SHM & Damped/Forced Harmonic Motion
From clean oscillation to damping and resonance — modelling motion with second-order differential equations.
Simple harmonic motion, revisited
A particle undergoes simple harmonic motion (SHM) if its acceleration satisfies $\ddot x=-\omega^2x$, where $x$ is displacement from a fixed centre and $\omega$ is a positive constant. This is a second-order homogeneous differential equation with constant coefficients.
Auxiliary equation: $m^2+25=0\Rightarrow m=\pm5i$, so $x=A\cos5t+B\sin5t$.
$\dot x=-5A\sin5t+5B\cos5t$.
At $t=0$: $x=A=4$; $\dot x=5B=0\Rightarrow B=0$.
So $x=4\cos5t$. Here $\omega=5$, giving period $\dfrac{2\pi}5$ and amplitude $4$.
Damped harmonic 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).
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
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).
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 $x=\dfrac56\cos2t-\dfrac56\cos4t=\dfrac56(\cos2t-\cos4t)$.
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 6 marks
A particle moves such that $\ddot x=-49x$. Given that $x=0$ and $\dot x=14$ when $t=0$, find $x$ in terms of $t$, and state the amplitude of the motion.
$m^2+49=0\Rightarrow m=\pm7i\Rightarrow x=A\cos7t+B\sin7t$. (M1, A1)
$\dot x=-7A\sin7t+7B\cos7t$. At $t=0$: $x=A=0$; $\dot x=7B=14\Rightarrow B=2$. (M1, A1)
$x=2\sin7t$; amplitude $2$. (A1, A1)
Question 2 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 3 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)