| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | December |
| Marks | 14 |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a standard Further Maths second-order differential equations question covering simple harmonic motion, damped oscillations, and critical damping. While it requires multiple techniques (auxiliary equation, applying initial conditions, interpreting physical meaning), these are all routine procedures from the FP2 syllabus with no novel problem-solving required. The multi-part structure and context add some complexity but this remains a typical textbook-style question. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\frac{d^2\theta}{dt^2} = -\left(\frac{5}{2}\right)^2 \theta\) | M1 | |
| \(\theta = A\cos\omega t + B\sin\omega t\) or \(R\cos(\omega t + \phi)\) with any positive value for \(\omega\) | A1 | If M0 then SC1 for \(\theta = A\cos\frac{5}{2}t\) or \(\theta = A\sin\frac{5}{2}t\) |
| (a)(ii) The model predicts infinite oscillations of the same amplitude; in practice the amplitude must decrease over time. | E1 | |
| (b)(i) AE: \(4m^2 + 16m + 25 = 0\) | M1 | Writing down the AE correctly or using \(\theta = Ae^{mt}\) and substituting into (*) to derive a three term quadratic AE. |
| \(-2 \pm \frac{3}{2}i\) | A1 | |
| \(\theta = e^{-2t}\left(A\cos\frac{3}{2}t + B\sin\frac{3}{2}t\right)\) | A1ft | Their e'' (\(A\cos qt + B\sin qt\)) for solution of AE = \(p \pm qi\) |
| (b)(ii) \(t = 0, \theta = 0.9 \Rightarrow A = 0.9\) | B1 | |
| \(\frac{d\theta}{dt} = -2e^{-2t}\left(A\cos\frac{3}{2}t + B\sin\frac{3}{2}t\right) + e^{-2t}\left(-\frac{3}{2}A\sin\frac{3}{2}t + \frac{3}{2}B\cos\frac{3}{2}t\right)\) | M1 | Attempt to differentiate using the product and chain rules (A may be replaced by a number). |
| \(t = 0, \frac{d\theta}{dt} = 0 \Rightarrow -2A + \frac{3}{2}B = 0\) | M1 | Substituting \(t = 0\) into \(\frac{d\theta}{dt}\) to derive an equation in (A and) B |
| \(B = 1.2\) | A1 | |
| \(\theta = e^{-2t}\left(0.9\cos\frac{3}{2}t + 1.2\sin\frac{3}{2}t\right)\) | A1 | |
| (b)(iii) In the modified model \(\theta \to 0\) as \(t \to \infty\) oe | B1 | ie the amplitude decays etc |
| This is the behaviour we would expect to observe with a real swing door and so the model is an improvement. | B1 | |
| (c) Need \(4m^2 + \lambda m + 25 = 0\) to have repeated solutions so \(\lambda^2 - 4 \times 4 \times 25 = 0\) | M1 | Using "b² - 4ac" = 0 directly or \(2m + \frac{\lambda}{4}\right)^2 + 25 - \frac{\lambda^2}{16} = 0\) and equating part outside brackets to 0 |
| \(\lambda > 0 \Rightarrow \lambda = 20\) | A1 | Not -20 or ±20 |
**(a)(i)** $\frac{d^2\theta}{dt^2} = -\left(\frac{5}{2}\right)^2 \theta$ | M1 |
$\theta = A\cos\omega t + B\sin\omega t$ or $R\cos(\omega t + \phi)$ with any positive value for $\omega$ | A1 | If M0 then SC1 for $\theta = A\cos\frac{5}{2}t$ or $\theta = A\sin\frac{5}{2}t$
**(a)(ii)** The model predicts infinite oscillations of the same amplitude; in practice the amplitude must decrease over time. | E1 |
**(b)(i)** AE: $4m^2 + 16m + 25 = 0$ | M1 | Writing down the AE correctly or using $\theta = Ae^{mt}$ and substituting into (*) to derive a three term quadratic AE.
$-2 \pm \frac{3}{2}i$ | A1 |
$\theta = e^{-2t}\left(A\cos\frac{3}{2}t + B\sin\frac{3}{2}t\right)$ | A1ft | Their e'' ($A\cos qt + B\sin qt$) for solution of AE = $p \pm qi$
**(b)(ii)** $t = 0, \theta = 0.9 \Rightarrow A = 0.9$ | B1 |
$\frac{d\theta}{dt} = -2e^{-2t}\left(A\cos\frac{3}{2}t + B\sin\frac{3}{2}t\right) + e^{-2t}\left(-\frac{3}{2}A\sin\frac{3}{2}t + \frac{3}{2}B\cos\frac{3}{2}t\right)$ | M1 | Attempt to differentiate using the product and chain rules (A may be replaced by a number).
$t = 0, \frac{d\theta}{dt} = 0 \Rightarrow -2A + \frac{3}{2}B = 0$ | M1 | Substituting $t = 0$ into $\frac{d\theta}{dt}$ to derive an equation in (A and) B
$B = 1.2$ | A1 |
$\theta = e^{-2t}\left(0.9\cos\frac{3}{2}t + 1.2\sin\frac{3}{2}t\right)$ | A1 |
**(b)(iii)** In the modified model $\theta \to 0$ as $t \to \infty$ oe | B1 | ie the amplitude decays etc
This is the behaviour we would expect to observe with a real swing door and so the model is an improvement. | B1 |
**(c)** Need $4m^2 + \lambda m + 25 = 0$ to have repeated solutions so $\lambda^2 - 4 \times 4 \times 25 = 0$ | M1 | Using "b² - 4ac" = 0 directly or $2m + \frac{\lambda}{4}\right)^2 + 25 - \frac{\lambda^2}{16} = 0$ and equating part outside brackets to 0
$\lambda > 0 \Rightarrow \lambda = 20$ | A1 | Not -20 or ±20
---A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it.
The extent to which the door is open at any time, $t$ seconds, is measured by the angle at the hinge, $\theta$, which the plane of the door makes with the plane of the equilibrium position. See the diagram below.
\begin{tikzpicture}[thick]
% --- Parameters ---
\def\rodlen{6} % rod length
\def\rodw{0.25} % rod half-width
\def\ang{35} % angle theta (degrees)
% --- Wall bracket (small rectangle on the left) ---
\draw[fill=white] (-1.2, -0.55) rectangle (-0.25, 0.55);
% --- Pivot point (black dot) ---
\fill (0, 0) circle (0.18cm);
% --- Rod (rotated rectangle) ---
\begin{scope}[rotate=\ang]
\draw[fill=white]
(0, -\rodw) -- (\rodlen, -\rodw)
-- (\rodlen, \rodw) -- (0, \rodw) -- cycle;
\end{scope}
% --- Dashed equilibrium line ---
\draw[dashed, very thick] (0.3, 0) -- (9.5, 0);
% --- Angle arc with label theta ---
\draw (1.4, 0) arc[start angle=0, end angle=\ang-10, radius=1.4];
\node at ({1.85*cos(\ang/2)}, {1.85*sin(\ang/2)+0.1})
{$\theta$};
% --- "Equilibrium position" label with bracket ---
\draw (7.2, 0) -- (7.2, -0.6) -- (11.2, -0.6);
\node[anchor=north west, font=\large] at (7.25, -0.05)
{Equilibrium position};
\end{tikzpicture}
In an initial model of the motion of a certain swing door it is suggested that $\theta$ satisfies the following differential equation.
$$4\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} + 25\theta = 0 \quad (*)$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the general solution to (*). [2]
\item With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (*) is unlikely to be realistic. [1]
\end{enumerate}
\end{enumerate}
In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes
$$4\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} + \lambda\frac{\mathrm{d}\theta}{\mathrm{d}t} + 25\theta = 0 \quad (\dagger)$$
where $\lambda$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item In the case where $\lambda = 16$ the door is held open at an angle of $0.9$ radians and then released from rest at time $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item Find, in a real form, the general solution of ($\dagger$). [3]
\item Find the particular solution of ($\dagger$). [4]
\item With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ($\dagger$) improves the model. [2]
\end{enumerate}
\item Find the value of $\lambda$ for which the door is critically damped. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q10 [14]}}