PH 125 / LeClair
Spring 2014
Problem Set 6 Solutions
1. An object of mass m is dropped from a height h above the surface of a planet of mass M and radius R.
Assume the planet has no atmosphere so that friction can be ignored.
(a) What is the speed of the mass just before it strikes the surface of the planet? Do not assume that h is
small compared with R.
√
(b) Show that the expression from (a) reduces to v = 2gh for h R.
(c) How long does it take for the object to fall to the surface for an arbitrary value of h? Use any means
necessary to evaluate the integral required. Bonus points (10%) for code submissions.
Solution: (a) Conservation of energy, and no more. The only tricky point is to remember that since we
must use general gravitation (since we may not assume h R), we must take into account the potential
energy at the planet’s surface too. For general gravitation, potential energy is by convention only zero for
an infinite separation of two objects.
GM m
R+h
1
GM m
Ef = Kf + Uf = mv 2 −
2
R
r
√
1
1
v = 2GM
−
R R+h
Ei = Ui = −
=⇒
(1)
(2)
(3)
(b) If h R, we may use the approximation (1 + h/R)n ≈ 1 + nh/R.
√
v=
r
2GM
1
1
−
=
R R+h
r
2GM
R
s
1
≈
1−
1 + h/R
r
2GM
R
s
r
h
GM
= 2 2 h
1− 1−
R
R
(4)
√
Since g = GM/R2 , this reduces to v ≈ 2gh.
(c) You may recall from exam 2 that we can find the time it takes to go from xi to xf from the total and
potential energy of a particle. Start by solving the conservation of energy equation for kinetic energy, and
subsequently dx/dt.
1
1
K = E − U (x) = mv 2 = m
2
2
r
dx
2
=
(E − U (x))
dt
m
dx
dt
2
Now integrate both sides of Eq. 6 with respect to x.
(5)
(6)
r
dx
2
=
(E − U (x))
dt
m
r
m
dt
1
p
=
dx
2 E − U (x)
Zxf
Zxf r
dt
m
1
p
dx =
dx
2 E − U (x)
xi
(7)
(8)
(9)
xi
Zxf r
tf − ti ≡ t =
xi
m
1
p
2 E − U (x)
(10)
For the last line, we basically said ti = 0 and tf = t. In the present case, the total energy E is simply the
starting potential energy of the particle at height h above the planet’s surface, and the potential energy at
any position x above the planet’s surface is
U (x) = −
GM m
R+x
(11)
The time is then found from Eq. 10 above, and the resulting integral is known (e.g., Wolfram Alpha).
Zxf r
t=
xi
m
1
p
=
2 E − U (x)
Zxf r
xi
m
1
q
2
GM m
−
R
r
=
GM m
R+x
1
2GM
Zxf r
R(R + x)
dx
x
(12)
xi
h
√
√ p
1
R3/2 ln
R + x + x + Rx(R + x)
2GM
0
r
√ !
√
p
1
R
+
h
+
h
√
=
R3/2 ln
+ Rh(R + h)
2GM
R
r
r
r ! r
h
R3/2
h
1 p 2
1+ +
Rh + R2 h
=
ln
+
2GM
R
R
2GM
r
=
(13)
(14)
(15)
This is rather ugly. Is it correct? We can check the result for h R, since we know in that limit that
p
√
t = 2h/g. First, we know 1 + x ≈ 1 + x/2 and ln (1 + x) ≈ x for x 1. This gives for the first term
r
ln
h
1+ +
R
r
h
R
!
h
≈ ln 1 +
+
2R
r
h
R
!
h
≈
+
2R
r
h
R
(16)
If h/R 1, then the first term is much smaller than the second, and we may neglect it. This reduces our
overall expression to
r
t≈
R3
2GM
r
h
+
R
r
1 p 2
Rh + R2 h =
2GM
r
R2 h
+
2GM
r
1 p 2
Rh + R2 h
2GM
(17)
In the second term, if h R, we may neglect Rh2 compared to R2 h. Thus,
r
t≈
R2 h
+
2GM
r
1 √ 2
R h=
2GM
r
2R2 h
GM
(18)
value
dly to
M as
disk is
disk is
mbly is
sition
of the
is
(a)
T ! 2#
√
m(k 1 $ k 2)
k 1k 2
√
k1 $ k2
Noting again that g = GM/R2 , we havem
(b)
s
t≈
T ! 2#
2h
g
(19)
In the limit of small heights above the planet’s surface, we again recover the usual result that assumes
k1
k2
constant acceleration.
m
2. A block of mass m is connected to two springs of force constants k1 and k2 as shown below. The block
moves on a frictionless table after it is displaced from equilibrium and released. Determine the period of
simple harmonic motion. (Hint:
(a) what is the total force on the block if it is displaced by an amount x?
k1
k2
m
Solution: Say we displace (b)
the block to the right by an amount x. Both springs will try to bring the block
Figure
back toward equilibrium - one P15.71
will pull, one will push, but both will act in the same direction. That means
the net force is
72. A lobsterman’s
solid
cylinder of radius r
Fnet = −k1 xbuoy
− k2 xis=a −
(k1 +wooden
k2 ) x = ma
(20)
and mass M. It is weighted at one end so that it floats upright
calm the
sea same
water,
'. Aforpassing
This isinexactly
as having
what wedensity
would find
a singleshark
spring, except the spring constant has become
tugs
on
the
slack
rope
mooring
the
buoy
to
a
lobster
trap,
k1 + k2 rather than just k. The solution must be
pulling the buoy down a distance x from its equilibrium
r
position and
2π releasingmit. Show that the buoy will execute
T =
= 2π
(21)
simple harmonic
motion
ω
k1 + k2 if the resistive effects of the
water are neglected, and determine the period of the
oscillations.
73. 3. A Consider
a bob on to
a light
stiff rod,
forming
a simple
mass m is connected
two springs
in series
as shown
below. What is the period of simple harmonic
pendulum
length
L ! 1.20
It is displaced
frommust
the be true of the displacement of each spring
motion if theof
mass
is displaced
fromm.
equilibrium.
Hint: what
vertical
by
an
angle
"
and
then
released.
Predict
the
max
if the total displacement is
∆x?
subsequent angular positions if "max is small or if it is large.
Proceed as follows: Set up and carry out a numerical
k
k
method to integrate the equation of motion for the simple
pendulum:
m
d 2"
g
! % sin "
2
dt
L
Solution: We have two springs in series. Do we just add the spring constants, like in the previous problem,
or is it more complicated? We should expect at least that like in the previous problem we could replace the
combination of two springs by a single effective spring with a different spring constant - physically the two
springs together should still just act like a single spring, the question is what should the spring constant be.
Best to start from scratch.
When the block is displaced, each spring will experience some displacement, and they need not be the same.
If you imagine one spring is much “stiffer” than the other, you would expect that the stiffer spring is extended
or contracted less than the other. If the springs have displacements x1 and x2 , the total displacement would
be x1 +x2 . What we expect is that the net force balance for both springs together should look like that of a
single spring, just with a different spring constant. A single spring with a spring constant keq displaced by
that amount would have a force equation
Feq = −keq (x1 + x2 )
(22)
We can also say something about the individual springs: by Newton’s third law, both must have the same
force, and the force on the block must also be the same. What we wish to find is the spring constant keq for
a single spring that also gives the same force, given a displacement of this equivalent spring of x1 +x2 . The
force on each spring is readily found given their displacements:
F1 = −k1 x1 = F2 = −k2 x2 = Feq
(23)
Solving for x2 , we find x2 = (k1 /k2 )x2 . Using this result in our expression for Feq ,
k2 + k1
k1
Feq = −keq x1 + x1 = −keq
x1
k2
k2
(24)
We are specifically interested in finding the keq for which Feq = F1 , so equating the two forces we have
−keq
k2 + k1
k2
x1 = −k1 x1
keq =
=⇒
k1 k2
k1 + k2
(25)
(26)
Thus, two springs in series behave the same as a single spring with spring constant keq = k1 k2 /(k1 + k2 ), or
equivalently 1/keq = 1/k1 + 1/k2 . The period is thus
T =
2π
= 2π
ω
r
m
keq
(27)
4. A horizontal plank of mass m and length L is pivoted at one end. The plank’s other end is supported by
a spring of force constant k. The moment of inertia of the plank about the pivot is I = 13 mL2 . The plank
is displaced by a small angle θ from horizontal equilibrium and released. Find the angular frequency ω of
simple harmonic motion. (Hint: consider the torques about the pivot point.)
the mass is 5.00 kg and the spring has a force constant of
100 N/m.
note that the mass of a segment of the
is dm " (m/!)dx. Find (a) the kinetic energy o
system when the block has a speed v and (b) the p
of oscillation.
Pivot
v
θ
dx
k
x
M
Figure P15.61
Figure
Solution: The presence of a pivot point – even labeled as such – immediately suggests
the useP15.66
of torque
to solve this problem. First, we need to find the compression of the spring at equilibrium, i.e., θ = 0. Since
62. Review problem.
A particle
mass
4.00
kg is attached
the plank has
non-zeroof
mass,
even
without
an angularto
displacement the spring must be compressed by some
67.
A ball of mass m is connected to two rubber ba
a spring with
a force
constant Once
of 100
It is oscillating
amount
at equilibrium.
weN/m.
have found
the equilibrium position, we can worry about the torques when
length L, each under tension T, as in Figure P15.6
on a horizontal
frictionless
surfaceθ with
an amplitude of
a small angular
displacement
is applied.
ball is displaced by a small distance y perpendicular
2.00 m. A 6.00-kg object is dropped vertically on top of the
length of the rubber bands. Assuming that the t
4.00-kg object
as it passes through
itsdefined
equilibrium
point.
Let counterclockwise
rotations be
as positive,
and let the equilibrium
of theshow
spring that
correspond
does notposition
change,
(a) the restoring
The two objects
stick
together.
(a)
By
how
much
does
the
to the tip of the plank being at vertical position xo relative to its unstretched
length.
The
sum
of
the
torques
is $ (2T/L)y and (b) the system exhibits simple har
amplitudeabout
of the
system
change as
a 0)
result
of the
thevibrating
pivot point
at equilibrium
(θ =
is given
by considering the weight of the plank acting about its
motion with an angular frequency # " √2T/mL .
collision?center
(b) of
Bymass
howand
much
does the
change?
the restoring
forceperiod
of the spring.
The plank may be treated as a point mass a distance
(c) By how
much does the energy change? (d) Account for
L/2 from the pivot point, while the spring force acts at a distance L. At equilibrium, the net torque must
the change in energy.
be zero.
63. A simple pendulum with alength
of 2.23 m and a mass of
X
L
6.74 kg is given
an
initial
speed
of
m/s at its equilibτ = −mg
+ kx2.06
oL = 0
2
rium position. Assume it undergoes simple harmonic
mg
motion, and determine
xo = its (a) period, (b) total energy, and
(c) maximum angular2k
displacement.
y
L
(28)
L
(29)
Figure P15.67
64. Review problem.
endthe
of equilibrium
a light spring
with we
force
Now thatOne
we have
position,
canconfind the torques when the plank makes an angle θ with
stant 100 the
N/m
is attached
vertical
wall.change
A lightitsstring
vertical.
Since to
thea plank
cannot
lengthis (we assume),
the
amount
that the
springM,
stretches
68.
When
a block
of mass
connected to the en
tied to the
other
end
of
the
horizontal
spring.
The
string
should correspond to the arc length that the tip of the plank moves
through,
Lθ,mifs "
the7.40
angleg is
relatively
spring
of mass
and
force constant k, is s
i
changes from
it passesbyover
a solid Lθ − xo from
small.horizontal
The springtois vertical
thereforeasdisplaced
an amount
its harmonic
unstretchedmotion,
length. This
gives usof its motion is
simple
the period
pulley of the
diameter
cm. The
to turn
on a to the plank, the spring force always acts perpenspring’s4.00
restoring
force.pulley
Since is
thefree
spring
is attached
fixed smooth
axle.
The
vertical
section
of
the
string
sup-the torque is easily found.
dicularly to the length of the plank at distance L, and
M ' (m s/3)
% " 2&
ports a 200-g object. The string does not slip at its contact
k
with the pulley. Find the frequency of oscillation of the
The plank’s weight still acts at a distance L/2 from the center of mass, but now at an angle 90 − θ relative
object if the mass of the pulley is (a) negligible, (b) 250 g,
A two-part experiment is conducted with the u
to the axis of the plank. The overall torque is then
and (c) 750 g.
blocks of various masses suspended vertically fro
√
X
L
1
τ = −mg
sin (90 − θ) − k (Lθ − xo ) L = − mgL cos θ − kL2 θ + kxo L
2
2
Since the angle θ is small, we may approximate cos θ ≈ 1, and we may also make use of our earlier expression
for xo . Finally, out of equilibrium the torques must give the moment of inertia times the angular acceleration.
i Small
enough such that we don’t have to worry about the spring bending to the left, for one.
1
1
1
τ = − mgL cos θ − kL2 θ + kxo L ≈ − mgL − kL2 θ + mgL = −kL2 θ = Iα
2
2
2
d2 θ
2
−kL θ = I 2
dt
X
(30)
(31)
Noting that I = 31 mL2 for a thin plank, we can put the last equation in the desired form for simple harmonic
motion in terms of known quantities:
3k
d2 θ
=− θ
dt2
rm
3k
ω=
m
(32)
(33)
Note that the length of the plank does not matter at all.
5. Assume the earth to be a solid sphere of uniform density. A hole is drilled through the earth, passing
through its center, and a ball is dropped into the hole. Neglect friction.
(a) Calculate the time for the ball to return to the release point. Hint: what sort of motion results?
(b) Compare the result of part a to the time required for the ball to complete a circular orbit of radius RE
about the earth
Solution: We consider a spherically-symmetric mass of uniform density, radius RE , and mass ME . If we
are at a radius r < RE from the center of the earth, only the mass contained within r < RE gives rise to a
net gravitational force – the gravitational interactions for bits of mass at r > RE all cancel each other. If
we assume the mass of material removed to make the hole is negligible, then the mass contained within a
radius r is simply
M (r) = ME
r
RE
3
The force acting on a body of mass m at a distance r from earth’s center is then
F (r) = −
GME (r)m
GME mr
=−
3
r2
RE
Noting that the gravitational acceleration at earth’s surface is
g=
GME
2
RE
we can write this compactly as
F (r) = −gm
r
d2 r
= ma = m 2
RE
dt
d2 r
g
=−
r
dt2
RE
We have established that acceleration is a negative constant times position, which means that we have simple
harmonic motion. Since the constant of proportionality between position and acceleration is g/RE , we will
have periodic motion with
s
T = 2π
RE
≈ 84.4 min
g
Clearly, the time for the ball return to its original position is then T . We can compare this time to that
required for an orbit around the earth at radius RE (which would be an orbit skimming the surface of the
earth, a bit silly . . . ) via Kepler’s law, and they are the same:
s
Torbit = 2π
3
RE
= 2π
GME
s
RE
g
See http://www.askamathematician.com/2012/08/q- if- you- could- drill- a- tunnel- through- the- whole- planet- and- then- jumped- down- this- tunnel- how- would- you- fall/ for an interesting discussion of this problem.
6. A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because
A = πr2 . Determine how the following properties depend on r: (a) period, (b) kinetic energy, (c) angular
momentum, and (d) speed.
Solution: Period is given by Kepler’s law:
T 2 ∝ r3
T ∝ r3/2
=⇒
(34)
For a circular orbit, we can show that the kinetic energy is exactly − 12 times the potential. Briefly, for a
circular orbit the gravitational force must provide the centripetal force to maintain the circular path. Thus,
GM m
mv 2
=
2
r
r
r GM m
r mv 2
=
2 r2
2 r
1
1
GM m
= − U = mv 2 = K
2r
2
2
(35)
(36)
(37)
√
Thus, we can see K ∝ 1/r. Since K ∝ v 2 , it also follows immediately that v ∝ 1/ r. Since angular
√
√
momentum is, in magnitude, L = mvr, it must be the case that L ∝ vr ∝ (1/ r)(r), or L ∝ r.
7. Here are some functions:
2
f1 (x, t) = Ae−b(x−vt)
2
f3 (x, t) = Ae−b(bx
+vt)
f2 (x, t) =
A
2
b (x − vt) + 1
f4 (x, t) = A sin (bx) cos (bvt)
3
(a) Which ones satisfy the wave equation? Justify your answer with explicit calculations.
(b) For those functions that satisfy the wave equation, write down the corresponding functions g(x, t)
representing a wave of the same shape traveling in the opposite direction.
Solution: This is just grinding through the math, and showing that
1 ∂2y
∂2y
= 2 2
2
∂x
v ∂t
It is seriously tedious. If you do this, you will find that f1 and f2 satisfy the wave equation, while f3 and f4
do not.
We found earlier that general solutions to the wave equation must have the form
h(x, t) = af (x + vt) + bg(x − vt)
where a and b are constants, and g and f represent forward- and backward-traveling wave solutions, respectively. Knowing this, the fact that only f1 and f2 are solutions is clear by inspection (though you were
unfortunately asked to provide explicit calculations). Both f1 and f2 have the functional form g(x − vt),
meaning they are ‘forward-traveling’ solutions. The corresponding ’backward-traveling’ solutions are found
by simply changing the arguments to x + vt from x − vt:
f10 (x, t) = Ae−b(x+vt)
2
f20 (x, t) =
A
2
b (x + vt) + 1
8. The equation for a driven damped oscillator is
q
d2 x
dx
+ 2γωo
+ ωo2 x = E(t)
dt2
dt
m
(38)
(a) Explain the significance of each term.
(b) Let E = Eo eiωt and x = xo ei(ωt−α) where Eo and xo are real quantities. Substitute into the above
expression and show that
xo = q
qEo /m
(ωo2
−
2
ω2 )
(39)
2
+ (2γωωo )
(c) Derive an expression for the phase lag α, and sketch it as a function of ω, indicating ωo on the sketch.
Solution: The significance of each term is probably more apparent if we re-arrange and multiply by mass:
m
d2 x
dx
= −mωo2 x − 2γmωo
+ qE(t)
dt2
dt
(40)
The term on the right is the net force on the oscillator. The first term on the left is the restoring force, the
second the viscous damping term, and the last the driving force of the oscillator.
First, we find the derivatives of x, noting i2 = −1:
dx
= iωxo ei(ωt−α)
dt
d2 x
= −ω 2 xo ei(ωt−α)
dt2
(41)
(42)
Substituting into the original equaiton,
q
Eo eiωt
m
q
Eo eiωt
m
q
Eo eiωt
m
qEo iα
e
m
= −ω 2 xo ei(ωt−α) + 2γωo iωxo ei(ωt−α) + ωo2 xo ei(ωt−α)
= ei(ωt−α) −ω 2 xo + 2iγωo ωxo + ωo2 xo
= eiωt e−iα −ω 2 xo + 2iγωo ωxo + ωo2 xo
(43)
= −ω 2 xo + 2iγωo ωxo + ωo2 xo
(46)
(44)
(45)
To proceed, we use the Euler identity
eiθ = cos θ + i sin θ
(47)
Giving
qEo
(cos α + i sin α) = −ω 2 xo + 2iγωo ωxo + ωo2 xo
m
(48)
We now have two separate equations if we equate the purely real and purely imaginary parts:
qEo
cos α = ωo2 xo − ω 2 xo
m
qEo
sin α = 2γωωo xo
m
(49)
(50)
We can square both equations and add them together:
2
q 2 Eo2
2
cos2 α + sin2 α = ωo2 − ω 2 x2o + (2γωωo ) x2o
2
m
q 2 Eo2
1
x2o =
m2 (ωo2 − ω 2 )2 x2o + (2γωωo )2
qEo
1
q
xo =
m
2
2
(ωo2 − ω 2 ) x2o + (2γωωo )
(51)
(52)
(53)
This is the desired amplitude of vibration. Going back to the preceding two equations, we can divide the
second equation by the first to find the phase angle:
tan α =
2γωωo
ωo2 − ω 2
(54)
9. (a) A diatomic molecule has only one mode of vibration, and we may treat it as a pair of masses connected by a spring (figure (a) below). Find the vibrational frequency, assuming that the masses of A and B
are different. Call them ma and mb , and let the spring have constant k.
(b) A diatomic molecule adsorbed on a solid surface (figure (b) below) has more possible modes of vibration.
Presuming the two springs and masses to be equivalent this time, find their frequencies.
Figure 1: From http: // prb. aps. org/ abstract/ PRB/ v19/ i10/ p5355_ 1 .
Solution: Just because we can, we will solve the more general problem of three different springs shown
below (k1 , k2 , and k3 from left to right) and two different masses m1 and m2 . Though it requires a bit more
algebra, it solves both of our problems posed and several others. By setting k1 = k3 = 0 we solve problem (a),
and setting k3 = 0 we solve problem (b). By setting k1 = k2 = k3 we solve the simplest case of two coupled
oscillators, a problem you will no doubt encounter again. So, for the purposes of illustration, we will drag
this problem out in quite some detail.
Figure 2: From http: // en. wikipedia. org/ wiki/ Normal_ mode .
Let mass m1 be displaced from equilibrium by an amount x1 and mass m2 by an amount x2 , with positive
x running to the right.ii Mass m1 is connected to springs k1 and k2 . Spring k1 is compressed (or elongated)
only by mass m1 due to its displacement x1 , and it reacts with a force −k1 x1 on mass m1 . Similarly, spring
3 is compressed only by mass 2, so it reacts with a force −k3 x2 on mass m2 . Spring 2 is connected to both
masses m1 and m2 , and its net change in length from equilibrium is the difference between the displacements
of masses m1 and m2 , x2−x1 . If both masses move in the same direction by the same amount, the net change
in length is zero, whereas if both masses move in opposite directions in the same amount, the net change in
length is twice as much. Spring 2 thus pushes back on both masses m1 and m2 with a force k2 (x2 −x1 ).
Putting all this together, we can write the net force on masses m1 and m2 , making note of the fact that for
mass m1 the force from k1 is opposite in direction to that of k2 , and similarly for the forces from k3 and k2
on mass m2 .
ii It
makes no difference which direction we call +x, so long as we are consistent.
d2 x1
= −k1 x1 + k2 (x2 − x1 )
dt2
d2 x2
F2 = m2 2 = −k3 x2 + k2 (x1 − x2 )
dt
F1 = m1
(55)
(56)
Now, what are the possible modes of oscillation? First, we seek only steady-state solutions. Since we have
not included any damping, that means ones that involve both masses oscillating freely in a sinusoidal fashion.
The symmetry of the problem dictates that only two modes should be possible: a symmetric one where both
masses move in the same direction, and an antisymmetric one where the masses move in opposite directions.
In the symmetric mode, in the limiting case that k1 = k3 and m1 = m2 , we would have the masses moving
in unison and the central spring k2 would remain at its equilibrium length (and in this case the frequency
should not depend on k2 ). In the antisymmetric mode, a higher frequency vibration occurs where the masses
move toward and away from each other. In any case: if we seek steady-state sinusoidal solutions, symmetric
or antisymmetric, there is a single frequency governing each mode, and we may choose
x1 = A1 eiωt
(57)
x2 = A2 eiωt
(58)
Plugging this trial solution into our equations of motion above,
−m1 ω 2 A1 eiωt = −k1 A1 eiωt + k2 (A2 − A1 ) eiωt
2
−m2 ω A2 e
iωt
= −k3 A2 e
iωt
+ k2 (A1 − A2 ) e
iωt
(59)
(60)
Simplifying, and canceling the common factor of eiωt
−m1 ω 2 A1 = −k1 A1 + k2 (A2 − A1 )
(61)
2
−m2 ω A2 = −k3 A2 + k2 (A1 − A2 )
(62)
We may write this as a system of two equations in terms of the two unknown amplitudes A1 and A2 :
m1 ω 2 − k1 − k2 A1 + k2 A2 = 0
k2 A1 + m2 ω 2 − k3 − k2 A2 = 0
(63)
(64)
Of course, we do not really wish to find the amplitudes, we wish to find ω. We may find ω by investigating
the conditions under which a solution to the above equations exists. First, we write the equation above in
matrix form:
"
m1 ω 2 − k1 − k2
k2
k2
2
m2 ω − k3 − k2
#" # " #
A1
0
=
A2
0
(65)
This system of equations has a solution only if the matrix of coefficients has a determinant of zero:
m ω2 − k − k 1
1
2
k2
k2
= 0 = m1 ω 2 − k1 − k2 m2 ω 2 − k3 − k2 − k22
2
m2 ω − k3 − k2
(66)
Expanding,
m1 m2 ω 4 − [(k2 + k3 ) m1 + (k1 + k2 ) m2 ] ω 2 + (k1 + k2 ) (k2 + k3 ) − k22 = 0
(67)
This is a quadratic in ω 2 , which we can readily solve:
2
ω =
ω2 =
(k2 + k3 ) m1 + (k1 + k2 ) m2 ±
q
2
((k2 + k3 ) m1 + (k1 + k2 ) m2 ) − 4m1 m2 [(k1 + k2 ) (k2 + k3 ) − k22 ]
2m1 m2
(k2 + k3 ) m1 + (k1 + k2 ) m2 ±
q
2
((k2 + k3 ) m1 − (k1 + k2 ) m2 ) + 4m1 m2 k22
2m1 m2
(68)
It doesn’t simplify a lot more than this in the general case. Let us examine then the cases of interest.
First, it instructive to keep the more general setup with three springs but consider the special case of identical
masses and springs by letting k1 = k2 = k3 ≡ k and m1 = m2 ≡ m. Our expression above then simplifies to
4km ± 2km
=
ω =
2m2
2
3k k
,
m m
(69)
p
Physically, this makes sense. We have the symmetric mode (ω = k/m) in which the two masses move in
unison back and forth and the central spring remains uncompressed at all times. The second is an antisymmetric mode which has the two masses moving out of phase, both moving outward at the same time or both
moving inward at the same time. The exterior springs are being compressed by each mass during half a cycle
of oscillation, and during the other half the central spring is compressed by both masses (so twice as much),
almost as if three springs are acting on each mass. This leads to the higher frequency of the antisymmetric
p
ω = 3k/m mode.
(b) For the diatomic molecule, we set k1 = k3 = 0 and k2 ≡ k in the general solution, leading to
ω2 =
ω2 =
km1 + km2 ±
q
2
(km1 − km2 ) + 4m1 m2 k 2
2m1 m2
k (m1 + m2 )
k
=
m1 m2
µ
=
k (m1 + m2 ) ± k (m1 + m2 )
2m1 m2
(70)
(71)
Here µ = m1 m2 /(m1 + m2 ) is the reduced mass of the system. The diatomic molecule has only one mode
of vibration, the antisymmetric one, which is the same as that of a mass µ connected to a fixed point by a
spring k. The symmetric mode in this case would correspond to a translation of the whole molecule, since
it isn’t anchored to anything. If the molecule is symmetric, m1 = m2 , we have ω 2 = 2k/m – since the only
mode is the one in which both atoms compress the spring together, we would expect the frequency to be
twice as high as that of a single mass connected by a spring to a fixed point.
(c) For the symmetric diatomic molecule on a surface, we set k3 = 0 and m1 = m2 ≡ m in the general solution:
ω2 =
(k1 + 2k2 ) m ±
q
2
(k2 m − (k1 + k2 ) m) + 4m2 k22
2m
p
2 + k2
4k
k
+
2k
±
1
2
2
1
ω2 =
2m
2
=
(k1 + 2k2 ) ±
q
2
(k2 − (k1 + k2 )) + 4k22
2m
(72)
If the springs are equal – not very realistic for a molecule adsorbed on a surface – this simplifies to
√ 3± 5 k
ω =
2
m
2
(73)
While our free diatomic molecule has only a single mode of vibration, after bonding to the surface the system again has two vibrational modes, corresponding to symmetric and antisymmetric vibrations of the two
masses.
Under the more realistic assumption that the “spring” coupling the molecule to the surface is much weaker
than the interatomic bond, k1 k2 ,
q
p
k12
k
+
2k
±
2k
2
2
k
+
2k
±
2k
1
+
1
2
2 1+
1
2
2
k1 + 2k2 ± 4k2 + k1
4k22
=
≈
ω2 =
2m
2m
2m


 k1 − k12 k1 + 4k2 + k12  k 2k
k1
1
2
4k2
4k2
ω2 ≈
,
≈
,
+
 2m

2m
2m m
2m
k12
8k22
(74)
(75)
Here we used the approximation (1 + x)n ≈ 1 + nx for x 1 to simplify the radical in the first line. If we
p
write the isolated diatomic molecule’s vibrational frequency as ωo = 2k2 /m,
2
ω =
k1 2k2
k1
,
+
2m m
2m
= {δω, ωo + δω}
(76)
p
Thus, for weak coupling to the surface, the fundamental mode is shifted upward by an amount δω = k1 /2m,
and a new low-frequency mode is introduced at δω. Spectroscopically, one can use this upward shift of the
fundamental mode to detect the absorption of molecules on a surface and estimate the adsorption energy.
10. Energetics of diatomic systems An approximate expression for the potential energy of two ions as a
function of their separation is
PE = −
ke2
b
+ 9
r
r
(77)
The first term is the Coulomb interaction representing the electrical attraction of the two ions, while the
second term is introduced to account for the repulsive effect of the two ions at small distances. (a) Find
b as a function of the equilibrium spacing ro . (What must be true at equilibrium?) (b) For KCl, with an
equilibrium spacing of ro = 0.279 nm, calculate the frequency of small oscillations about r = ro . Hint: do a
Taylor expansion of the potential energy to make it look like a harmonic oscillator for small r = ro .
Solution: The equilibrium spacing will be characterized by the net force between the ions being zero, or
equivalently, the potential energy being zero:
F (ro ) = −
ke2
9b
dU = 0 = 2 − 10
dr r=ro
ro
ro
ke2 ro8 = 9b
1
b = ke2 ro8
9
(78)
(79)
(80)
Substituting this result back into our potential energy expression, we can find the potential energy at
equilibrium, how much energy is gained by the system of ions condensing into a crystal. First, the potential
energy as a function of spacing:
P E = U (r) = −
ke2
ke2 ro8
+
r
9r9
(81)
Evaluating at equilibrium, ro = 0.279 nm,
U (ro ) = −
ke2
ke2
8ke2
+
=−
≈ −4.59 eV
ro
9ro
9ro
(82)
The frequency of small oscillations can be found by Taylor expanding the potential about equilibrium for
small displacements from equilibrium:
1
2
U (r − ro ) ≈ U (ro ) + U 0 (ro ) (r − ro ) + U 00 (ro ) (r − ro )
2
(83)
The first term in the expansion is just the potential energy at equilibrium which we found above. The second
term, linear in displacement, must vanish at equilibrium (which is exactly the condition we enforced to find
b, after all). The third term is quadratic in displacement, just as it would be for a simple harmonic oscillator,
2
U = 12 k (r − ro ) . Thus, the coefficient of the quadratic term must be 12 k, which means the frequency of small
p
oscillations is ω = k/µ, where µ is the reduced mass of the system (see the last problem for a derivation).
That is, the diatomic molecule looks like two masses coupled by a spring.
1
1
k = U 00 (ro )
2
2
2ke2 90b
8ke2
k = U 00 (ro ) = − 3 11 = 3 ≈ 84.9 N/m
ro ro
ro
s
k
ω=
= 2πf
µ
(84)
(85)
(86)
The reduced mass of the molecule in terms of the K and Cl atomic masses is
µ=
mK mCl
≈ 3.09 × 10−26 kg
mK + mCl
(87)
which gives the frequency of oscillation f as
1
f=
2π
s
k
≈ 8.35 × 1012 Hz ≈ 278 cm−1
µ
(88)
The accepted valueiii is 281 cm−1 , in excellent agreement with our simple model.
iii NIST,
see http://cccbdb.nist.gov/compvibs3.asp?casno=7447407&charge=0&method=14&basis=9