The mechanics of two pendulums coupled by a stressed spring is discussed, and the behavior for small oscillations is described. When the system is in its highest symmetry configuration, the pendulums are independent and the normal frequencies are degenerate.

1.
N. W.
Ashcroft
and
N. D.
Mermin
,
Solid State Physics
(
Harcourt College
,
Fort Worth, TX
,
1976
), Chap. 22.
2.
F. S.
Crawford
, Jr.
,
Waves
(
McGraw-Hill
,
New York
,
1968
), Chap. 1.
3.
R. P.
Feynman
,
R. B.
Leighton
, and
M.
Sands
,
The Feynman Lectures on Physics
(
,
San Francisco, CA
,
2006
), Vol.
1
, Chap. 49.
4.
L. O.
Olsen
, “
Coupled pendulums: An advanced laboratory experiment
,”
Am. J. Phys.
13
,
321
324
(
1945
).
5.
M. J.
Moloney
, “
String coupled pendulum oscillators: Theory and experiment
,”
Am. J. Phys.
46
,
1245
1246
(
1978
).
6.
Reference 2, Chap. 2.
7.
In the large stretch approximation the rest length of the spring can be neglected.
8.
A system satisfying this condition must have small $kc$ and large $|d⃗0|$ so that their product equals $|F⃗0|$. The small displacements of the bodies during the oscillations (compared with the large $|d⃗0|$) implies that the force of the spring remains approximately $F⃗0$ .
9.
In this pendulum the constant force $S⃗$$(|S⃗|=mg/cos θ0)$ plays the role of the weight $mg⃗$ in ordinary pendulums. The difference can be seen as a replacement of the gravitational field $g$ by $g/cos θ0$. The frequency $ωp$ is obtained by replacing $g$ by $g/cos θ0$ in the relation for an ordinary pendulum $(ω=g/L)$. The value $θ0=90°$ is not attainable because it would require infinite tension in the spring and imply that $ωp→+∞$.
10.
For example, when $θ0=0°$ the motion of the center of mass is only horizontal.
11.
The rectilinear trajectories (segments) can be obtained from the initial conditions of $ẋ1=ẋ2=0$.
12.
The experimental devices were by Pasco and the sensors were managed by SCIENCE WORKSHOP software (www.pasco.com/).
13.
The angular displacements from $θ0$ were set using the sensors and chosen to be $<5°$.
14.
The mass of the pendulum rods was neglected in deriving Eqs. (12), (14), and (15). Better agreement with experiment is obtained using the equations of motion $θ̈1=−(lcmMg/I cos θ0)θ1−(kcl2/I)[(θ1+θ2)cos2 θ0+(θ1−θ2)sin2 θ0]$ and $θ̈2=−(lcmMg/I cos θ0)θ2−(kcl2/I)[(θ2+θ1)cos2 θ0+(θ2−θ1)sin2 θ0]$. Here $θ1$ and $θ2$ are the angular displacements of the pendulums from $θ0$; $lcm$ is the distance between the pivot and the center of mass of each pendulum; $l$ is the distance between the pivot and the point where the spring is jointed; and $M$ and $I$ are the mass and moment of inertia (relative to the pivot) of each pendulum. The result for the normal frequencies is $ωs=(lcmMg/I cos θ0)+2(kcl2/I)cos2 θ0$ and$ωd=(lcmMg/I cos θ0)+2(kcl2/I)sin2 θ0$ [see Fig. 7(b)]. The values of the parameters were $lcm=30.1 cm$, $l=34.5 cm$, $M=103.3 g$, and $I=102 300 g cm2$.
15.
Reference 3, Chap. 48.
16.
The derivation is as follows. An initial position with $ycm=xcm$ is chosen.Because the circular motion must have the angular velocity $ωp2+ωc2$ (the degeneracy frequency), the initial vertical component of the velocity must be the product of this angular velocity and $xcm$: $ẏcm=ωp2+ωc2xcm$ (having chosen the anticlockwise direction). From geometry it follows that $ẋcm=−ẏcm$.
17.
This procedure comes from the fact that in this case the pendulums are independent and oscillate with frequency $ωp2+ωc2$. Therefore, if they start at the same distance $(x1=x2)$ from the origin, pendulum 2, after 3/4 of a oscillation, passes the origin with the velocity $ωp2+ωc2x1$.