Math 142-2, Homework 1
Solutions
Problem 5.8
(a) Show that x = c1 cos ωt + c2 sin ωt is the general solution of mx′′ = −kx. What is the value of ω?
x = c1 cos ωt + c2 sin ωt
x′ = −c1 ω sin ωt + c2 ω cos ωt
x′′ = −c1 ω 2 sin ωt − c2 ω 2 sin ωt
mx′′ = −c1 mω 2 sin ωt − c2 mω 2 sin ωt
mx′′ + kx = (c1 sin ωt + c2 sin ωt)(k − mω 2 )
Then, mx′′ + kx = 0 provided mω 2 = k. This occurs when
r
k
.
ω=
m
(b) Show that an equivalent expression for the general solution is x = B cos(ωt + θ0 ). How do B and
θ0 depend on c1 and c2 ?
x = B cos(ωt + θ0 )
= B cos θ0 cos ωt − B sin θ0 sin ωt
c1 = B cos θ0
c2 = −B sin θ0
q
B = c21 + c22
c2
θ0 = − tan−1
c1
Problem 6.2
Suppose a quantity y having dimensions of time is only a function of k and m. Do not use constants
with units.
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(a) Give an example of a possible dependence of y on k and m.
F = −kx
[kg ms
−2
] = k[m]
k = [kg s−2 ]
m = [kg]
m
= [s2 ]
k
r
m
= [s]
k
r
m
y=
k
(b) Can you describe the most general dependence that y can have on k and m?
The most general dependence, using only dimensionless constants
r
m
y=c
,
k
where c is a dimensionless constant. Of course, if we are allowed to use constants with dimensions (g, G, R,
¯ , ǫ0 , me , etc.), the possibilities are endless. If such constants are included, the dependence could potentially
h
include funny terms like sin(m/me ), where me is the mass of an electron.
Problem 7.2
A weight (of unknown mass) is placed on a vertical spring (of unknown spring constant) compressing it
by 2.5 cm. What is the natural frequency of oscillation of this spring-mass system?
Let m be the mass and k be the spring constant. The force on the spring due to gravity is mg, and the
2
force due to compressing by ∆x is k∆x. Thus,
k∆x = mg
g
k
=
m r
∆x
k
ω=
m
r
g
=
∆x
r
9.8 ms−2
=
0.025 m
≈ 19.8 s−1
ω
f=
2π
≈ 3.15 s−1
Problem 9.3
Suppose that a mass m were attached between two walls a distance d apart (see figures in book).
(a) Briefly explain why it is not necessary for d = ℓ1 + ℓ2 .
The springs need not be at their relaxed length when the system is in equilibrium. Many systems are
always kept under tension or compression.
(b) What position of the mass would be called the equilibrium position of the mass? If both springs
are identical, where should the equilibrium position be? Show that your formula is in agreement.
Let x be the position of the mass relative to the left wall. The left spring is displaced from its natural
length by ∆x1 = x − ℓ1 . The right spring is displaced from its natural length by ∆x2 = (d − x) − ℓ2 . Since
the system is in equilibrium, the forces balance, and
k1 ∆x1 = k2 ∆x2
k1 (x − ℓ1 ) = k2 (d − x − ℓ2 )
(k1 + k2 )x = k1 ℓ1 + k2 (d − ℓ2 )
x=
k1 ℓ1 + k2 (d − ℓ2 )
k1 + k2
If both springs are identical, the mass should be in the middle at x =
ℓ1 = ℓ2 = ℓ then,
k1 ℓ1 + k2 (d − ℓ2 )
k1 + k2
kℓ + k(d − ℓ)
=
k+k
d
=
2
x=
3
d
2.
Indeed, if k1 = k2 = k and
(c) Show that the mass executes simple harmonic motion about its equilibrium position.
The net force on the mass is f = −k1 ∆x1 + k2 ∆x2 . Let x0 be the equilibrium position, so that k1 ℓ1 +
k2 (d − ℓ2 ) = (k1 + k2 )x0 . Then,
mx′′ = −k1 ∆x1 + k2 ∆x2
mx′′ = −k1 (x − ℓ1 ) + k2 ((d − x) − ℓ2 )
mx′′ = k1 ℓ1 + k2 (d − ℓ2 ) − k1 x − k2 x
mx′′ = (k1 + k2 )x0 − k1 x − k2 x
m(x − x0 )′′ = −(k1 + k2 )(x − x0 )
mz ′′ = −ks z
where z = x − x0 is the displacement from equilibrium and ks = k1 + k2 is the effective spring constant. The
resulting equation is the equation for one spring, which undergoes simple harmonic motion.
(d) What is the period of the oscillation?
r
ks
m
r
k1 + k2
=
m
ω=
(e) How does the period of oscillation depend on d?
The period of oscillation does not depend on d.
Problem 9.4
Suppose that a mass m were attached to two springs in parallel (see figure in book).
(a) What position of the mass would be called the equilibrium position of the mass?
Let x be the distance of the mass relative from the right wall. The springs are displaced from their
natural lengths by ∆x1 = x − ℓ1 and ∆x2 = x − ℓ2 . Since the system is in equilibrium, the forces balance,
and
0 = k1 ∆x1 + k2 ∆x2
= k1 (x − ℓ1 ) + k2 (x − ℓ2 )
= (k1 + k2 )x − (k1 ℓ1 + k2 ℓ2 )
k1 ℓ1 + k2 ℓ 2
x=
k1 + k2
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(b) Show that the mass executes simple harmonic motion about its equilibrium position.
The net force on the mass is f = −k1 ∆x1 −k2 ∆x2 . Let x0 be the equilibrium position, so that k1 ℓ1 +k2 ℓ2 =
(k1 + k2 )x0 . Then,
mx′′ = −k1 ∆x1 − k2 ∆x2
mx′′ = −k1 (x − ℓ1 ) − k2 (x − ℓ2 )
mx′′ = k1 ℓ1 + k2 ℓ2 − k1 x − k2 x
mx′′ = (k1 + k2 )x0 − k1 x − k2 x
m(x − x0 )′′ = −(k1 + k2 )(x − x0 )
mz ′′ = −ks z
where z = x − x0 is the displacement from equilibrium and ks = k1 + k2 is the effective spring constant. The
resulting equation is the equation for one spring, which undergoes simple harmonic motion.
(c) What is the period of the oscillation?
r
ks
m
r
k1 + k2
=
m
ω=
(d) If the two springs were to be replaced by one spring, what would be the unstretched length and
spring constant of the new spring such that the motion would be equivalent?
The new spring would have ks = k1 + k2 and ℓs = x0 =
k1 ℓ1 +k2 ℓ2
k1 +k2 .
Problem 9.5
Suppose that a mass m were attached to two springs in series (see figure in book).
(a) What position of the mass would be called the equilibrium position of the mass?
The system is at rest when the net force on the mass is zero, which in this case means both springs are
at their rest length. x0 = ℓ1 + ℓ2 .
(b) Show that the mass executes simple harmonic motion about its equilibrium position.
Let ∆x = ∆x1 + ∆x2 be the total displacement of the spring system. If we consider there to be a
point of very small mass ǫ connecting the springs at y, ǫy ′′ = k1 ∆x1 − k2 ∆x2 . Since ǫ is arbitrarily small,
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k1 ∆x1 = k2 ∆x2 .
∆x = ∆x1 + ∆x2
k1 ∆x1 = k2 ∆x2
k2
∆x1 = ∆x2
k1
k2
∆x = ∆x2 + ∆x2
k1
k1 + k2
∆x =
∆x2
k1
k1
∆x
∆x2 =
k1 + k2
k2
∆x1 =
∆x
k1 + k2
m∆x′′ = −k1 ∆x1
k1 k2
=−
∆x
k1 + k2
= −ks ∆x
This again looks like a single spring and thus undergoes simple harmonic motion.
(c) What is the period of the oscillation?
ω=
=
r
s
ks
m
k1 k2
m(k1 + k2 )
(d) If the two springs were to be replaced by one spring, what would be the unstretched length and
spring constant of the new spring such that the motion would be equivalent?
The new spring would have ks =
k1 k2
k1 +k2
and ℓs = ℓ1 + ℓ2 .
Problem 9.6
Consider two masses each of mass m attached between two walls a distance d apart (see figure in book).
Assume that all three springs have the same spring constant and unstretched length.
(a) Suppose that the left mass is a distance x from the left wall and the right mass a distance y from
the right wall. What position of each mass would be called the equilibrium position of the system of
masses?
The forces of the three springs are k(x−ℓ), k(d−x−y −ℓ), and k(y −ℓ). Since the masses are not moving,
these must all be equal. From the first and last being equal, x = y. From k(x − ℓ) = k(d − x − y − ℓ) =
k(d − 2x − ℓ), x = y = d3 .
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(b) Show that the distance between the masses oscillates. What is the period of the that oscillation?
Let s = d − x − y be the distance between the masses. Then,
mx′′ = −k(x − ℓ) + k(d − x − y − ℓ)
my ′′ = −k(y − ℓ) + k(d − x − y − ℓ)
m(x + y)′′ = −k((x + y) − 2ℓ) + 2k(d − x − y − ℓ)
−ms′′ = −k((d − s) − 2ℓ) + 2k(s − ℓ)
−ms′′ = k(3s − d)
m(s − d/3)′′ = −3k(s − d/3)
mˆ
s′′ = −3kˆ
s
This looks like a single spring and thus undergoes simple harmonic motion. The period of oscillation is
r
3k
ω=
m
(c) Show that x − y executes simple harmonic motion with a period of oscillation different from b.
mx′′ = −k(x − ℓ) + k(d − x − y − ℓ)
my ′′ = −k(y − ℓ) + k(d − x − y − ℓ)
m(x − y)′′ = −k(x − y)
The oscillation frequency is
ω=
r
k
m
(d) If the distance between the two masses remained constant, describe the motion that could take
place both qualitatively and quantitatively.
If the distance between the masses does not change, it is as though the masses were connected by a
rigid bar. The setup looks exactly like the two-spring, one-mass setup of 9.3, where kˆ1 = kˆ2 = k, m
ˆ = m,
dˆ = d − s, and ℓˆ1 = ℓˆ2 = ℓ. As in that case, the rigid bar undergoes harmonic motion.
(e) If x = y, describe the motion that could take place both qualitatively and quantitatively.
In this case, the center of the middle spring will be fixed, so the system behaves as though there was a
wall here. The setup looks like 9.3, with kˆ1 = kˆ2 = k, m
ˆ = m, dˆ = d2 , ℓˆ1 = ℓ, and ℓˆ2 = 2ℓ . The second mass
mirrors the first. Both undergo harmonic motion.
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Additional Problem 1
Maxwell’s equations are
ρ
ε0
∇·B=0
∂B
∇×E=−
∂t
∇·E=
∇ × B = µ 0 J + ε0
∂E
∂t
A particle with charge q and velocity v in this field will experience a Lorentz force
f = q(E + (v × B))
Additionally, the total charge Q in some volume of space Ω is
Z
Q=
ρ dV
Ω
If q = [C] and Q = [C] both have units of charge (the unit C is the Coulomb), deduce the units of the
quantities in the table below. You may assume units for f and v.
var
E
B
ρ
J
ε0
µ0
Q
q
v
f
meaning
electric field
magnetic field
charge density
current density
permittivity of free space
permeability of free space
total charge
particle charge
particle velocity
force on particle
The units for f , q, Q, and v are given or known. From the Lorentz force we can deduce units for E and
B.
f = q(E + (v × B))
[kg ms
−2
] = [C](E + [ms−1 ]B)
E = [kg ms−2 C −1 ]
B = [kg s−1 C −1 ]
We get units for ρ from the total charge
Q=
Z
ρ dV
Ω
[C] = [m3 ]ρ
ρ = [m−3 C]
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Next for ε0
ρ
ε0
−3
[m
C]
[m−1 ][kg ms−2 C −1 ] =
ε0
[m−3 C]
ε0 =
−1
[m ][kg ms−2 C −1 ]
∇·E=
= [kg −1 m−3 s2 C 2 ]
And now the rest are obtained from
∇ × B = µ0
∂E
J + ε0
∂t
[m−1 ][kg s−1 C −1 ] = µ0 J + [kg −1 m−3 s2 C 2 ][s−1 ][kg ms−2 C −1 ]
[kg m−1 s−1 C −1 ] = µ0 J + [m−2 s−1 C]
J = [m−2 s−1 C]
µ0 =
[kg m−1 s−1 C −1 ]
[m−2 s−1 C]
= [kg mC −2 ]
These units are summarized in the table below.
var
E
B
ρ
J
ε0
µ0
Q
q
v
f
units
[kg ms−2 C −1 ]
[kg s−1 C −1 ]
[m−3 C]
[m−2 s−1 C]
[kg −1 m−3 s2 C 2 ]
[kg mC −2 ]
[C]
[C]
[ms−1 ]
[kg ms−2 ]
Additional Problem 2
A particle of mass m is connected to a circle of radius r by a very large number of identical and equally
spaced springs. The particle is free to move around in three dimensions, but the circle is fixed.
9
m
(a) The springs have zero rest length but are free to rotate at both ends. Formulate a force law for one
of the springs. That is, express f in terms of the mass position x, assuming the other end of the spring
is fixed at z. (If your force does not work with the calculations below, choose a new one.)
f = −k(x − z)
(b) A force is conservative if it comes from an energy. That is, there exists a scalar function φ(x) such
that f = −∇φ. Determine φ for your force.
φ=
k
kx − zk2
2
(c) What is the total potential energy if there are n equally spaced identical springs, as n → ∞?
Remember that the mass moves in three dimensions.
φ=
Z
=
Z
=
2π
0
2π
0
 2

r cos θ k
x −  r sin θ 
dθ
2
0
k
((x − r cos θ)2 + (y − r sin θ)2 + z 2 ) dθ
2
k 2
((r + x2 + y 2 + z 2 )θ − 2xr sin θ + 2kyr cos θ)
2
= kπ(r2 + x2 + y 2 + z 2 )
= kπ(r2 + kxk2 )
10
2π
0
Since a constant shift on potential energy does not matter, we can use φ = kπkxk2 .
(d) What is the corresponding force?
f = −∇φ = −2πkx
(e) Solve for the path of the mass, if initially at location x0 with velocity v0 .
Each dimension is an independent spring with the same frequency
r
2πk
ω=
m
so the solution is
x = x0 cos ωt +
11
1
v0 sin ωt
ω