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Topic 7 of 11

Coupled Oscillators

When two or more oscillators are coupled, energy transfers between them periodically — this is the origin of beats. In the limit of many coupled oscillators, the normal modes become the modes of a continuous string or solid, connecting discrete mechanics to wave physics.

Key Concepts

Weak Coupling

When the coupling spring constant

k_c\ll k

, the two normal frequencies are nearly equal:

\omega_\pm\approx\omega_0\pm k_c/(2m\omega_0)

. The system exhibits beats with frequency

\Delta\omega=\omega_+-\omega_-

Beat Frequency

If both modes are excited with equal amplitude, the motion of each mass modulates with beat frequency

\omega_{\rm beat}=(\omega_+-\omega_-)/2

. Energy oscillates back and forth between the two oscillators at this frequency.

Normal Coordinates

\eta_1=(x_1+x_2)/\sqrt{2}

(in-phase) and

\eta_2=(x_1-x_2)/\sqrt{2}

(out-of-phase) are the normal coordinates. Each satisfies a simple harmonic equation with its own frequency.

Continuum Limit

Taking

N\to\infty

identical masses and springs with

N\Delta x=L

fixed converts the

N

normal modes into standing waves on a string with dispersion relation

\omega=2\sqrt{k/m}\sin(k_\text{wave}\Delta x/2)

Key Equations

Normal frequencies (3-spring system)

\omega_\pm = \sqrt{\frac{k \pm k_c}{m}}

Two masses m, outer springs k, coupling spring kc between the masses.

Beat frequency

\omega_{\rm beat} = \frac{\omega_+ - \omega_-}{2}

Frequency at which energy shuttles between the two coupled oscillators.

Dispersion relation (1D lattice)

\omega = 2\sqrt{\frac{k}{m}}\left|\sin\frac{k_{\rm wave}a}{2}\right|

Dispersion relation for a 1D monatomic chain with lattice constant a.

Worked Example

Energy Exchange Between Coupled Oscillators

Problem

Two masses ( $m=1.0$ kg) are each attached to a wall by springs ( $k=9.0$ N/m) and connected to each other by a weak coupling spring ( $k_c=1.0$ N/m). Find the two normal frequencies and the beat period.

Solution

\omega_- = \sqrt{k/m} = 3.0\text{ rad/s}, \quad \omega_+ = \sqrt{(k+k_c)/m} = \sqrt{10.0} = 3.162\text{ rad/s}

Beat frequency: $\omega_{\rm beat}=(\omega_+-\omega_-)/2=(3.162-3.000)/2=0.0811$ rad/s.

T_{\rm beat} = \frac{2\pi}{\omega_{\rm beat}} = \frac{2\pi}{0.0811} = 77.5\text{ s}

Answer

\omega_-=3.00

rad/s,

\omega_+=3.16

rad/s, beat period ≈ 77.5 s.

Practice

Exercises

7 problems

1 of 7

Two masses ( $m=1.0$ kg) on springs ( $k=9.0$ N/m) are coupled by $k_c=1.0$ N/m. Find $\omega_+$ (in rad/s).

Your answer

rad/s

2 of 7

Same system. Find the beat angular frequency $\omega_{\rm beat}$ (in rad/s).

Your answer

rad/s

3 of 7

Same system. Find the beat period $T_{\rm beat}$ (in s).

Unlock Exercise 3