Two coupled oscillators

Setting up the system. Take two identical masses $m$, each held to a wall by its own spring of stiffness $k$, and joined to each other by a coupling spring of stiffness $k_c$. Let $x_1, x_2$ be their displacements from equilibrium. The coupling spring's stretch is $x_1 - x_2$; it pulls mass 1 back and pushes mass 2 forward by the same amount, so Newton's second law on each mass gives

where: $m$ is each mass, $k$ the stiffness of the two identical end springs, $k_c$ the stiffness of the coupling spring, and $x_1, x_2$ the displacements of mass 1 and mass 2 from equilibrium. Each equation contains both coordinates — that is exactly what "coupled" means, and it is why neither mass on its own executes SHM.

Decoupling by symmetry. The system has a left–right symmetry (swap mass 1 and mass 2 and the equations swap too), and that symmetry is the clue: add the two equations, then subtract them.

Naming $q_1 = x_1+x_2$ and $q_2 = x_1-x_2$ turns these into two independent SHM equations, one per coordinate:

where: $q_1, q_2$ are the normal coordinates, and $\omega_1, \omega_2$ are the normal-mode angular frequencies. Physically, $\omega_1=\sqrt{k/m}$ is the frequency of the mode where both masses move together, in phase (the coupling spring never stretches, so $k_c$ drops out); $\omega_2=\sqrt{(k+2k_c)/m}$ is the mode where the masses move in antiphase, mirror-imaging each other (the coupling spring stretches maximally, stiffening the restoring force, hence the higher frequency — always $\omega_2>\omega_1$ for $k_c>0$).

Recovering the real motion. $q_1,q_2$ are pure cosines, $q_1(t)=C_1\cos(\omega_1 t+\phi_1)$ and $q_2(t)=C_2\cos(\omega_2 t+\phi_2)$, and inverting the definitions gives the physical displacements as an equal mixture of both modes:

The four constants $C_1,\phi_1,C_2,\phi_2$ come from the four initial conditions $x_1(0),x_2(0),\dot x_1(0),\dot x_2(0)$. Unless the system is launched into a single pure mode, each mass's motion is a superposition of two frequencies, not SHM.

Two identical masses $m=0.50\ \text{kg}$ are connected by end springs $k=20\ \text{N/m}$ and a coupling spring $k_c=5.0\ \text{N/m}$. The system is released from rest with mass 1 displaced by $x_1(0)=A_0=0.02\ \text{m}$ and mass 2 undisturbed, $x_2(0)=0$. Find the normal-mode frequencies and the resulting $x_1(t), x_2(t)$.

Step 1. Normal-mode frequencies follow directly from the formulas derived above:

Step 2. Convert the initial conditions to normal coordinates: $q_1(0)=x_1(0)+x_2(0)=A_0$ and $q_2(0)=x_1(0)-x_2(0)=A_0$. Released from rest means $\dot x_1(0)=\dot x_2(0)=0$, so $\dot q_1(0)=\dot q_2(0)=0$ too — each normal coordinate starts at maximum displacement with zero velocity, so both are pure cosines with zero phase:

Step 3. Invert to the physical displacements using $x_1=\tfrac12(q_1+q_2)$, $x_2=\tfrac12(q_1-q_2)$:

with $A_0=0.02\ \text{m}$ and $t$ in seconds.

Step 4. Using the sum-to-product identity, $x_1(t)=A_0\cos\!\left(\tfrac{\omega_2-\omega_1}{2}t\right)\cos\!\left(\tfrac{\omega_1+\omega_2}{2}t\right)$ — a fast carrier oscillation at the average frequency $\approx 7.03\ \text{rad/s}$, modulated by a slow envelope at half the frequency difference, $\approx 0.71\ \text{rad/s}$. This is the beat phenomenon: energy sloshes from mass 1 to mass 2 and back, with mass 1's amplitude dying as mass 2's grows, at beat frequency $|\omega_2-\omega_1|\approx 1.43\ \text{rad/s}$.

• Assuming each mass individually undergoes SHM. It does not, in general — only the normal coordinates $q_1,q_2$ are simple harmonic; $x_1(t)$ and $x_2(t)$ are sums of two different frequencies unless a single pure mode is excited.
• Getting the coupling-spring sign wrong: it must pull mass 1 and push mass 2 in opposite senses, $-k_c(x_1-x_2)$ and $+k_c(x_1-x_2)$ — using the same sign in both equations of motion breaks Newton's third law and gives unphysical results.
• Confusing normal coordinates with physical displacements. $q_1,q_2$ are algebraic constructs chosen to decouple the maths; only $x_1,x_2=\tfrac12(q_1\pm q_2)$ are what an observer actually measures moving.
• Forgetting that $\omega_2>\omega_1$ is a consequence of this particular spring layout ($k+2k_c>k$), not a universal law — a different way of attaching the coupling spring changes the formula for $\omega_2$.

Ask the tutor: if you want to see how the setup changes for a coupling spring attached differently, or want the initial-condition algebra worked through for a different starting displacement, ask the AI tutor to walk through it step by step.