ry
jfj
-.ld"rtt
ffisp
ri.'L+
*,*o
y tj 1.0
#*fri
[#
4J
,t
fd b[o^o#-r-t''srtrg caJo tt**Vn f,va):0
'${y'd'
$,, 4 yn *l,r* ,,M* /i * 6 r 0
g, n^'1,tngC$olt= *'Fq i,v,A
f
$
=) { oS 9 = WJLL- :#
s;yt#/ #
€r
er {rW&;0
) en= ?rTt, o rr 6 rt 5€o'dh{*
gr r 5in6 caS843
c
+
#a:0
W
:
*,W*)
q-"
o(JL,
a
S *w&o-*,llf*
fu^ I 5wr&h&*, -ts1/n?0
:) lA# lT/n
f,
Cr*5{-ff
er{l
=) # s#
*ffa
'*ff^)
{*Jf
*fr* *#a )po
,*\r& tJ n* w{fteo *'
=)..* r^/*=
Co',s+
ca?#r, fue$u dd=&n=Hb)*. ,-
ffr,
n
5;,nU.-
-t'''t*-*"
(toOnJ.Y.r 0
llrPad)
;$oy& ,vr{h S ws*rnff ryrua**wn
*{'sS{'r s'W{
:
c{tSv
*,ti.
ft
q**v f{A{wvr}
6n*p^ **l) : *#5tn **$v*5tmE^Lnh{
*tuprrs?sint6+6 coS&o
co| [0o4\ X
#s - t {vt$s
La,9
gb-
g
,,5" =€*nff"+S$J%) {ws** *5*-o=6J f?f
*
P/^
{rtuffa{*rwst*)
= tsrn 4 @sao + S€oJbo -s,kb;h):
*&$*
#noSd#4)
-#,t*Y
=$f-,6.qrJfr
'lL
r**;h,
ou/Li
?
)01
7
ffis W{
u
,.,t*- fr *-$-- l^1, Plxdy
*)
s d {/L)
n p^ rL&<, r,y* *ht;n
k
o"**i
*
ta E =,$, fu6*[;r l,s
Y
fro-*-,#-#+t
'')
*w'iLt*-fr,x,i
*;L?{i
ry*J
b
../ f
-=
t {.fftw **d*ffis#trw#}# s#
.imq -8w '3 * aUf tth*oa6*
?
,) {)o=# =ftiar1 j A r f"/*o=
-\:
;0 +l=f*A -L
,,h^shbrL'**ff**ffi
#*i
*hrdu
rffWy *S"f,*"$ Fq"rrySJ
&a th{S r-g (,aw4te*{: fr* rs *tq* T-l/,
Elizabeth Wicks
Phys 200A
November 18th, 2013
Problem Set 5
Problem 7 (Fetter and Walecka 4.13)
Refer to Figure 23.1 on page 101 of Fetter and Walecka for a picture. We have a long chain of identical
pendulums connected by springs. In equilibrium, all the pendulums are vertical and the springs unstretched
at length a. Therefore a is the horizontal distance separating the tops of the pendulums. Define θi as the angle
from the vertical for the ith pendulum, and ηi as the transverse displacement from the equilibrium position.
Now we can write a Lagrangian. Using the small oscillation approximation, we will express everything in
terms of ηi .
For small displacements, we can see from the picture that
ηi
= sin θi ≈ θi .
l
This also gives
η˙i = lθ˙i .
Using these approximations, we obtain:
T =
N
X
1
i=1
Vgrav =
N
X
i=1
2
mvi2 =
N
X
1
i=1
mgl(1 − cos θi ) =
N
X
mgl
i=1
Vspr =
2
N
X
i=1
m(lθ˙i )2 =
N
X
1
i=1
1 2
θ
2 i
=
N
X
i=1
2
mη˙i 2 .
mgl
N
η X
mg 2
i
η .
=
2
2l
2l i
i=1
1
k(ηi+1 − ηi )2 .
2
Note: The displacement from equilibrium of a spring is (d − d0 )i = (ηi+1 − ηi ) + O(η 4 ), so we simply
ignore the higher order terms since ηi is small.
Now we have our Lagrangian completely in terms of the coordinates ηi .
L=
N
X
1
i=1
2
1
mg 2
mη˙i 2 − k(ηi+1 − ηi )2 −
η .
2
2l i
We obtain the equations of motion for each ηi :
mg mη¨i + 2k +
ηi − k(ηi+1 + ηi−1 ) = 0.
l
As suggested in Fetter and Walecka (p.115), the reasonable solution to try is
ηj = Aei(qxj −ωt) ,
where xj = ja and A is some constant. This is the plane wave normal mode with wavenumber q and
frequency ω that propagates along the chain. Plugging this in to the equation of motion yields
mg −mω 2 ηj + 2k +
ηj − k(eiqa + e−iqa )ηj = 0.
l
1
Since ηj 6= 0, we have
g 2k
+
(1 − cos(qa))
l
m
qa g 2k = +
2 sin2
l
m
2
qa g 4k
.
= +
sin2
l
m
2
ω2 =
This gives us the dispersion relation
ω+ (q) =
r
ω− (q) = −
qa g 4k
+
sin2
l
m
2
r
qa g 4k
.
+
sin2
l
m
2
See attached sheet for plots of the dispersion relation.
If we assume periodic boundary conditions, we put restrictions on q. Periodic boundary conditions require
that ηj = ηj+N for all j. Using the plane wave definition for ηj , we have the requirement
eiqN a = 1 ⇒ q =
2πn
Na
for some n ∈ Z.
Now the allowed frequencies are
ω 2 (n) =
πn g 4k
+
sin2
.
l
m
N
We can see that if n = 0 we have the lowest frequency
g
,
l
which is the oscillation frequency of a single pendulum without any springs attached. pThis frequency
corresponds to the physical situation of all the pendulums swinging in unison at frequency g/l.
ω02 =
2