How to Find the Resonant Amplitude of a Forced Oscillator

A resonant amplitude is the size of an oscillator’s response when a periodic force drives it near a natural frequency. To calculate it, you need the oscillator’s mass, stiffness, damping, and the driving frequency. The key distinction is that an ideal undamped oscillator has no finite steady-state amplitude at exact resonance, whereas damping makes the steady-state response finite.

Start with the forced-oscillator equation

For a mass $m$ attached to a spring with stiffness $k$, subject to damping coefficient $c$ and driving force $F_0\cos(\omega t)$, the equation is

$$m\ddot{x}+c\dot{x}+kx=F_0\cos(\omega t).$$

Here $F_0$ is the force amplitude and $\omega$ is the driving angular frequency, measured in radians per second. The undamped natural angular frequency is

$$\omega_0=\sqrt{\frac{k}{m}}.$$

After any initial transient has died away, a damped oscillator moves at the driving frequency. Its steady-state displacement can be written $x_{\mathrm{ss}}(t)=A\cos(\omega t-\delta)$, where the amplitude is

$$A(\omega)=\frac{F_0}{\sqrt{(k-m\omega^2)^2+(c\omega)^2}}.$$

This is the formula to use when the oscillator has damping. Substitute the given values of $m$, $k$, $c$, $F_0$, and $\omega$ to find the steady-state amplitude.

Evaluate the response at resonance

In many introductory problems, “resonance” means driving at the undamped natural frequency, $\omega=\omega_0$. Substitution gives

$$A(\omega_0)=\frac{F_0}{c\omega_0}$$

when $c>0$, since $k=m\omega_0^2$ and the stiffness and inertia terms cancel. The formula shows why reducing damping increases the resonant response.

Worked example. Let $m=1\,\mathrm{kg}$, $k=100\,\mathrm{N/m}$, $c=1\,\mathrm{kg/s}$, and $F_0=2\,\mathrm{N}$. The undamped natural frequency is

$$\omega_0=\sqrt{\frac{100}{1}}=10\,\mathrm{rad/s}.$$

Driving at $10\,\mathrm{rad/s}$ gives the steady-state amplitude

$$A(\omega_0)=\frac{2}{1\cdot 10}=0.20\,\mathrm{m}.$$

This is the steady-state displacement amplitude, not necessarily the oscillator’s total displacement while its initial transient is still present.

Why an undamped oscillator has no finite resonant amplitude

If $c=0$, the steady-state formula away from resonance reduces to

$$A(\omega)=\frac{F_0}{|k-m\omega^2|}.$$

This works when $\omega\ne\omega_0$, but its denominator becomes zero at $\omega=\omega_0$. That does not mean the oscillator reaches an infinitely large displacement instantly. It means there is no bounded sinusoidal steady-state solution at exact resonance. Instead, the resonant response grows with time.

For example, if the oscillator starts from rest and is driven exactly at $\omega_0$, its displacement is

$$x(t)=\frac{F_0}{2m\omega_0}\,t\sin(\omega_0t).$$

The oscillation’s envelope grows linearly as $F_0t/(2m\omega_0)$. Thus, to give an amplitude for an ideal undamped oscillator at resonance, you must specify a time or another physical effect that limits the growth, such as damping.

For the example above with damping removed, driving at $8\,\mathrm{rad/s}$ instead gives a finite amplitude:

$$A(8)=\frac{2}{|100-1\cdot 8^2|}=\frac{2}{36}\approx 0.0556\,\mathrm{m}.$$

At $10\,\mathrm{rad/s}$, the undamped steady-state formula no longer gives a finite answer; the time-dependent resonant solution applies instead.

The frequency of maximum response can shift

With damping, “resonant frequency” can mean the undamped natural frequency $\omega_0$, or the frequency at which the displacement amplitude is largest. These are not always identical. For the amplitude formula above, a nonzero peak occurs at

$$\omega_{\mathrm{peak}}=\sqrt{\omega_0^2-\frac{c^2}{2m^2}}$$

when $c^2<2mk$. For light damping, this peak lies close to $\omega_0$. If a problem explicitly defines resonance as driving at the natural frequency, use $\omega_0$; if it asks for the frequency of maximum steady-state displacement, use $\omega_{\mathrm{peak}}$ when the stated condition holds.

Common mistakes

• Using the undamped formula at exact resonance. The denominator is zero, so that formula is not valid there. The ideal oscillator’s response grows over time.
• Leaving damping out of a real oscillator. A finite resonant amplitude depends on damping; include the $c\dot{x}$ term when the model specifies it.
• Confusing angular frequency and ordinary frequency. If frequency is given in hertz, convert using $\omega=2\pi f$ before substituting.
• Calling every amplitude a steady-state amplitude. Initial conditions add a transient response. The formula for $A(\omega)$ describes the steady-state part after that transient has decayed.
• Assuming the peak always occurs at $\omega_0$. With damping, the frequency of maximum displacement can differ slightly from the undamped natural frequency.