Forced Oscillator — Resonant and Non-Resonant Forcing Together: 04-BS-1, December 2016, Q2

Question 2: Forced Oscillator — Resonant and Non-Resonant Forcing Together 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An undamped oscillator with natural frequency $\omega_0=2$, forced simultaneously at its own frequency ($3\cos2t$, resonant) and at a different frequency ($4\cos3t$, non-resonant).

Find. The general solution $x(t)$.

Approach. Solve the homogeneous equation, then find a particular solution for each forcing term separately (superposition), using the resonant trial $t(A\cos2t+B\sin2t)$ for the $2\cos t$-frequency term and the ordinary trial $C\cos3t+D\sin3t$ for the other.

Homogeneous solution. $r^2+4=0\Rightarrow r=\pm2i\Rightarrow x_h(t)=C_1\cos2t+C_2\sin2t$.
Resonant particular solution (for $3\cos2t$). Since $\cos2t$ solves the homogeneous equation, try $x_{p1}=t(A\cos2t+B\sin2t)$. Substituting and simplifying (the $t\cos2t,\,t\sin2t$ terms cancel automatically, leaving only the terms produced by differentiating the envelope $t$ itself): $$x_{p1}''+4x_{p1}=-4A\sin2t+4B\cos2t=3\cos2t\ \Rightarrow\ A=0,\ B=\tfrac34.$$ So $x_{p1}=\tfrac34\,t\sin2t$.
Non-resonant particular solution (for $4\cos3t$). Try $x_{p2}=C\cos3t+D\sin3t$: $x_{p2}''+4x_{p2}=(-9+4)(C\cos3t+D\sin3t)=-5(C\cos3t+D\sin3t)=4\cos3t\Rightarrow C=-\tfrac45,\ D=0$. So $x_{p2}=-\tfrac45\cos3t$.
Superpose. $$x(t)=\boxed{C_1\cos2t+C_2\sin2t+\dfrac34\,t\sin2t-\dfrac45\cos3t}$$ (Check: direct substitution of the full expression into $x''+4x$ reproduces $3\cos2t+4\cos3t$ exactly.)

Quantity	Result
Homogeneous solution	$C_1\cos2t+C_2\sin2t$
Resonant term ($3\cos2t$)	$\tfrac34\,t\sin2t$
Non-resonant term ($4\cos3t$)	$-\tfrac45\cos3t$
$x(t)$	$C_1\cos2t+C_2\sin2t+\tfrac34\,t\sin2t-\tfrac45\cos3t$

Concept & Theory

Topic. Superposition of resonant and non-resonant forcing in a linear constant-coefficient ODE.

Key relations. By linearity, if $x''+\omega_0^2x=F_1(t)+F_2(t)$, the particular solution is $x_{p1}+x_{p2}$ where each $x_{pi}$ solves the equation with only $F_i$ on the right. A forcing term at the natural frequency $\omega_0$ needs the resonant trial $t(A\cos\omega_0t+B\sin\omega_0t)$; any other frequency uses the plain trial.

Why this works. Superposition holds because the differential operator $D^2+4$ is linear. The resonance multiplier $t$ is required specifically for the $\cos2t$ term because $\cos2t,\sin2t$ already solve the homogeneous equation — an un-multiplied trial would be linearly dependent on $x_h$ and could never balance the equation; the $\cos3t$ term needs no such fix since $3\ne\omega_0=2$.

Common pitfall. Applying the resonant multiplier $t$ to both forcing terms instead of only the one matching $\omega_0$; sign or coefficient-matching slips when separately equating the $\cos2t$ and $\sin2t$ coefficients that arise from differentiating $t(A\cos2t+B\sin2t)$ twice.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.8 Modeling: Forced Oscillations, Resonance).

Question 2 of 8: Forced Oscillator — Resonant and Non-Resonant Forcing Together

Question 2: Forced Oscillator — Resonant and Non-Resonant Forcing Together 20 marks