Core Pure · Modelling with Differential Equations · Year 2

Damped & Forced Harmonic Motion

From decaying oscillation to resonance — modelling resistance and driving forces with second-order differential equations.

13 min read 2 worked examples

Damped harmonic motion

Setting up the equation of motion

When a resistive force proportional to velocity acts alongside a restoring force proportional to displacement, Newton's second law $F=m\ddot x$ gives an equation of the form

$$m\ddot x=-kx-c\dot x \quad\Longrightarrow\quad \ddot x+\frac cm\dot x+\frac km x=0$$

a second-order homogeneous equation whose auxiliary equation's discriminant determines the qualitative behaviour: complex roots give oscillatory (underdamped) motion decaying in amplitude; a repeated real root gives critical damping (fastest return to equilibrium with no oscillation); distinct real roots give overdamping (slow, non-oscillatory return).

Underdamped (complex roots) — the motion oscillates while its amplitude decays, trapped between the envelope $\pm x_0e^{-kt}$.

Critically damped (repeated real root) — the particle returns to equilibrium as fast as possible without oscillating, never crossing $x=0$.

Overdamped (distinct real roots) — the particle also returns without oscillating, but more sluggishly than the critically damped case.

Example 1 · Setting up and solving a damped motion problem

A particle P of mass $2$ kg moves on the $x$-axis. A restoring force of magnitude $18x$ N and a resistance of magnitude $12\dot x$ N act on P. Given that $x=3$ and $\dot x=0$ when $t=0$, find $x$ in terms of $t$, and describe the motion.

Equation of motion: $2\ddot x=-18x-12\dot x$, i.e. $\ddot x+6\dot x+9x=0$.

Auxiliary equation: $m^2+6m+9=0\Rightarrow(m+3)^2=0\Rightarrow m=-3$ (repeated).

General solution: $x=(A+Bt)e^{-3t}$, so $\dot x=Be^{-3t}-3(A+Bt)e^{-3t}$.

At $t=0$: $x=A=3$; $\dot x=B-3A=0\Rightarrow B=9$.

So $x=(3+9t)e^{-3t}$. Since the auxiliary equation has a repeated root, this is critically damped motion — the particle returns to equilibrium as quickly as possible without oscillating (it never crosses $x=0$ for $t>0$, since $3+9t>0$ throughout).

Forced harmonic motion

Setting up with an external driving force

If an additional periodic external force $F_0\cos pt$ (or $\sin pt$) is applied, the equation of motion becomes non-homogeneous:

$$\ddot x+\frac cm\dot x+\frac km x=\frac{F_0}m\cos pt$$

The general solution is complementary function (the "transient" response, which typically decays if damping is present) plus a particular integral (the "steady-state" oscillation driven by the forcing term).

Example 2 · Forced motion without damping

A particle P of mass $1$ kg moves on the $x$-axis. A restoring force of magnitude $16x$ N acts on P, together with an applied force of $10\cos2t$ N. Given that $x=0$ and $\dot x=0$ when $t=0$, find $x$ in terms of $t$.

Equation of motion: $\ddot x+16x=10\cos2t$.

Complementary function: $m^2+16=0\Rightarrow m=\pm4i\Rightarrow x_{CF}=A\cos4t+B\sin4t$.

Particular integral: try $x_{PI}=\lambda\cos2t+\mu\sin2t$ (no clash with the CF, since $2\neq4$). Then $\ddot x_{PI}=-4\lambda\cos2t-4\mu\sin2t$.

Substitute: $(-4\lambda+16\lambda)\cos2t+(-4\mu+16\mu)\sin2t=10\cos2t\Rightarrow12\lambda=10,\;12\mu=0$.

So $\lambda=\dfrac56$, $\mu=0$, giving $x_{PI}=\dfrac56\cos2t$.

General solution: $x=A\cos4t+B\sin4t+\dfrac56\cos2t$, so $\dot x=-4A\sin4t+4B\cos4t-\dfrac53\sin2t$.

At $t=0$: $x=A+\dfrac56=0\Rightarrow A=-\dfrac56$; $\dot x=4B=0\Rightarrow B=0$.

So $x=\dfrac56\cos2t-\dfrac56\cos4t=\dfrac56(\cos2t-\cos4t)$.

Resonance

If the driving frequency $p$ exactly matches the natural frequency $\omega$ of an undamped system, the naive particular-integral trial clashes with the complementary function, forcing a trial of the form $t(\lambda\cos\omega t+\mu\sin\omega t)$ instead (as in the second-order non-homogeneous technique from the previous lesson). The resulting particular integral grows without bound as $t$ increases — this is resonance, and it is the reason engineers avoid designing structures with a natural frequency close to an expected driving frequency.

Exam-style questions

Question 1 9 marks

A particle P of mass $1$ kg moves on the $x$-axis, subject to a restoring force of magnitude $13x$ N and a resistance of magnitude $4\dot x$ N. Given that $x=2$ and $\dot x=0$ when $t=0$, find $x$ in terms of $t$, and describe the motion.

Equation of motion: $\ddot x+4\dot x+13x=0$. (M1, A1)

$m^2+4m+13=0\Rightarrow m=-2\pm3i$, so $x=e^{-2t}(A\cos3t+B\sin3t)$. (M1, A1)

$\dot x=-2e^{-2t}(A\cos3t+B\sin3t)+e^{-2t}(-3A\sin3t+3B\cos3t)$. At $t=0$: $x=A=2$. (M1, A1)

$\dot x=-2A+3B=0\Rightarrow B=\dfrac43$. So $x=e^{-2t}\left(2\cos3t+\dfrac43\sin3t\right)$. (A1)

Since the roots are complex, this is underdamped (oscillatory) motion, with amplitude decaying as $e^{-2t}$. (A1)

Question 2 10 marks

A particle P of mass $1$ kg moves on the $x$-axis. A restoring force of magnitude $9x$ N acts on P, together with an applied force of $12\sin3t$ N (matching the natural frequency). Given that $x=0$ and $\dot x=0$ when $t=0$, find $x$ in terms of $t$.

Equation of motion: $\ddot x+9x=12\sin3t$. C.F.: $m^2+9=0\Rightarrow x_{CF}=A\cos3t+B\sin3t$. (M1, A1)

Since $\sin3t$ clashes with the C.F. (resonance), try $x_{PI}=t(\lambda\cos3t+\mu\sin3t)$. (M1)

Differentiating twice (product rule) and substituting gives, after collecting terms, $-6\lambda\sin3t+6\mu\cos3t=12\sin3t$, so $\mu=0,\lambda=-2$. (M1, A1, A1)

General solution: $x=A\cos3t+B\sin3t-2t\cos3t$; $\dot x=-3A\sin3t+3B\cos3t-2\cos3t+6t\sin3t$. (A1)

At $t=0$: $x=A=0$; $\dot x=3B-2=0\Rightarrow B=\dfrac23$. (A1)

$x=\dfrac23\sin3t-2t\cos3t$ — note the amplitude grows without bound as $t$ increases (resonance). (A1)

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Simple Harmonic Motion

Coupled Simultaneous Equations