PH 125 / LeClair Spring 2014

Problem Set 6 Solutions1. 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 issmall 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 meansnecessary 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 wemust use general gravitation (since we may not assume h� R), we must take into account the potentialenergy at the planet’s surface too. For general gravitation, potential energy is by convention only zero foran infinite separation of two objects.

Ei = Ui = −GMm

R+ h(1)

Ef = Kf + Uf = 12mv

2 − GMm

R(2)

=⇒ v =√

2GM√

1R− 1R+ h

(3)

(b) If h�R, we may use the approximation (1 + h/R)n≈1 + nh/R.

v =√

2GM√

1R− 1R+ h

=√

2GMR

√1− 1

1 + h/R≈√

2GMR

√1−

(1− h

R

)=√

2GMR2 h (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 andpotential energy of a particle. Start by solving the conservation of energy equation for kinetic energy, andsubsequently dx/dt.

K = E − U(x) = 12mv

2 = 12m

(dx

dt

)2(5)

dx

dt=√

2m

(E − U(x)) (6)

Now integrate both sides of Eq. 6 with respect to x.

dx

dt=√

2m

(E − U(x)) (7)

dt

dx=√m

21√

E − U(x)(8)

xf∫xi

dt

dxdx =

xf∫xi

√m

21√

E − U(x)(9)

tf − ti ≡ t =xf∫xi

√m

21√

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 thestarting potential energy of the particle at height h above the planet’s surface, and the potential energy atany position x above the planet’s surface is

U(x) = −GMm

R+ x(11)

The time is then found from Eq. 10 above, and the resulting integral is known (e.g., Wolfram Alpha).

t =xf∫xi

√m

21√

E − U(x)=

xf∫xi

√m

21√

GMmR − GMm

R+x

=√

12GM

xf∫xi

√R(R+ x)

xdx (12)

=√

12GM

[R3/2 ln

(√R+ x+

√x)

+√Rx(R+ x)

]h0

(13)

=√

12GM

[R3/2 ln

(√R+ h+

√h√

R

)+√Rh(R+ h)

](14)

=√R3/2

2GM ln(√

1 + h

R+√h

R

)+√

12GM

√Rh2 +R2h (15)

This is rather ugly. Is it correct? We can check the result for h� R, since we know in that limit thatt=√

2h/g. First, we know√

1 + x≈1 + x/2 and ln (1 + x)≈x for x�1. This gives for the first term

ln(√

1 + h

R+√h

R

)≈ ln

(1 + h

2R +√h

R

)≈ h

2R +√h

R(16)

If h/R� 1, then the first term is much smaller than the second, and we may neglect it. This reduces ouroverall expression to

t ≈√

R3

2GM

√h

R+√

12GM

√Rh2 +R2h =

√R2h

2GM +√

12GM

√Rh2 +R2h (17)

In the second term, if h�R, we may neglect Rh2 compared to R2h. Thus,

t ≈√

R2h

2GM +√

12GM

√R2h =

√2R2h

GM(18)

Noting again that g=GM/R2, we have

t ≈

√2hg

(19)

In the limit of small heights above the planet’s surface, we again recover the usual result that assumesconstant acceleration.

2. A block of mass m is connected to two springs of force constants k1 and k2 as shown below. The blockmoves on a frictionless table after it is displaced from equilibrium and released. Determine the period ofsimple harmonic motion. (Hint: what is the total force on the block if it is displaced by an amount x?

m

(a)

k1 k2

(b)

k1 k2

m

Figure P15.71

Problems 483

spring, as shown in Figure P15.68. (a) Static extensionsof 17.0, 29.3, 35.3, 41.3, 47.1, and 49.3 cm are measuredfor M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, re-spectively. Construct a graph of Mg versus x, and per-form a linear least-squares fit to the data. From the slopeof your graph, determine a value for k for this spring.(b) The system is now set into simple harmonic motion,and periods are measured with a stopwatch. WithM ! 80.0 g, the total time for 10 oscillations is measuredto be 13.41 s. The experiment is repeated with M valuesof 70.0, 60.0, 50.0, 40.0, and 20.0 g, with correspondingtimes for 10 oscillations of 12.52, 11.67, 10.67, 9.62, and7.03 s. Compute the experimental value for T from eachof these measurements. Plot a graph of T 2 versus M, anddetermine a value for k from the slope of the linear least-squares fit through the data points. Compare this valueof k with that obtained in part (a). (c) Obtain a value forms from your graph and compare it with the given valueof 7.40 g.

A smaller disk of radius r and mass m is attached rigidly tothe face of a second larger disk of radius R and mass M asshown in Figure P15.69. The center of the small disk islocated at the edge of the large disk. The large disk ismounted at its center on a frictionless axle. The assembly isrotated through a small angle " from its equilibrium positionand released. (a) Show that the speed of the center of thesmall disk as it passes through the equilibrium position is

(b) Show that the period of the motion is

T ! 2# ! (M $ 2m)R 2 $ mr 2

2mgR "1/2

v ! 2 ! Rg(1 % cos ")(M/m) $ (r/R)2 $ 2 "

1/2

69.

70. Consider a damped oscillator as illustrated in Figures15.21 and 15.22. Assume the mass is 375 g, the springconstant is 100 N/m, and b ! 0.100 N & s/m. (a) Howlong does it takes for the amplitude to drop to half itsinitial value? (b) What If? How long does it take for themechanical energy to drop to half its initial value? (c) Show that, in general, the fractional rate at which theamplitude decreases in a damped harmonic oscillator ishalf the fractional rate at which the mechanical energydecreases.

71. A block of mass m is connected to two springs of force con-stants k1 and k2 as shown in Figures P15.71a and P15.71b.In each case, the block moves on a frictionless table after itis displaced from equilibrium and released. Show that inthe two cases the block exhibits simple harmonic motionwith periods

(b) T ! 2# ! mk1 $ k2

(a) T ! 2# ! m(k1 $ k2)k1k2

72. A lobsterman’s buoy is a solid wooden cylinder of radius rand mass M. It is weighted at one end so that it floats up-right in calm sea water, having density '. A passing sharktugs on the slack rope mooring the buoy to a lobster trap,pulling the buoy down a distance x from its equilibriumposition and releasing it. Show that the buoy will executesimple harmonic motion if the resistive effects of thewater are neglected, and determine the period of theoscillations.

73. Consider a bob on a light stiff rod, forming a simplependulum of length L ! 1.20 m. It is displaced from thevertical by an angle "max and then released. Predict thesubsequent angular positions if "max is small or if it is large.Proceed as follows: Set up and carry out a numericalmethod to integrate the equation of motion for the simplependulum:

d 2"

dt2 ! %gL

sin "

R

M

""

mv

Figure P15.69

m

Figure P15.68

Solution: Say we displace the block to the right by an amount x. Both springs will try to bring the blockback toward equilibrium - one will pull, one will push, but both will act in the same direction. That meansthe net force is

Fnet = −k1x− k2x = − (k1 + k2)x = ma (20)

This is exactly the same as what we would find for a single spring, except the spring constant has becomek1 + k2 rather than just k. The solution must be

T = 2πω

= 2π√

m

k1 + k2(21)

3. A mass m is connected to two springs in series as shown below. What is the period of simple harmonicmotion if the mass is displaced from equilibrium. Hint: what must be true of the displacement of each springif the total displacement is ∆x?

mkk

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 thecombination of two springs by a single effective spring with a different spring constant - physically the twosprings 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 extendedor contracted less than the other. If the springs have displacements x1 and x2, the total displacement wouldbe x1+x2. What we expect is that the net force balance for both springs together should look like that of asingle spring, just with a different spring constant. A single spring with a spring constant keq displaced bythat 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 sameforce, and the force on the block must also be the same. What we wish to find is the spring constant keq fora single spring that also gives the same force, given a displacement of this equivalent spring of x1+x2. Theforce on each spring is readily found given their displacements:

F1 = −k1x1 = F2 = −k2x2 = Feq (23)

Solving for x2, we find x2 =(k1/k2)x2. Using this result in our expression for Feq,

Feq = −keq

(x1 + k1

k2x1

)= −keq

(k2 + k1

k2

)x1 (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 = −k1x1 (25)

=⇒ keq = k1k2

k1 + k2(26)

Thus, two springs in series behave the same as a single spring with spring constant keq =k1k2/(k1 + k2), orequivalently 1/keq =1/k1 + 1/k2. The period is thus

T = 2πω

= 2π√

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 bya spring of force constant k. The moment of inertia of the plank about the pivot is I = 1

3mL2. The plank

is displaced by a small angle θ from horizontal equilibrium and released. Find the angular frequency ω ofsimple harmonic motion. (Hint: consider the torques about the pivot point.)

482 C H A P T E R 15 • Oscillatory Motion

of the system for small values of the amplitude (small !).Assume the vertical suspension of length L is rigid, but ig-nore its mass.

60. A particle with a mass of 0.500 kg is attached to a springwith a force constant of 50.0 N/m. At time t " 0 theparticle has its maximum speed of 20.0 m/s and is movingto the left. (a) Determine the particle’s equation of motion, specifying its position as a function of time. (b) Where in the motion is the potential energy threetimes the kinetic energy? (c) Find the length of a simplependulum with the same period. (d) Find the minimumtime interval required for the particle to move from x " 0to x " 1.00 m.

61. A horizontal plank of mass m and length L is pivoted at oneend. The plank’s other end is supported by a spring of forceconstant k (Fig P15.61). The moment of inertia of the plankabout the pivot is mL2. The plank is displaced by a small an-gle ! from its horizontal equilibrium position and released.(a) Show that it moves with simple harmonic motion with anangular frequency . (b) Evaluate the frequency ifthe mass is 5.00 kg and the spring has a force constant of100 N/m.

# " !3k/m

13

62. Review problem. A particle of mass 4.00 kg is attached toa spring with a force constant of 100 N/m. It is oscillatingon a horizontal frictionless surface with an amplitude of2.00 m. A 6.00-kg object is dropped vertically on top of the4.00-kg object as it passes through its equilibrium point.The two objects stick together. (a) By how much does theamplitude of the vibrating system change as a result of thecollision? (b) By how much does the period change?(c) By how much does the energy change? (d) Account forthe change in energy.

A simple pendulum with a length of 2.23 m and a mass of6.74 kg is given an initial speed of 2.06 m/s at its equilib-rium position. Assume it undergoes simple harmonicmotion, and determine its (a) period, (b) total energy, and(c) maximum angular displacement.

64. Review problem. One end of a light spring with force con-stant 100 N/m is attached to a vertical wall. A light string istied to the other end of the horizontal spring. The stringchanges from horizontal to vertical as it passes over a solidpulley of diameter 4.00 cm. The pulley is free to turn on afixed smooth axle. The vertical section of the string sup-ports a 200-g object. The string does not slip at its contactwith the pulley. Find the frequency of oscillation of theobject if the mass of the pulley is (a) negligible, (b) 250 g,and (c) 750 g.

63.

65. People who ride motorcycles and bicycles learn to look outfor bumps in the road, and especially for washboarding, acondition in which many equally spaced ridges are worninto the road. What is so bad about washboarding? A mo-torcycle has several springs and shock absorbers in its sus-pension, but you can model it as a single spring supportinga block. You can estimate the force constant by thinkingabout how far the spring compresses when a big biker sitsdown on the seat. A motorcyclist traveling at highwayspeed must be particularly careful of washboard bumpsthat are a certain distance apart. What is the order of mag-nitude of their separation distance? State the quantitiesyou take as data and the values you measure or estimatefor them.

66. A block of mass M is connected to a spring of mass mand oscillates in simple harmonic motion on a horizon-tal, frictionless track (Fig. P15.66). The force constant ofthe spring is k and the equilibrium length is !. Assumethat all portions of the spring oscillate in phase and thatthe velocity of a segment dx is proportional to the dis-tance x from the fixed end; that is, vx " (x/!)v. Also,note that the mass of a segment of the springis dm " (m/!)dx. Find (a) the kinetic energy of thesystem when the block has a speed v and (b) the periodof oscillation.

Pivot

"

k

Figure P15.61

x

dx

M

v

Figure P15.66

A ball of mass m is connected to two rubber bands oflength L, each under tension T, as in Figure P15.67. Theball is displaced by a small distance y perpendicular to thelength of the rubber bands. Assuming that the tensiondoes not change, show that (a) the restoring forceis $ (2T/L)y and (b) the system exhibits simple harmonic motion with an angular frequency .# " !2T/mL

67.

y

L L

Figure P15.67

68. When a block of mass M, connected to the end of aspring of mass ms " 7.40 g and force constant k, is set intosimple harmonic motion, the period of its motion is

A two-part experiment is conducted with the use ofblocks of various masses suspended vertically from the

% " 2& ! M ' (ms/3)k

Solution: The presence of a pivot point – even labeled as such – immediately suggests the use of torqueto solve this problem. First, we need to find the compression of the spring at equilibrium, i.e., θ=0. Sincethe plank has non-zero mass, even without an angular displacement the spring must be compressed by someamount at equilibrium. Once we have found the equilibrium position, we can worry about the torques whena small angular displacement θ is applied.

Let counterclockwise rotations be defined as positive, and let the equilibrium position of the spring correspondto the tip of the plank being at vertical position xo relative to its unstretched length. The sum of the torquesabout the pivot point at equilibrium (θ=0) is given by considering the weight of the plank acting about itscenter of mass and the restoring force of the spring. The plank may be treated as a point mass a distanceL/2 from the pivot point, while the spring force acts at a distance L. At equilibrium, the net torque mustbe zero.

∑τ = −mg

(L

2

)+ kxoL = 0 (28)

xo = mg

2k (29)

Now that we have the equilibrium position, we can find the torques when the plank makes an angle θ withthe vertical. Since the plank cannot change its length (we assume), the amount that the spring stretchesshould correspond to the arc length that the tip of the plank moves through, Lθ, if the angle is relativelysmall.i The spring is therefore displaced by an amount Lθ − xo from its unstretched length. This gives usthe spring’s restoring force. Since the spring is attached to the plank, the spring force always acts perpen-dicularly to the length of the plank at distance L, and the torque is easily found.

The plank’s weight still acts at a distance L/2 from the center of mass, but now at an angle 90− θ relativeto the axis of the plank. The overall torque is then

∑τ = −mg

(L

2

)sin (90− θ)− k (Lθ − xo)L = −1

2mgL cos θ − kL2θ + kxoL

Since the angle θ is small, we may approximate cos θ≈1, and we may also make use of our earlier expressionfor xo. Finally, out of equilibrium the torques must give the moment of inertia times the angular acceleration.

iSmall enough such that we don’t have to worry about the spring bending to the left, for one.

∑τ = −1

2mgL cos θ − kL2θ + kxoL ≈ −12mgL− kL

2θ + 12mgL = −kL2θ = Iα (30)

−kL2θ = Id2θ

dt2(31)

Noting that I= 13mL

2 for a thin plank, we can put the last equation in the desired form for simple harmonicmotion in terms of known quantities:

d2θ

dt2= −3k

mθ (32)

ω =√

3km

(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, passingthrough 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 REabout the earth

Solution: We consider a spherically-symmetric mass of uniform density, radius RE , and mass ME . If weare at a radius r < RE from the center of the earth, only the mass contained within r < RE gives rise to anet gravitational force – the gravitational interactions for bits of mass at r > RE all cancel each other. Ifwe assume the mass of material removed to make the hole is negligible, then the mass contained within aradius 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)mr2 = −GMEmr

R3E

Noting that the gravitational acceleration at earth’s surface is

g = GME

R2E

we can write this compactly as

F (r) = −gm r

RE= ma = m

d2r

dt2

d2r

dt2= − g

REr

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 willhave periodic motion with

T = 2π

√REg≈ 84.4min

Clearly, the time for the ball return to its original position is then T . We can compare this time to thatrequired for an orbit around the earth at radius RE (which would be an orbit skimming the surface of theearth, a bit silly . . . ) via Kepler’s law, and they are the same:

Torbit = 2π

√R3E

GME= 2π

√REg

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 in-teresting 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 becauseA=πr2. Determine how the following properties depend on r: (a) period, (b) kinetic energy, (c) angularmomentum, 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,

GMm

r2 = mv2

r(35)

r

2GMm

r2 = r

2mv2

r(36)

GMm

2r = −12U = 1

2mv2 = K (37)

Thus, we can see K ∝ 1/r. Since K ∝ v2, 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:

f1(x, t) = Ae−b(x−vt)2

f2(x, t) = A

b (x− vt)2 + 1

f3(x, t) = Ae−b(bx2+vt) 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

∂2y

∂x2 = 1v2∂2y

∂t2

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, re-spectively. Knowing this, the fact that only f1 and f2 are solutions is clear by inspection (though you wereunfortunately 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 foundby simply changing the arguments to x+ vt from x− vt:

f ′1(x, t) = Ae−b(x+vt)2f ′2(x, t) = A

b (x+ vt)2 + 1

8. The equation for a driven damped oscillator is

d2x

dt2+ 2γωo

dx

dt+ ω2

ox = q

mE(t) (38)

(a) Explain the significance of each term.(b) Let E = Eoe

iωt and x = xoei(ωt−α) where Eo and xo are real quantities. Substitute into the above

expression and show that

xo = qEo/m√(ω2o − ω2)2 + (2γωωo)2

(39)

(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:

md2x

dt2= −mω2

ox− 2γmωodx

dt+ qE(t) (40)

The term on the right is the net force on the oscillator. The first term on the left is the restoring force, thesecond the viscous damping term, and the last the driving force of the oscillator.

First, we find the derivatives of x, noting i2 =−1:

dx

dt= iωxoe

i(ωt−α) (41)

d2x

dt2= −ω2xoe

i(ωt−α) (42)

Substituting into the original equaiton,

q

mEoe

iωt = −ω2xoei(ωt−α) + 2γωoiωxoei(ωt−α) + ω2

oxoei(ωt−α) (43)

q

mEoe

iωt = ei(ωt−α) (−ω2xo + 2iγωoωxo + ω2oxo)

(44)q

mEoe

iωt = eiωte−iα(−ω2xo + 2iγωoωxo + ω2

oxo)

(45)

qEom

eiα = −ω2xo + 2iγωoωxo + ω2oxo (46)

To proceed, we use the Euler identity

eiθ = cos θ + i sin θ (47)

Giving

qEom

(cosα+ i sinα) = −ω2xo + 2iγωoωxo + ω2oxo (48)

We now have two separate equations if we equate the purely real and purely imaginary parts:

qEom

cosα = ω2oxo − ω2xo (49)

qEom

sinα = 2γωωoxo (50)

We can square both equations and add them together:

q2E2o

m2

(cos2 α+ sin2 α

)=(ω2o − ω2)2

x2o + (2γωωo)2

x2o (51)

x2o = q2E2

o

m21

(ω2o − ω2)2

x2o + (2γωωo)2 (52)

xo = qEom

1√(ω2o − ω2)2

x2o + (2γωωo)2

(53)

This is the desired amplitude of vibration. Going back to the preceding two equations, we can divide thesecond equation by the first to find the phase angle:

tanα = 2γωωoω2o − ω2 (54)

9. (a) A diatomic molecule has only one mode of vibration, and we may treat it as a pair of masses con-nected 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 shownbelow (k1, k2, and k3 from left to right) and two different masses m1 and m2. Though it requires a bit morealgebra, 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 coupledoscillators, a problem you will no doubt encounter again. So, for the purposes of illustration, we will dragthis 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 positivex 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 −k1x1 on mass m1. Similarly, spring3 is compressed only by mass 2, so it reacts with a force −k3x2 on mass m2. Spring 2 is connected to bothmasses m1 and m2, and its net change in length from equilibrium is the difference between the displacementsof masses m1 and m2, x2−x1. If both masses move in the same direction by the same amount, the net changein length is zero, whereas if both masses move in opposite directions in the same amount, the net change inlength 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 formass 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.iiIt makes no difference which direction we call +x, so long as we are consistent.

F1 = m1d2x1

dt2= −k1x1 + k2 (x2 − x1) (55)

F2 = m2d2x2

dt2= −k3x2 + k2 (x1 − x2) (56)

Now, what are the possible modes of oscillation? First, we seek only steady-state solutions. Since we havenot 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 bothmasses 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 movingin unison and the central spring k2 would remain at its equilibrium length (and in this case the frequencyshould not depend on k2). In the antisymmetric mode, a higher frequency vibration occurs where the massesmove toward and away from each other. In any case: if we seek steady-state sinusoidal solutions, symmetricor antisymmetric, there is a single frequency governing each mode, and we may choose

x1 = A1eiωt (57)

x2 = A2eiωt (58)

Plugging this trial solution into our equations of motion above,

−m1ω2A1e

iωt = −k1A1eiωt + k2 (A2 −A1) eiωt (59)

−m2ω2A2e

iωt = −k3A2eiωt + k2 (A1 −A2) eiωt (60)

Simplifying, and canceling the common factor of eiωt

−m1ω2A1 = −k1A1 + k2 (A2 −A1) (61)

−m2ω2A2 = −k3A2 + k2 (A1 −A2) (62)

We may write this as a system of two equations in terms of the two unknown amplitudes A1 and A2:

(m1ω

2 − k1 − k2)A1 + k2A2 = 0 (63)

k2A1 +(m2ω

2 − k3 − k2)A2 = 0 (64)

Of course, we do not really wish to find the amplitudes, we wish to find ω. We may find ω by investigatingthe conditions under which a solution to the above equations exists. First, we write the equation above inmatrix form:[(

m1ω2 − k1 − k2

)k2

k2(m2ω

2 − k3 − k2)] [A1

A2

]=[

00

](65)

This system of equations has a solution only if the matrix of coefficients has a determinant of zero:∣∣∣∣∣(m1ω

2 − k1 − k2)

k2

k2(m2ω

2 − k3 − k2)∣∣∣∣∣ = 0 =

(m1ω

2 − k1 − k2) (m2ω

2 − k3 − k2)− k2

2 (66)

Expanding,

m1m2ω4 − [(k2 + k3)m1 + (k1 + k2)m2]ω2 + (k1 + k2) (k2 + k3)− k2

2 = 0 (67)

This is a quadratic in ω2, which we can readily solve:

ω2 =(k2 + k3)m1 + (k1 + k2)m2 ±

√((k2 + k3)m1 + (k1 + k2)m2)2 − 4m1m2 [(k1 + k2) (k2 + k3)− k2

2]2m1m2

ω2 =(k2 + k3)m1 + (k1 + k2)m2 ±

√((k2 + k3)m1 − (k1 + k2)m2)2 + 4m1m2k2

2

2m1m2(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 identicalmasses and springs by letting k1 =k2 =k3≡k and m1 =m2≡m. Our expression above then simplifies to

ω2 = 4km± 2km2m2 =

{3km,k

m

}(69)

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 antisym-metric mode which has the two masses moving out of phase, both moving outward at the same time or bothmoving inward at the same time. The exterior springs are being compressed by each mass during half a cycleof 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 antisymmetricω=

√3k/m mode.

(b) For the diatomic molecule, we set k1 =k3 = 0 and k2≡k in the general solution, leading to

ω2 =km1 + km2 ±

√(km1 − km2)2 + 4m1m2k2

2m1m2= k (m1 +m2)± k (m1 +m2)

2m1m2(70)

ω2 = k (m1 +m2)m1m2

= k

µ(71)

Here µ=m1m2/(m1 + m2) is the reduced mass of the system. The diatomic molecule has only one modeof vibration, the antisymmetric one, which is the same as that of a mass µ connected to a fixed point by aspring k. The symmetric mode in this case would correspond to a translation of the whole molecule, sinceit isn’t anchored to anything. If the molecule is symmetric, m1 =m2, we have ω2 = 2k/m – since the onlymode is the one in which both atoms compress the spring together, we would expect the frequency to betwice 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±

√(k2m− (k1 + k2)m)2 + 4m2k2

2

2m2 =(k1 + 2k2)±

√(k2 − (k1 + k2))2 + 4k2

2

2m

ω2 = k1 + 2k2 ±√

4k22 + k2

12m (72)

If the springs are equal – not very realistic for a molecule adsorbed on a surface – this simplifies to

ω2 =(3±√

5)

2k

m(73)

While our free diatomic molecule has only a single mode of vibration, after bonding to the surface the sys-tem again has two vibrational modes, corresponding to symmetric and antisymmetric vibrations of the twomasses.

Under the more realistic assumption that the “spring” coupling the molecule to the surface is much weakerthan the interatomic bond, k1�k2,

ω2 = k1 + 2k2 ±√

4k22 + k2

12m =

k1 + 2k2 ± 2k2

√1 + k2

14k2

2

2m ≈k1 + 2k2 ± 2k2

(1 + k2

18k2

2

)2m (74)

ω2 ≈

k1 − k21

4k2

2m ,k1 + 4k2 + k2

14k2

2m

≈{k1

2m,2k2

m+ k1

2m

}(75)

Here we used the approximation (1 + x)n≈ 1 + nx for x� 1 to simplify the radical in the first line. If wewrite the isolated diatomic molecule’s vibrational frequency as ωo=

√2k2/m,

ω2 ={k1

2m,2k2

m+ k1

2m

}= {δω, ωo + δω} (76)

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 thefundamental 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 afunction of their separation is

PE = −ke2

r+ b

r9 (77)

The first term is the Coulomb interaction representing the electrical attraction of the two ions, while thesecond term is introduced to account for the repulsive effect of the two ions at small distances. (a) Findb as a function of the equilibrium spacing ro. (What must be true at equilibrium?) (b) For KCl, with anequilibrium spacing of ro = 0.279nm, calculate the frequency of small oscillations about r= ro. Hint: do aTaylor 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, orequivalently, the potential energy being zero:

F (ro) = −dUdr

∣∣∣∣r=ro

= 0 = ke2

r2o

− 9br10o

(78)

ke2r8o = 9b (79)

b = 19ke

2r8o (80)

Substituting this result back into our potential energy expression, we can find the potential energy atequilibrium, how much energy is gained by the system of ions condensing into a crystal. First, the potentialenergy as a function of spacing:

PE = U(r) = −ke2

r+ ke2r8

o

9r9 (81)

Evaluating at equilibrium, ro=0.279nm,

U(ro) = −ke2

ro+ ke2

9ro= −8ke2

9ro≈ −4.59 eV (82)

The frequency of small oscillations can be found by Taylor expanding the potential about equilibrium forsmall displacements from equilibrium:

U(r − ro) ≈ U(ro) + U ′(ro) (r − ro) + 12U′′(ro) (r − ro)2 (83)

The first term in the expansion is just the potential energy at equilibrium which we found above. The secondterm, linear in displacement, must vanish at equilibrium (which is exactly the condition we enforced to findb, after all). The third term is quadratic in displacement, just as it would be for a simple harmonic oscillator,U= 1

2k (r − ro)2. Thus, the coefficient of the quadratic term must be 12k, which means the frequency of small

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.

12k = 1

2U′′(ro) (84)

k = U ′′(ro) = −2ke2

r3o

90br11o

= 8ke2

r3o

≈ 84.9N/m (85)

ω =

√k

µ= 2πf (86)

The reduced mass of the molecule in terms of the K and Cl atomic masses is

µ = mKmCl

mK +mCl≈ 3.09× 10−26 kg (87)

which gives the frequency of oscillation f as

f = 12π

√k

µ≈ 8.35× 1012 Hz ≈ 278 cm−1 (88)

The accepted valueiii is 281 cm−1, in excellent agreement with our simple model.

iiiNIST, see http://cccbdb.nist.gov/compvibs3.asp?casno=7447407&charge=0&method=14&basis=9