Forced oscillations

Leave a damped oscillator alone and it eventually stops — but keep pushing it with a periodic force, and something more interesting happens. After an initial messy transient dies away, the system settles into a steady rhythm that oscillates at the driver's frequency, not its own natural frequency. This is forced (driven) oscillation, and it is the direct sequel to damped oscillations in the PH11001 Oscillations and Vibrations block: you take the damped-oscillator equation, add a sinusoidal force $F_0\cos\omega t$ , and solve for the steady-state amplitude and phase lag as functions of the driving frequency $\omega$ . That amplitude formula is the same object whose peak defines resonance in the very next module, so getting comfortable with it here — how it behaves for slow driving, fast driving, and driving near the natural frequency — pays off immediately.

The core insight

A damped oscillator driven by a sinusoidal force settles into steady-state motion at the driving frequency $\omega$ , with amplitude $A(\omega)$ and phase lag $\delta(\omega)$ set by how far $\omega$ sits from the natural frequency $\omega_0$ .

The core idea

Adding a sinusoidal drive $F_0\cos\omega t$ to the damped-oscillator equation gives

m\ddot x + b\dot x + kx = F_0\cos\omega t \quad\Longrightarrow\quad \ddot x + 2\gamma\dot x + \omega_0^2 x = \frac{F_0}{m}\cos\omega t,

where: $m$ is the mass, $b$ the damping constant, $k$ the spring constant, $\gamma=b/2m$ the damping rate, and $\omega_0=\sqrt{k/m}$ the natural frequency. The full solution is transient plus steady state. The transient is just the homogeneous damped solution, decaying as $e^{-\gamma t}$ ; once several damping times have passed, only the steady state survives, oscillating at the drive frequency $\omega$ :

x(t) = A(\omega)\cos(\omega t-\delta).

Solving for $A(\omega)$ and $\delta$ (worked below) gives

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}}, \qquad \tan\delta = \frac{2\gamma\omega}{\omega_0^2-\omega^2},

where: $A(\omega)$ is the steady-state amplitude, $\delta$ is the phase by which displacement lags the driving force, and $F_0/m$ is the force per unit mass. Three regimes are worth holding in your head. At low drive frequency ( $\omega\ll\omega_0$ ), $A\to F_0/(m\omega_0^2)=F_0/k$ , the plain static spring stretch, and $\delta\to0$ — the mass tracks the force almost instantly, as if the inertia and damping terms barely mattered. At high drive frequency ( $\omega\gg\omega_0$ ), $A\to F_0/(m\omega^2)\to0$ and $\delta\to\pi$ — the mass simply cannot keep up, and ends up moving opposite to the force at every instant, so the force spends most of each cycle fighting the motion rather than driving it. Near $\omega=\omega_0$ the amplitude rises sharply (more so the smaller the damping) and $\delta$ passes through $\pi/2$ ; at exactly that point the velocity is in phase with the force, which is why power transfer peaks there regardless of damping strength. It helps to keep the two independent frequencies straight from the start. $\omega_0=\sqrt{k/m}$ is a property of the oscillator alone — it never changes once $m$ and $k$ are fixed. $\omega$ is a knob you (or the driving agent) control, swept across a range of values in an experiment. The amplitude curve $A(\omega)$ is what you get by holding the oscillator fixed and scanning $\omega$ ; it is not a statement about how $\omega_0$ itself varies. This distinction is exactly what the resonance article builds on: resonance is the special value of $\omega$ (close to, but for finite damping not always exactly at, $\omega_0$ ) where $A(\omega)$ is maximised.

Figure

Amplitude response curve rising from a low plateau at small driving frequency to a peak near the natural frequency and falling off at high driving frequency, with a lower flatter curve for stronger damping

Derivation ASteady-state amplitude and phase lag by the phasor method

The goal is to solve the driven-damped equation $\ddot x + 2\gamma\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ for its steady-state part only, extracting the amplitude $A(\omega)$ and phase lag $\delta(\omega)$ as clean closed-form functions of the driving frequency, using the complex-exponential trick rather than brute-force trigonometric substitution.

Step 1. Write the drive as the real part of a complex exponential: $F_0\cos\omega t=\mathrm{Re}(F_0e^{i\omega t})$ . Since the equation is linear, look for a steady-state solution of the same complex form, $x=\mathrm{Re}(\tilde x_0e^{i\omega t})$ , where $\tilde x_0$ is a complex constant carrying both the amplitude and the phase.

Step 2. Substitute into the equation. Each derivative of $e^{i\omega t}$ brings down a factor of $i\omega$ , so $\ddot x\to -\omega^2\tilde x_0e^{i\omega t}$ and $\dot x\to i\omega\tilde x_0e^{i\omega t}$ . The equation becomes $\tilde x_0(-\omega^2+2i\gamma\omega+\omega_0^2)e^{i\omega t}=(F_0/m)e^{i\omega t}$ , and cancelling the common $e^{i\omega t}$ leaves an algebraic equation for $\tilde x_0$ instead of a differential one.

Step 3. Solve for $\tilde x_0$ : $\tilde x_0=\dfrac{F_0/m}{(\omega_0^2-\omega^2)+2i\gamma\omega}$ . This is a single complex number whose magnitude gives the real amplitude and whose argument gives the phase lag, because $x(t)=\mathrm{Re}(\tilde x_0 e^{i\omega t})=|\tilde x_0|\cos(\omega t+\arg\tilde x_0)$ .

Step 4. Take the magnitude of the denominator's complex conjugate pair to get $A(\omega)=|\tilde x_0|=\dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}}$ , and read off the phase from the denominator's real and imaginary parts: since $\tilde x_0$ has a negative argument when the imaginary part $2\gamma\omega$ is positive, the displacement lags the force by $\delta$ with $\tan\delta=2\gamma\omega/(\omega_0^2-\omega^2)$ , matching the steady-state form $x(t)=A(\omega)\cos(\omega t-\delta)$ . Result: $A(\omega) = \dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}}$ , $\tan\delta = \dfrac{2\gamma\omega}{\omega_0^2-\omega^2}$ .

Worked example

A driven oscillator has $m=0.20\ \text{kg}$ , $k=80\ \text{N/m}$ , and damping constant $b=1.0\ \text{kg/s}$ , driven by $F_0=2.0\ \text{N}$ at $\omega=15\ \text{rad/s}$ . Find the steady-state amplitude and phase lag, then find the driving frequency at which the phase lag is exactly $90^\circ$ and the amplitude there.

Step 1. Find $\omega_0$ and $\gamma$ : $\omega_0=\sqrt{k/m}=\sqrt{80/0.20}=\sqrt{400}=20\ \text{rad/s}$ , and $\gamma=b/2m=1.0/0.40=2.5\ \text{s}^{-1}$ .

Step 2. Compute the two pieces of the amplitude denominator: $\omega_0^2-\omega^2=400-225=175\ \text{rad}^2/\text{s}^2$ , and $2\gamma\omega=2\times2.5\times15=75\ \text{rad}^2/\text{s}^2$ . The denominator is $\sqrt{175^2+75^2}=\sqrt{30625+5625}=\sqrt{36250}\approx190.4$ , and with $F_0/m=2.0/0.20=10\ \text{N/kg}$ ,

A(\omega)=\frac{10}{190.4}\approx0.0525\ \text{m}\approx5.25\ \text{cm}.

Step 3. Phase lag: $\tan\delta=75/175=0.4286$ , so $\delta=\arctan(0.4286)\approx23.2^\circ\approx0.405\ \text{rad}$ . Since $\omega<\omega_0$ , $\delta$ stays below $90^\circ$ — the mass lags only modestly behind the force.

Step 4. For $\delta=90^\circ$ , $\tan\delta$ must diverge, which needs the denominator of the tangent to vanish: $\omega_0^2-\omega^2=0$ , i.e. $\omega=\omega_0=20\ \text{rad/s}$ , independent of $\gamma$ . There, $A(\omega_0)=\dfrac{F_0/m}{2\gamma\omega_0}=\dfrac{10}{2\times2.5\times20}=\dfrac{10}{100}=0.10\ \text{m}$ .

The amplitude has grown from 5.25 cm at $\omega=15\ \text{rad/s}$ to 10.0 cm at $\omega=\omega_0=20\ \text{rad/s}$ , where the phase lag is exactly $90^\circ$ — the velocity-resonance point where power transfer is maximum.

Common traps

Reading off the amplitude too soon: right after the drive switches on, the motion is a mix of the decaying transient (at the damped frequency $\omega_1$ ) and the steady state; only after several damping times $1/\gamma$ does the pure $A(\omega)\cos(\omega t-\delta)$ form apply.
Confusing $\omega_0$ (fixed natural frequency) with $\omega$ (the driving frequency you are varying) — $A$ and $\delta$ are functions of $\omega$ , not of $\omega_0$ .
Assuming the steady state runs at the damped natural frequency $\omega_1=\sqrt{\omega_0^2-\gamma^2}$ — it never does; the steady state always locks to the drive frequency $\omega$ .
Forcing $\delta$ into the range $-90^\circ$ to $90^\circ$ : when $\omega>\omega_0$ the denominator $\omega_0^2-\omega^2$ is negative, so $\delta$ must be taken in the second quadrant, between $90^\circ$ and $180^\circ$ .
Treating $F_0/k$ (the low-frequency amplitude limit) as if it were the amplitude at every driving frequency — it is only the special value the formula approaches as $\omega\to0$ , not a general property of driven motion.

Ask the tutor: Want a walkthrough deriving $A(\omega)$ from scratch by direct trigonometric substitution instead of the phasor method, or more practice plugging in different $m,k,b,F_0,\omega$ values?

Why does the steady-state motion ignore the system's own natural frequency completely?

Because the natural frequency only governs the transient part of the solution, which is the homogeneous response to the initial disturbance and decays as $e^{-\gamma t}$ . The particular (steady-state) solution is forced by the driving term $F_0\cos\omega t$ on the right-hand side, and a linear differential equation with a sinusoidal forcing term always responds at that same forcing frequency once the transient has died out. The system's own $\omega_0$ still shapes how large the response is at a given $\omega$ , but not what frequency it oscillates at.

Why does the phase lag reach exactly 90 degrees at omega equals omega0, no matter how much damping there is?

Look at $\tan\delta=2\gamma\omega/(\omega_0^2-\omega^2)$ : the denominator vanishes exactly when $\omega=\omega_0$ , regardless of the value of $\gamma$ , so the tangent diverges and $\delta=90^\circ$ there for any damping strength. Physically, at $\delta=90^\circ$ the displacement lags the force by a quarter cycle, which means the velocity (which itself leads displacement by $90^\circ$ ) is exactly in phase with the force — this is why $\omega=\omega_0$ is the frequency of maximum power delivery, independent of $\gamma$ .

Is the amplitude always largest exactly at omega equals omega0?

Not quite for the displacement amplitude — the exact peak of $A(\omega)$ shifts slightly below $\omega_0$ for finite damping, landing precisely at $\omega_0$ only in the light-damping limit; that fine detail is examined in the resonance article that follows this one. What is always true at $\omega=\omega_0$ , for any damping, is that the phase lag is exactly $90^\circ$ and the velocity amplitude (not displacement amplitude) peaks there — this is called velocity resonance, and it is the safer fact to quote if asked "what happens exactly at $\omega_0$ ."

Forced oscillations

The core idea

Worked example

Common traps

Before this

Where this leads