dy
ds
dy
y
H = −r
= −r
2 − y 1 − ds
2
dy
1 − dy
1
−
ds
ds
in terms of dy/dx gives
y2
1− 2 =
H
dy
ds
2
whose solution is
y = Hsin(s)
which is the equation of a circle of radius H, related to length by l = 2πH.
(6) 6.24 We want to minimize
Z
T =
Z
nds = a
1
ds = a
r2
s
Z
dr
Thus the Euler-Lagrange equation is

d δL
d 

=
dr δ dφ
dr 
dr
dφ
dr

r2 dφ
dr
r
r2
1 + r2
1 + r2
dφ
dr


2  = 0
2
1
r2
6
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
implying
dφ
dr
r
2 = c
dφ
dr
dφ
dr
1 + r2
or
2
d −r
2
2
=1
with d = 1/c.
We can rearrange this equation to
dr
d2 − r2
writing r = dsinψ we conclude dψ = dφ so
dφ = √
r = dsinφ
Thus
d
d
sin2φ
y = rsinψ = dsin2 φ = (1 − cos2φ)
2
2
so
2
d
d2
2
x +
−y =
2
4
which is the equation of a circle centered at 0, d2 and touching the origin on one
side.
As light approaches the origin (moving along the circle, it is progressively slowed
down and its angle changes more.
x = rcosφ = dcosφsinφ =
(7) 7.8
In the general case
L=
1 2 k
1 2
x˙ 1 +
x˙ − (x1 − x2 − l)2
2m1
2m2 2 2
We have
1
x1 = X + x
2
1
x2 = X − x
2
Specializing to m1 = m2 = m we find
1
1 2 k
L = X˙ 12 +
x˙ − (x − l)2
m
4m
2
Thus because L does not depend on X the equations of motion are
¨ = 0
X
x
¨ = k (x − l)
PHY 3003 SPRING 2015: WEEKS OF MARCH 9 AND 16
7
so that X simply moves at a velocity determined by the initial conditions while the
relative coordinate executes simple harmonic motion around the value x = l.
(8) 7.14
The angular velocity is Rω = x˙ so
m
m
m 2 mR2 2
x˙ +
ω˙ − mgx = x˙ 2 + x˙ 2 − mgx
2
4
2
4
The Euler-Lagrange equation is thus
d m mx˙ + x˙ = −mg
dt
2
implying
2g
x
¨=−
3
L=
(9) 7.22 Here the velocity of the pendulum bob is the sum of the velocity relative to
the suspension point |v| = Lφ˙ plus the velocity due to the upward acceleration
vaccel = at; the position is −Lcosφ + at2 /2; thus
2
2
m ˙
at
Lφ + at − mg
− Lcosφ
L=
2
2
so the Euler-Lagrange equation is (note the factor of L which comes from the chain
rule)
d ˙
Lφ + at = −mgLsin(φ)
mL
dt
or
g+a
φ¨ = −
sinφ
L
For small oscillations we may replace sinφp→ φ, obtaining a harmonic oscillator
equation. The mass is 1 so the frequency is (g + a)/L