| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Challenging +1.2 This is a multi-part SHM question requiring understanding of the acceleration formula a = ω²(x-d) and standard techniques (finding centre, using v² = ω²(a²-x²), and time integration). Part (i) requires setting up an equation from the acceleration ratio, part (ii) is straightforward application of the velocity formula, and part (iii) involves integrating to find time between positions. While it requires multiple steps and careful algebra, these are standard Further Maths techniques without requiring novel insight or particularly complex problem-solving. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega^2(7.5 - d) = 2\omega^2(6.5 - d)\), \(d = 5.5\) | M1 A1 | Find \(d\) by relating accelerations at \(A\) and \(B\) (\(\omega^2\) may be omitted) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = 2\pi/T = 2\) | B1 | Find \(\omega\) from period \(T\) (may be implied) |
| \(a = 10/\omega^2 = 2.5\) [m] (ignore sign; FT on \(\omega\)) | B1\(\sqrt{}\) | Find amplitude \(a\) from max. acceleration 10 |
| \(v^2 = \omega^2(a^2 - x^2)\), \(x = 7 - d\) [= 1.5] | M1, M1 | Find speed \(v\) when \(OP = 7\) (2nd M1 dep. on 1st M1) |
| \(= 2^2(2.5^2 - 1.5^2) = 16\), \(v = 4\) [m s\(^{-1}\)] | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega^{-1}\sin^{-1}(x_B/a) - \omega^{-1}\sin^{-1}(x_A/a)\) or \(\omega^{-1}\cos^{-1}(x_A/a) - \omega^{-1}\cos^{-1}(x_B/a)\) | M1 | Find time from \(A\) to \(B\) from \(x = a\sin\omega t\) or \(a\cos\omega t\) (all terms must be correct; FT on \(a\), \(\omega\) for M1) |
| \(= \frac{1}{2}\sin^{-1}(2/2.5) - \frac{1}{2}\sin^{-1}(1/2.5)\) or \(\frac{1}{2}\cos^{-1}(1/2.5) - \frac{1}{2}\cos^{-1}(2/2.5)\) \(= \frac{1}{2}(0.9273 - 0.4115)\) or \(0.4636 - 0.2058\) or \(\frac{1}{2}(1.1593 - 0.6435)\) or \(0.5796 - 0.3218\) | A1 | To 3 d.p., AEF throughout |
| \(= 0.258\) [s] | A1 |
## Question 3:
### Part 3(i):
$\omega^2(7.5 - d) = 2\omega^2(6.5 - d)$, $d = 5.5$ | M1 A1 | Find $d$ by relating accelerations at $A$ and $B$ ($\omega^2$ may be omitted)
### Part 3(ii):
$\omega = 2\pi/T = 2$ | B1 | Find $\omega$ from period $T$ (may be implied)
$a = 10/\omega^2 = 2.5$ [m] (ignore sign; FT on $\omega$) | B1$\sqrt{}$ | Find amplitude $a$ from max. acceleration 10
$v^2 = \omega^2(a^2 - x^2)$, $x = 7 - d$ [= 1.5] | M1, M1 | Find speed $v$ when $OP = 7$ (2nd M1 dep. on 1st M1)
$= 2^2(2.5^2 - 1.5^2) = 16$, $v = 4$ [m s$^{-1}$] | A1 |
### Part 3(iii):
$\omega^{-1}\sin^{-1}(x_B/a) - \omega^{-1}\sin^{-1}(x_A/a)$ or $\omega^{-1}\cos^{-1}(x_A/a) - \omega^{-1}\cos^{-1}(x_B/a)$ | M1 | Find time from $A$ to $B$ from $x = a\sin\omega t$ or $a\cos\omega t$ (all terms must be correct; FT on $a$, $\omega$ for M1)
$= \frac{1}{2}\sin^{-1}(2/2.5) - \frac{1}{2}\sin^{-1}(1/2.5)$ or $\frac{1}{2}\cos^{-1}(1/2.5) - \frac{1}{2}\cos^{-1}(2/2.5)$ $= \frac{1}{2}(0.9273 - 0.4115)$ or $0.4636 - 0.2058$ or $\frac{1}{2}(1.1593 - 0.6435)$ or $0.5796 - 0.3218$ | A1 | To 3 d.p., AEF throughout
$= 0.258$ [s] | A1 |
---3 A particle $P$ moves on the positive $x$-axis in simple harmonic motion. The centre of the motion is a distance $d \mathrm {~m}$ from the origin $O$, where $0 < d < 6.5$. The points $A$ and $B$ are on the positive $x$-axis, with $O A = 6.5 \mathrm {~m}$ and $O B = 7.5 \mathrm {~m}$. The magnitude of the acceleration of $P$ when it is at $B$ is twice the magnitude of the acceleration of $P$ when it is at $A$.\\
(i) Find $d$.\\
The period of the motion is $\pi \mathrm { s }$ and the maximum acceleration of $P$ during the motion is $10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Find the speed of $P$ when it is 7 m from $O$.\\
(iii) Find the time taken by $P$ to travel directly from $A$ to $B$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q3 [10]}}