Forced Mechanical Vibrations
Differential Equations · Section 3.8 · generated practice set
Transient versus steady state, the amplitude formula, beats, and why resonance needs zero damping.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
Add an external force: $$mu'' + gamma u' + ku = F_0cosomega t.$$ The answer is $u = u_h + U$, and which part matters depends on whether there is damping.
With damping ($gamma > 0$)
Every homogeneous solution decays, so $u_h$ is the transient and $U$ is the steady state — the only thing left for large $t$. Its amplitude is $$R = rac{F_0}{sqrt{(k - momega^2)^2 + gamma^2omega^2}}.$$ $R$ is largest near $omega^2 = dfrac{k}{m} - dfrac{gamma^2}{2m^2}$, just below the undamped natural frequency, and stays finite because $gamma > 0$.
Without damping, $omega eq omega_0$
$$U = rac{F_0}{omega_0^2 - omega^2}cosomega t.$$ The sign flips as $omega$ crosses $omega_0$: below it the mass moves with the force, above it against. Adding the free oscillation at $omega_0$ gives two nearby frequencies superposed — beats, an envelope oscillating at $dfrac{|omega_0-omega|}{2}$.
Without damping, $omega = omega_0$
Resonance. The forcing already solves the homogeneous equation, so the guess needs a factor of $t$: $$U = rac{F_0}{2omega_0},tsinomega_0 t.$$ The amplitude grows linearly forever. Real systems have damping, so the growth is bounded — but it can still be large enough to destroy the system.
Definitions and theorems
Worked example
Find the steady-state amplitude for $u'' + 2u' + 5u = 3cos(2t)$.
- Read off $m = 1$, $gamma = 2$, $k = 5$, $F_0 = 3$, $omega = 2$.
- Damping is present, so the homogeneous part dies out and the steady state is the particular solution.
- $k - momega^2 = 5 - 4 = 1$ and $gammaomega = 4$.
- $R = dfrac{3}{sqrt{1^2 + 4^2}} = dfrac{3}{sqrt{17}} approx 0.728$.
- Compare: the undamped natural frequency is $sqrt{5} approx 2.24$, close to the forcing frequency 2 — which is why the response is comparable to the forcing rather than much smaller.
Common mistakes
- Looking for resonance when damping is present. With $gamma > 0$ the amplitude has a finite maximum; unbounded resonance requires $gamma = 0$ exactly.
- Forgetting the factor of $t$ when $omega = omega_0$. The plain cosine guess gives $0 = F_0cosomega_0t$.
- Confusing beats with resonance. Beats happen when $omega$ is near $omega_0$ and the amplitude is modulated but bounded. Resonance is $omega = omega_0$ exactly.
- Calling the transient the steady state. The steady state is the particular solution; the homogeneous part is the transient, whatever the initial conditions were.
- Dividing by $omega^2 - omega_0^2$ instead of $omega_0^2 - omega^2$. The sign matters and is what tells you whether the response is in phase.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.