| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.3 This is a standard second-order differential equation with constant coefficients requiring the auxiliary equation method to find complex roots, leading to the form e^(αt)(A cos βt + B sin βt). Applying initial conditions and finding the maximum displacement are routine procedures covered in Core Pure 2. The question is slightly easier than average because it follows a predictable template with no novel insights required, though it does require careful algebraic manipulation. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4m^2+4m+37=0 \Rightarrow m = -\frac{1}{2}\pm 3i\) | M1 | Uses model to form and solve auxiliary equation \(4m^2+4m+37=0\) |
| \(h = e^{-0.5t}(A\cos3t+B\sin3t)\) | A1 | Correct general solution including \(h=\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0\), \(h=-20 \Rightarrow A=-20\) | M1 | Applies initial condition |
| \(\frac{dh}{dt} = -0.5e^{-0.5t}(A\cos3t+B\sin3t)+e^{-0.5t}(-3A\sin3t+3B\cos3t)\); \(t=0\), \(\frac{dh}{dt}=55 \Rightarrow B=15\) | M1 | Differentiates and applies second condition |
| \(h = e^{-0.5t}(15\sin3t-20\cos3t)\) | A1 | Correct particular solution |
| \(-0.5e^{-0.5t}(15\sin3t-20\cos3t)+e^{-0.5t}(60\sin3t+45\cos3t)=0\) | M1 | Sets \(\frac{dh}{dt}=0\) and attempts to solve |
| \(\tan3t = -\frac{22}{21}\) or \(3t+\tan^{-1}\frac{22}{21}=0\) | A1, M1 on ePEN | Correct equation for \(t\) |
| \(t = 0.778\) s | A1 | Correct value of \(t\) |
| \(h = e^{-0.5\times0.778}(15\sin(3\times0.778)-20\cos(3\times0.778))\) | dM1 | Substitutes \(t\) value into \(h\) |
| \(= 16.7\) cm | A1 | Correct value of \(h\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Considers large values of \(t\) in model for \(h\), or states for large \(t\), \(h\) becomes smaller/zero | M1 | Must reference behaviour of model for large \(t\) |
| E.g. \(h\) is very small for large \(t\), suggesting model suitable as displacement gets smaller; board tends to equilibrium; but \(h\) never exactly zero so not fully appropriate for large \(t\) | A1, B1 on ePEN | Credit for valid comment about suitability; B1 for additional qualifying remark about limitation |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4m^2+4m+37=0 \Rightarrow m = -\frac{1}{2}\pm 3i$ | M1 | Uses model to form and solve auxiliary equation $4m^2+4m+37=0$ |
| $h = e^{-0.5t}(A\cos3t+B\sin3t)$ | A1 | Correct general solution including $h=$ |
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0$, $h=-20 \Rightarrow A=-20$ | M1 | Applies initial condition |
| $\frac{dh}{dt} = -0.5e^{-0.5t}(A\cos3t+B\sin3t)+e^{-0.5t}(-3A\sin3t+3B\cos3t)$; $t=0$, $\frac{dh}{dt}=55 \Rightarrow B=15$ | M1 | Differentiates and applies second condition |
| $h = e^{-0.5t}(15\sin3t-20\cos3t)$ | A1 | Correct particular solution |
| $-0.5e^{-0.5t}(15\sin3t-20\cos3t)+e^{-0.5t}(60\sin3t+45\cos3t)=0$ | M1 | Sets $\frac{dh}{dt}=0$ and attempts to solve |
| $\tan3t = -\frac{22}{21}$ or $3t+\tan^{-1}\frac{22}{21}=0$ | A1, M1 on ePEN | Correct equation for $t$ |
| $t = 0.778$ s | A1 | Correct value of $t$ |
| $h = e^{-0.5\times0.778}(15\sin(3\times0.778)-20\cos(3\times0.778))$ | dM1 | Substitutes $t$ value into $h$ |
| $= 16.7$ cm | A1 | Correct value of $h$ |
# Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Considers large values of $t$ in model for $h$, or states for large $t$, $h$ becomes smaller/zero | M1 | Must reference behaviour of model for large $t$ |
| E.g. $h$ is very small for large $t$, suggesting model suitable as displacement gets smaller; board tends to equilibrium; but $h$ never exactly zero so not fully appropriate for large $t$ | A1, B1 on ePEN | Credit for valid comment about suitability; B1 for additional qualifying remark about limitation |\begin{enumerate}
\item An engineer is investigating the motion of a sprung diving board at a swimming pool.
\end{enumerate}
Let $E$ be the position of the end of the diving board when it is at rest in its equilibrium position and when there is no diver standing on the diving board.\\
A diver jumps from the diving board.\\
The vertical displacement, $h \mathrm {~cm}$, of the end of the diving board above $E$ is modelled by the differential equation
$$4 \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} h } { \mathrm {~d} t } + 37 h = 0$$
where $t$ seconds is the time after the diver jumps.\\
(a) Find a general solution of the differential equation.
When $t = 0$, the end of the diving board is 20 cm below $E$ and is moving upwards with a speed of $55 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$.\\
(b) Find, according to the model, the maximum vertical displacement of the end of the diving board above $E$.\\
(c) Comment on the suitability of the model for large values of $t$.
\hfill \mbox{\textit{Edexcel CP2 2019 Q5 [12]}}