12. Coupled Pendulums — Normal Modes and Beats¶

Module

Oscillations, Damping & Acoustic Methods (PG) · Module 4

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Aim¶

To study the normal modes of two identical pendulums coupled by a light spring, and to observe and interpret the phenomenon of beats arising from energy transfer between them.

Theoretical Background¶

Two identical simple pendulums, each of length l, connected near their bob by a light coupling spring, form a system with two degrees of freedom and hence two natural (normal) modes of oscillation, each with its own characteristic frequency. In the symmetric mode, both pendulums swing exactly in phase, with the same amplitude at every instant; the coupling spring is never stretched or compressed at all, so this mode has exactly the same frequency as either pendulum would have on its own, f_s = (1/2π)√(g/l).

In the antisymmetric mode, the two pendulums swing exactly out of phase (mirror images of each other at every instant); the spring is now alternately stretched and compressed as they swing, adding an extra restoring force on top of gravity, so this mode has a higher frequency, f_a = (1/2π)√(g/l + 2k/m), where k is the spring constant and m the bob mass.

Any other initial condition — most strikingly, releasing just one pendulum from rest while the other hangs still — can be represented as an equal superposition of these two normal modes. Because the two modes have slightly different frequencies, they gradually drift out of phase with each other as time passes; the result is that the energy of oscillation appears to slosh back and forth between the two pendulums with a characteristic beat period, even though no energy is lost overall — the first pendulum gradually comes to rest as the second one, initially still, swings up to the full original amplitude, and then the process reverses.

This transfer-of-energy phenomenon, purely a consequence of two nearby but distinct normal-mode frequencies beating against each other, is a direct mechanical analogue of phenomena as varied as coupled LC oscillator circuits, vibrational energy transfer between coupled molecular bonds, and the coupled-cavity behaviour of some laser systems.

Governing Formulas¶

Symmetric (in-phase) mode frequency:

\[ f_{s} = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \]

Antisymmetric (out-of-phase) mode frequency:

\[ f_{a} = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}+\dfrac{2k}{m}} \]

Beat period (energy-transfer period between the two pendulums):

\[ T_{\text{beat}} = \dfrac{1}{f_{a}-f_{s}} \]

Labelled apparatus layout — original schematic prepared for this guide (not a photograph). Two identical simple pendulums of equal length hung from the same rigid horizontal support, with a light spring connecting the two bobs near their midpoints to provide the coupling.

Simulator Controls¶

Control	Symbol	Range	Unit
Pendulum length (both)	l	20 – 100	cm
Coupling spring constant	k	0.5 – 5.0	N/m
Bob mass (each)	m	50 – 300	g

Readouts:

Quantity	Symbol	Unit
Symmetric mode frequency	f_s	Hz
Antisymmetric mode frequency	f_a	Hz
Beat period	T_beat	s

Procedure¶

Launch the Coupled Pendulums panel; set the pendulum length l and bob mass m to their default values.
With the coupling spring constant k at its minimum (weakest coupling), note the readout values of f_s, f_a and T_beat.
Increase k in equal steps, holding l and m fixed, and tabulate f_s, f_a and T_beat at each step.
Verify that f_s stays constant (independent of k) while f_a increases with k, exactly as the formulas predict.
Plot T_beat against k and observe how strengthening the coupling shortens the beat period — energy transfers back and forth more rapidly as the coupling gets stronger.

Observation Table¶

Table: Normal mode frequencies and beat period with increasing coupling strength

S. No.	l (cm)	k (N/m)	m (g)	f_s (Hz)	f_a (Hz)	T_beat (s)
1
2
3
4
5
6

Graph¶

Plot Beat period, T_beat (y-axis) against Coupling spring constant, k (x-axis).

Expected trend: a monotonically decreasing curve — stronger coupling transfers energy between the pendulums more rapidly, shortening the beat period.

T_beat vs k, expected to decrease monotonically as coupling strengthens

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

Verify that the measured f_s matches (1/2π)√(g/l) using the set pendulum length, independent of k.
Verify that the measured f_a matches (1/2π)√(g/l + 2k/m) for each value of k and m used.
Compute T_beat = 1/(f_a − f_s) for each trial and compare with the readout value.

Result¶

Table: Normal mode frequencies and beat behaviour

l (cm)	k (N/m)	f_s (Hz)	f_a (Hz)	T_beat, calculated (s)	T_beat, observed (s)
1
2
3

Maximum Permissible Error¶

Because T_beat depends on the small difference (f_a − f_s), even modest uncertainties in either frequency are amplified when this difference is small (weak coupling).

Precautions¶

Release both pendulums from rest with only one displaced, to cleanly excite the beating pattern rather than a general mixture of modes.
Keep oscillation amplitudes small so both pendulums remain in the simple-harmonic regime throughout.
Ensure both pendulums are of exactly equal length, since any asymmetry mixes the two normal modes and complicates the simple beat picture.

Self-Check Quiz¶

1. In the symmetric normal mode of two identical coupled pendulums, the coupling spring:

A) Stretches and compresses maximally
B) Is never stretched or compressed, since both pendulums move identically
C) Breaks
D) Oscillates at twice the pendulum frequency

Answer: B — Since both pendulums swing exactly in phase with equal amplitude at every instant, the distance between their bobs (and hence the spring's length) never changes.

2. The antisymmetric mode always has a higher frequency than the symmetric mode because:

A) The antisymmetric mode has a shorter effective pendulum length
B) The coupling spring provides an additional restoring force only when the pendulums move out of phase
C) It doesn't — they are always equal
D) The antisymmetric mode ignores gravity

Answer: B — In the antisymmetric mode the spring is alternately stretched and compressed, contributing an extra restoring force on top of gravity, which raises the frequency above f_s = (1/2π)√(g/l).

3. If the coupling spring constant k is increased (all else fixed), the beat period T_beat:

A) Increases
B) Decreases
C) Stays exactly the same
D) Becomes negative

Answer: B — Stronger coupling increases f_a while f_s stays fixed, widening the frequency difference (f_a−f_s) and hence shortening T_beat = 1/(f_a−f_s).

Viva-Voce Questions¶

Explain physically why releasing just one pendulum excites an equal superposition of both normal modes.
Why does the symmetric mode frequency not depend on the coupling spring constant at all?
Derive, in outline, the beat period in terms of the two normal-mode frequencies.
What would happen to the beating pattern if the two pendulums had noticeably different lengths?
Name at least one other physical system (electrical, molecular, or optical) that shows an analogous normal-mode beating phenomenon.
How would friction/damping, if included, eventually affect the long-term behaviour of this coupled system?