Periodic Forcing Can Produce Resonance

A periodically forced oscillator satisfies

$mx''+cx'+kx=F_0\cos(\omega t).$

Its response combines a transient determined by initial conditions with a steady periodic motion caused by the forcing.

Transient and steady response

Damping makes the homogeneous transient fade. The remaining steady-state response oscillates at the forcing frequency $\omega$, with an amplitude and phase shift determined by $m$, $c$, $k$, and $\omega$.

Resonant amplification

With weak damping, forcing near the natural frequency $\sqrt{k/m}$ produces a large response. Damping lowers and broadens the resonance peak. In the ideal undamped case, exact resonance makes amplitude grow with time.

Derive the amplitude response

For a steady trial response $x_p=A\cos(\omega t-\phi)$, coefficient matching gives

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

The phase satisfies $\tan\phi=c\omega/(k-m\omega^2)$ with quadrant chosen correctly. The denominator shows the competition between stiffness, inertia, and damping.

Check your understanding

Why is the steady response at the forcing frequency rather than the natural frequency?

Show the reasoning

The particular solution inherits the periodic input’s frequency. Natural-frequency components belong to the homogeneous transient and decay when damping is present.