PHY 3221: Mechanics I
Fall Term 2010
Midterm Exam 2, October 27, 2009
• This is a closed book exam lasting 50 minutes.
• Since calculators are not allowed on this
√ test, if the problem asks for a numerical answer,
answering 2 + 2 is as good as 4, and 2 is as good as 1.4142....
• There are four problems worth a total of 20 pts. All four problems appear on the second
page of this test. Begin each problem on a fresh sheet of paper. Use only one side of the
paper. Avoid microscopic handwriting.
• Put your name, the problem number and the page number in the upper right-hand corner
of each sheet.
• To receive partial credit you must explain what you are doing. Carefully labelled figures
are important! Randomly scrawled equations are not helpful.
• Draw a box around important results (or at least results which you think might be important).
• Good luck!
1
Problem 1. [4 pts] In Einstein’s theory of general relativity the Schwarzschild radius Rs is
the radius at which a spherical body (such as a star or a planet) would become a black hole.
For example, if you take a star of mass M and squeeze all of its matter within a sphere of radius
Rs (or smaller), the gravitational pull on the surface becomes so strong that even light cannot
escape from it, and the star becomes a black hole.
Knowing nothing about general relativity, use dimensional analysis to derive a formula for
the Schwarzschild radius Rs in terms of the mass of the star M, Newton’s gravitational constant
G and the speed of light c.
Solution. First figure out the units of the relevant quantities:
[Rs ] = L,
[M] = M,
[G] = L3 M −1 T −2 ,
[c] = LT −1 .
Then write
Rs ∼ M α Gβ cγ
from where
[Rs ] = [M]α [G]β [c]γ
L = M α L3β M −β T −2β Lγ T −γ
giving three equations for α, β and γ:
1 = 3β + γ,
0 = α − β,
0 = −2β − γ.
Solving thse, we find
α = β = 1,
γ = −2.
The formula is therefore
GM
.
c2
Interestingly, this is only a factor of 2 off from the true answer.
Rs =
Problem 2. [6 pts.] A block of mass m is suspended vertically from a fixed support by a
spring of force constant k.
(a) [1 pt.] Find the extension ∆L ≡ L − L0 of the spring from its relaxed length L0 . You
can ignore the mass of the spring.
(b) [5 pts.] The block is now subjected to a vertical driving force F0 cos ωt. Given that the
spring will yield when its extension exceeds a certain value ∆Lmax , find the range of angular
frequencies which can safely be applied. You can ignore air resistance.
Solution. (a) The force of gravity is balanced by the force of the spring, therefore
mg = k∆L
from where
∆L =
2
mg
.
k
(b) The amplitude of a driven oscillator is given by (from formula sheet)
A
q
(ω02
− ω 2 )2 + 4ω 2β 2
q
k
, and, since there is no air-resistance, β = 0. Therefore the amplitude
Here A = Fm0 , ω0 = m
of oscillation about the equilibrium position from part (a) is given by
D(w) =
F0
− ω2|
m|ω02
and the maximum extension of the spring is now (don’t forget that the spring is already extended
a certain distance ∆L due to gravity - see part (a))
∆L + D(w)
which should not exceed ∆Lmax :
∆L + D(w) < ∆Lmax
F0
< ∆Lmax − ∆L
− ω 2|
m|ω02
|ω02 − ω 2 | >
F0
m(∆Lmax − ∆L)
1) Let ω < ω0 , then |ω02 − ω 2 | = ω02 − ω 2 and
ω02 − ω 2 >
F0
m(∆Lmax − ∆L)
and
ω 2 < ω02 −
ω < ω0
F0
m(∆Lmax − ∆L)
s
F0
=
1−
2
mω0 (∆Lmax − ∆L)
s
s
k
F0
1−
m
k(∆Lmax − ∆L)
2) Now let ω > ω0 , then |ω02 − ω 2 | = −(ω02 − ω 2 ) and
ω 2 − ω02 >
F0
m(∆Lmax − ∆L)
and solving for ω,
ω > ω0
s
F0
=
1+
2
mω0 (∆Lmax − ∆L)
3
s
s
F0
k
1+
m
k(∆Lmax − ∆L)
Therefore the allowed range of frequencies is
ω<
s
s
k
F0
1−
m
k(∆Lmax − ∆L)
or
s
ω>
s
k
F0
1+
m
k(∆Lmax − ∆L)
with ∆L given by the answer from part (a).
Problem 3. [5 pts.] A pendulum whose period in vacuum is T0 is now submerged in a
resistive medium. Its amplitude on each swing is observed to be half that on the previous
swing. What is the new period?
Solution. The amplitude of a damped oscillator decreases as e−βt , thus after one period
the damping factor is
1
e−βT =
2
where T is the new period that we want to find. On the other hand, the angular frequency of
a damped oscillator is
2π q 2
= ω0 − β 2
ω1 =
T
Here ω0 is known from the period in vacuum:
ω0 =
2π
T0
Therefore we have two equations
e−βT =
2π
=
T
1
2
s
2π
T0
2
− β2
for two unknowns: T and β. Solving for T , we find
T = T0
v
u
u
t
ln 2
1+
2π
!2
Problem 4. [5 pts.] An oscillator satisfies the equation
ẍ + 10ẋ + 16x = 0
(a) [1 pt.] Is this an underdamped, a critically damped, or an overdamped oscillator?
Explain your reasoning.
(b) [4 pts.] At time t = 0 the particle is projected from the point x = 1 towards the origin
with speed u, i.e. x(0) = 1 and ẋ(0) = −u. Find x(t).
4
Solution. (a) First identify the numerical coefficients in the oscillator equation
ω02 = 16
2β = 10,
from where
β = 5 > ω0 = 4,
which means that this is an overdamped oscillator.
(b) From the formula sheet
i
h
x(t) = e−βt A1 eω2 t + A2 e−ω2 t ,
where β = 5 from part (a) and
ω2 =
q
β 2 − ω02 =
√
52 − 42 =
√
25 − 16 =
√
9 = 3.
In order to determine the two constants A1 and A2 , use the initial conditions:
x(0) = 1 = A1 + A2
Then find the velocity
i
h
h
ẋ(t) = −βe−βt A1 eω2 t + A2 e−ω2 t + e−βt ω2 A1 eω2 t − ω2 A2 e−ω2 t
from where
i
ẋ(0) = −u = −β(A1 + A2 ) + ω2 (A1 − A2 ) = −β + ω2 (A1 − A2 )
Thus we get two equations for two unknowns:
A1 + A2 = 1
β−u
A1 − A2 =
ω2
from where
1 β−u
8−u
=
+
2
2ω2
6
u−2
1 β−u
=
−
=
2
2ω2
6
A1 =
A2
and
x(t) = e
−5t
8 − u 3t u − 2 −3t
e +
e
6
6
5
Formula sheet
A·(B × C) = B·(C × A) = C·(A × B) ≡ ABC
A×(B × C) = (A · C)B − (A · B)C
(A × B) · (C × D) = A · [B × (C × D)]
= A · [(B · D)C − (B · C)D]
= (A · C)(B · D) − (A · D)(B · C)
(A × B) × (C × D) = [(A × B) · D] C − [(A × B) · C] D
= (ABD)C − (ABC)D = (ACD)B − (BCD)A
v = ṙ er + r θ̇ eθ + r sin θφ̇ eφ
a =
+ 2ṙ φ̇ sin θ + 2r θ̇φ̇ cos θ + r θ̈ sin θ eφ
v = ṙ er + r φ̇ eφ + ż ez
a =
r̈ − r θ̇2 − r φ̇2 sin2 θ er + 2ṙ θ̇ + r θ̈ − r φ̇2 sin θ cos θ eθ
r̈ − r φ̇2 er + r φ̈ + 2ṙ φ̇ eφ + z̈ ez
X
k
εijk εlmk = δil δjm − δim δjl
X
εijk εljk = 2 δil
j,k
X
εijk εijk = 6
i,j,k
Time averages over one period T :
1 Z t+T
1
hsin ωti =
dt sin2 ωt =
T t
2
2
1
hcos ωti =
T
2
Z
t+T
t
6
dt cos2 ωt =
1
2
Simple harmonic oscillator:
mẍ + kx = 0
x(t) = A sin(ω0 t − δ)
x(t) = A cos(ω0 t − φ)
2π
ω0 = 2πν0 =
=
τ0
s
k
m
Damped oscillator:
b
ẍ + 2β ẋ + ω02 x = 0, 2β =
m
√ 2 2 √ 2 2
x(t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
Underdamped motion
x(t) = Ae−βt cos(ω1 t − δ),
ω1 =
q
ω02 − β 2
Critically damped motion
x(t) = (A + Bt)e−βt
Overdamped motion
i
h
x(t) = e−βt A1 eω2 t + A2 e−ω2 t ,
Driven oscillator
ω2 =
q
β 2 − ω02
F0
ẍ + 2β ẋ + ω02 x = A cos ωt, A =
m
√ 2 2
√ 2 2 xc (t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
A
cos(ωt − δ)
xp (t) = q
(ω02 − ω 2 )2 + 4ω 2β 2
2ωβ
2
ω0 − ω 2
δ = tan
−1
!
q
ω02 − 2β 2
ωR
Q=
2β
ωR =
RLC circuit
VL = L
dI
dt
VR = RI
7
VC =
q
C