| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.3 This is a standard Further Pure 1 second-order differential equations question testing classification of damped harmonic motion. Part (a) requires recognizing SHM occurs when λ=0 (trivial recall), part (b) tests understanding of overdamping conditions (solving λ²-12≥0), and part (c) requires sketching underdamped motion. All parts follow textbook templates with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (a) | (i) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | (ii) | The door should close, but in SHM the motion continues |
| indefinitely | B1 | 3.5b |
| Answer | Marks |
|---|---|
| (b) | Over- or critical- damping implies λ2 −12≥0 |
| So λ≥2 3 | M1 |
| A1 | 3.3 |
| 3.4 | Consider discriminant with ≥or> |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | e.g. | B1 |
Question 11:
11 | (a) | (i) | For SHM λ = 0 | B1 | 3.3
[1]
(a) | (ii) | The door should close, but in SHM the motion continues
indefinitely | B1 | 3.5b
[1]
(b) | Over- or critical- damping implies λ2 −12≥0
So λ≥2 3 | M1
A1 | 3.3
3.4 | Consider discriminant with ≥or>
Ignore λ≤-2 3
[2]
(c) | e.g. | B1 | 3.4 | Graph of under-damped system.
Start anywhere non-zero on θ-axis with zero
gradient.
Each peak must be lower than before
At least two peaks (not including start point)
The graph must look as though it is approaching the t
axis
[1]
PMT
Y540/01 Mark scheme October 2021
Appendix
8(b) Alternate solution
4sinhx+3coshx=5⇒4sinhx=5−3coshx
∴16sinh2x=16(cosh2x−1)=25−30coshx+9cosh2x
7cosh2x+30coshx−41=0
−30± 302−4×7×−41 −30± 2048
coshx= =
2×7 14
−30+32 2 −15+16 2
coshx≥1⇒coshx= =
14 7
⇒x=cosh−1 −15+16 2 =±ln −15+16 2 + −15+16 2 2 −1
7 7 7
−15+16 2 225+512−480 2 49
=±ln + −
7 49 49
−15+16 2 688−480 2 −15+16 2+4 43−30 2
=±ln + =±ln
7 49 7
But the negative root does not work in the original
equation since LHS would be negative while RHS
would be positive (but equal when squared).
−15+16 2+4 43−30 2
∴x=ln
7
( )2
NB 5−3 2 =25+18−30 2
=43−30 2 and 5−3 2>0
∴x=ln
−15+16 2+4 ( 5−3 2 )
=ln
5+4 2
7 7
18
PMT
Y540/01 Mark scheme October 2021
Question 2(a)(ii) Alternative solution
y =tan−1( 1+x )⇒1+x=tan y
dy
⇒1=sec2 y.
dx
dy 1 1 1
⇒ = = =
dx sec2 y 1+tan2 y 1+( 1+x )2
19
PMT
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recorded or monitoredThe displacement of a door from its equilibrium (closed) position is measured by the angle, $\theta$ radians, which the door makes with its closed position. The door can swing either side of the equilibrium position so that $\theta$ can take positive and negative values. The door is released from rest from an open position at time $t = 0$.
A proposed differential equation to model the motion of the door for $t \geqslant 0$ is
$$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} + \lambda \frac{\mathrm{d}\theta}{\mathrm{d}t} + 3\theta = 0$$ where $\lambda$ is a constant and $\lambda \geqslant 0$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item According to the model, for what value of $\lambda$ will the motion of the door be simple harmonic? [1]
\item Explain briefly why modelling the motion of the door as simple harmonic is unlikely to be realistic. [1]
\end{enumerate}
\item Find the range of values of $\lambda$ for which the model predicts that the door will never pass through the equilibrium position. [2]
\item Sketch a possible graph of $\theta$ against $t$ when $\lambda$ lies \textbf{outside} the range found in part (b) but the motion is not simple harmonic. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2021 Q11 [5]}}