| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Resonance cases requiring modified PI |
| Difficulty | Challenging +1.2 This is a standard resonance case differential equation requiring the modified particular integral θ = t(A sin 3t + B cos 3t). While resonance is conceptually important, the question heavily scaffolds part (a)(i) by giving the PI, reducing it to verification. Parts (b)-(d) involve routine application of initial conditions, numerical evaluation, and conceptual understanding of SHM vs forced motion. The topic is Further Maths level, but the execution is mostly procedural with minimal problem-solving demand. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d\theta}{dt} = \alpha\sin 3t + \beta t\cos 3t\) and \(\frac{d^2\theta}{dt^2} = \delta\cos 3t + \gamma t\sin 3t\) | M1 | Differentiates PI twice using product rule to achieve required form |
| \(\frac{d\theta}{dt} = \frac{1}{12}\sin 3t + \frac{1}{4}t\cos 3t\) and \(\frac{d^2\theta}{dt^2} = \frac{1}{2}\cos 3t - \frac{3}{4}t\sin 3t\) | A1 | Correct derivatives |
| \(\frac{1}{2}\cos 3t - \frac{3}{4}t\sin 3t + 9\left(\frac{1}{12}t\sin 3t\right) = \frac{1}{2}\cos 3t\) | dM1 | Substitutes into DE and simplifies. Depends on first M |
| \(= \frac{1}{2}\cos 3t\) so PI is \(\theta = \frac{1}{12}t\sin 3t\) | A1* | Achieves \(\frac{1}{2}\cos 3t\) and makes minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m^2 + 9 = 0 \Rightarrow m = \pm 3i\) | M1 | Forms and solves auxiliary equation |
| \(\theta = A\cos 3t + B\sin 3t\) | A1 | Correct complementary function in terms of \(t\) |
| \((\theta =) CF + PI\) | dM1 | Adds PI to CF for general solution |
| \(\theta = A\cos 3t + B\sin 3t + \frac{1}{12}t\sin 3t\) | A1 | Correct general solution including "\(\theta =\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, \, \theta = \frac{\pi}{3} \Rightarrow A = \frac{\pi}{3}\) | M1 | Uses \(t=0\), \(\theta = \frac{\pi}{3}\) to find a constant |
| \(t=0, \, \frac{d\theta}{dt} = -3A\sin 3t + 3B\cos 3t + \frac{1}{12}\sin 3t + \frac{1}{4}t\cos 3t = 0 \Rightarrow B = 0\) | M1 | Differentiates and uses \(t=0\), \(\frac{d\theta}{dt}=0\) |
| \(\alpha = \frac{\pi}{3}\cos(3\times10) + \frac{1}{12}(10)\sin(3\times10) = \ldots\) | ddM1 | Substitutes \(t=10\) into particular solution. Depends on both M marks |
| \(\alpha = \pm\) awrt \(0.662\) | A1 | Accept \(\pm 0.662\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.662\) is close to \(0.62\) so a good model (at \(t=10\)) | B1ft | Quantitative comparison with \(0.62\) and appropriate conclusion. Follow through on part (b) answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2\theta}{dt^2} + 9\theta = 0\) or equivalent | B1 | Refines the model; accept any constant on the right hand side |
# Question 10(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d\theta}{dt} = \alpha\sin 3t + \beta t\cos 3t$ and $\frac{d^2\theta}{dt^2} = \delta\cos 3t + \gamma t\sin 3t$ | M1 | Differentiates PI twice using product rule to achieve required form |
| $\frac{d\theta}{dt} = \frac{1}{12}\sin 3t + \frac{1}{4}t\cos 3t$ and $\frac{d^2\theta}{dt^2} = \frac{1}{2}\cos 3t - \frac{3}{4}t\sin 3t$ | A1 | Correct derivatives |
| $\frac{1}{2}\cos 3t - \frac{3}{4}t\sin 3t + 9\left(\frac{1}{12}t\sin 3t\right) = \frac{1}{2}\cos 3t$ | dM1 | Substitutes into DE and simplifies. Depends on first M |
| $= \frac{1}{2}\cos 3t$ so PI is $\theta = \frac{1}{12}t\sin 3t$ | A1* | Achieves $\frac{1}{2}\cos 3t$ and makes minimal conclusion |
# Question 10(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 9 = 0 \Rightarrow m = \pm 3i$ | M1 | Forms and solves auxiliary equation |
| $\theta = A\cos 3t + B\sin 3t$ | A1 | Correct complementary function in terms of $t$ |
| $(\theta =) CF + PI$ | dM1 | Adds PI to CF for general solution |
| $\theta = A\cos 3t + B\sin 3t + \frac{1}{12}t\sin 3t$ | A1 | Correct general solution including "$\theta =$" |
# Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, \, \theta = \frac{\pi}{3} \Rightarrow A = \frac{\pi}{3}$ | M1 | Uses $t=0$, $\theta = \frac{\pi}{3}$ to find a constant |
| $t=0, \, \frac{d\theta}{dt} = -3A\sin 3t + 3B\cos 3t + \frac{1}{12}\sin 3t + \frac{1}{4}t\cos 3t = 0 \Rightarrow B = 0$ | M1 | Differentiates and uses $t=0$, $\frac{d\theta}{dt}=0$ |
| $\alpha = \frac{\pi}{3}\cos(3\times10) + \frac{1}{12}(10)\sin(3\times10) = \ldots$ | ddM1 | Substitutes $t=10$ into particular solution. Depends on both M marks |
| $\alpha = \pm$ awrt $0.662$ | A1 | Accept $\pm 0.662$ |
# Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.662$ is close to $0.62$ so a good model (at $t=10$) | B1ft | Quantitative comparison with $0.62$ and appropriate conclusion. Follow through on part (b) answer |
# Question 10(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2\theta}{dt^2} + 9\theta = 0$ or equivalent | B1 | Refines the model; accept any constant on the right hand side |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f237de57-ed6d-4bea-8bb0-1b4e5b66d7da-28_428_301_246_881}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The motion of a pendulum, shown in Figure 3, is modelled by the differential equation
$$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 9 \theta = \frac { 1 } { 2 } \cos 3 t$$
where $\theta$ is the angle, in radians, that the pendulum makes with the downward vertical, $t$ seconds after it begins to move.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that a particular solution of the differential equation is
$$\theta = \frac { 1 } { 12 } t \sin 3 t$$
\item Hence, find the general solution of the differential equation.
Initially, the pendulum
\begin{itemize}
\end{enumerate}\item makes an angle of $\frac { \pi } { 3 }$ radians with the downward vertical
\item is at rest
\end{itemize}
Given that, 10 seconds after it begins to move, the pendulum makes an angle of $\alpha$ radians with the downward vertical,
\item determine, according to the model, the value of $\alpha$ to 3 significant figures.
Given that the true value of $\alpha$ is 0.62
\item evaluate the model.
The differential equation
$$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 9 \theta = \frac { 1 } { 2 } \cos 3 t$$
models the motion of the pendulum as moving with forced harmonic motion.
\item Refine the differential equation so that the motion of the pendulum is simple harmonic motion.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2022 Q10 [14]}}