Recipe

Find the amplitude of a driven system's steady-state response

Use when: you have a forced equation $mx''+cx'+kx=F_0\cos(\omega t)$ and need the size of the steady-state oscillation at one specific driving frequency $\omega$

Goal: the number $X(\omega)$ , the steady-state amplitude at that driving frequency, plus whether it sits near the resonance peak

Taught in Amplitude response and resonance

Quick recipe

Read off $m$ , $c$ , $k$ , $F_0$ , and $\omega$
Compute the natural frequency $\omega_0=\sqrt{k/m}$
Compute $k-m\omega^2$ and $c\omega$ separately
Combine into $X(\omega)$
Compare $X(\omega)$ to the resonance peak

1
Identify the mass, damping, stiffness, forcing amplitude, and driving frequency directly from the equation $mx''+cx'+kx=F_0\cos(\omega t)$ .

Example

For $x''+2x'+25x=10\cos(4t)$ : $m=1$ , $c=2$ , $k=25$ , $F_0=10$ , $\omega=4$ .

Why: every later step is built entirely from these five numbers; misreading one here propagates through the whole computation.
2
This is the frequency the unforced system would oscillate at on its own, and the reference point resonance is measured against.

Example

$\omega_0=\sqrt{25/1}=5$ .

Why: comparing $\omega$ to $\omega_0$ is what tells you how close the driving frequency sits to the peak, before the amplitude itself is even computed.
3
These are the two terms combined under the square root in Step 4; compute each one on its own first.

Example

$k-m\omega^2=25-1(4)^2=25-16=9$ ; $c\omega=2(4)=8$ .

Why: computing them separately catches a sign error or an order-of-operations slip (squaring $\omega$ before multiplying by $m$ ) before it's buried inside a square root.
4
Plug both building blocks from Step 3 into the amplitude formula: $X(\omega)=\frac{F_0}{\sqrt{(k-m\omega^2)^2+(c\omega)^2}}$

Example

$X=10/\sqrt{9^2+8^2}=10/\sqrt{145}\approx0.830$ .

⚡ Shortcut

If the system is undamped ( $c=0$ ), the formula collapses to $X=F_0/|k-m\omega^2|$ (no square root of a sum needed), and it blows up (undefined) exactly at $\omega=\omega_0$ , the true, unbounded resonance that damping otherwise keeps finite.

Why: this is the definition of the amplitude response, the size of the steady-state oscillation once every transient has died out.
5
Decide how close this driving frequency sits to the system's loudest response.
- $\omega$ close to $\omega_0$ , with $c$ small — $X(\omega)$ sits near its maximum, close to the resonance peak.
- $\omega$ far from $\omega_0$ , or $c$ large — $X(\omega)$ stays well below the peak, even for the same $F_0$ .
Why: the formula's whole point is this comparison; a raw number for $X(\omega)$ means little without knowing where it sits relative to $\omega_0$ .

Common mistakes

Using $\omega_0$ in the amplitude formula instead of the actual driving frequency $\omega$ .
Multiplying by $m$ before squaring $\omega$ (computing $(m\omega)^2$ instead of $m\omega^2$ ).
Forgetting to square $(k-m\omega^2)$ and $(c\omega)$ before adding them under the square root.
Treating a large but finite amplitude near $\omega=\omega_0$ as literally unbounded when $c\ne0$ . Only the undamped shortcut case truly blows up.