| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| 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 application of standard formulas (v² = ω²(a² - x²)) and integration/inverse trig for timing. Part (i) requires algebraic manipulation to find amplitude from speed ratio, parts (ii-iii) are routine applications of SHM formulas. More challenging than basic SHM but still follows standard Further Maths patterns without requiring novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_A^2 = \omega^2(a^2 - 1.6^2)\) and \(v_B^2 = \omega^2(a^2 - 1.2^2)\) | B1 | Use \(v^2 = \omega^2(a^2 - x^2)\) at \(A\) and \(B\) (may be implied) |
| \(\frac{9}{16} = \frac{(a^2 - 1.6^2)}{(a^2 - 1.2^2)}\) or \(\frac{(a^2 - 2.56)}{(a^2 - 1.44)}\) | M1 A1 | Find eqn. for \(a\) from ratio of \(v_A\) to \(v_B\) |
| \(7a^2 = 40.96 - 12.96\) or \(10 \times 2.8,\ a^2 = 4,\ a = 2\) [m] | A1 | and hence \(a\) (error in \(\frac{9}{16}\) will lose all A1s, so max \(\frac{4}{11}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = \frac{1}{3} \cdot \frac{\pi}{a} = \frac{\pi}{6},\ T = \frac{2\pi}{w} = 12\) [s] | M1 A1 | Find \(\omega\) and hence period \(T\) from \(v_{max} = \omega a\) and \(T = \frac{2\pi}{w}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\omega^{-1}\sin^{-1}\left(\frac{1.6}{a}\right) + \omega^{-1}\sin^{-1}\left(\frac{1.2}{a}\right) \left[= \omega^{-1}\frac{\pi}{2}\right]\) *or* \(\omega^{-1}\cos^{-1}\left(\frac{-1.6}{a}\right) - \omega^{-1}\cos^{-1}\left(\frac{1.2}{a}\right)\) *or* \(\frac{1}{2}T - \omega^{-1}\cos^{-1}\left(\frac{1.6}{a}\right) - \omega^{-1}\cos^{-1}\left(\frac{1.2}{a}\right)\) | 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 |
| \(= \omega^{-1}(0.9273 + 0.6435)\) *or* \(\omega^{-1}(2.498 - 0.9273)\) *or* \(6 - \omega^{-1}(0.6435 + 0.927)\) (AEF throughout) | A1 | |
| \(= 1.771 + 1.229\) *or* \(4.771 - 1.771\) *or* \(6 - 1.229 - 1.771\) | A1 | *OR* use geometry of circular motion with radius \(a\) and angular velocity \(\omega\) |
| 3 |
**Question 2(i):**
$v_A^2 = \omega^2(a^2 - 1.6^2)$ and $v_B^2 = \omega^2(a^2 - 1.2^2)$ | B1 | Use $v^2 = \omega^2(a^2 - x^2)$ at $A$ and $B$ (may be implied)
$\frac{9}{16} = \frac{(a^2 - 1.6^2)}{(a^2 - 1.2^2)}$ or $\frac{(a^2 - 2.56)}{(a^2 - 1.44)}$ | M1 A1 | Find eqn. for $a$ from ratio of $v_A$ to $v_B$
$7a^2 = 40.96 - 12.96$ or $10 \times 2.8,\ a^2 = 4,\ a = 2$ [m] | A1 | and hence $a$ (error in $\frac{9}{16}$ will lose all A1s, so max $\frac{4}{11}$)
**Total: 4 marks**
---
**Question 2(ii):**
$\omega = \frac{1}{3} \cdot \frac{\pi}{a} = \frac{\pi}{6},\ T = \frac{2\pi}{w} = 12$ [s] | M1 A1 | Find $\omega$ and hence period $T$ from $v_{max} = \omega a$ and $T = \frac{2\pi}{w}$
**Total: 2 marks**
## Question 2(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega^{-1}\sin^{-1}\left(\frac{1.6}{a}\right) + \omega^{-1}\sin^{-1}\left(\frac{1.2}{a}\right) \left[= \omega^{-1}\frac{\pi}{2}\right]$ *or* $\omega^{-1}\cos^{-1}\left(\frac{-1.6}{a}\right) - \omega^{-1}\cos^{-1}\left(\frac{1.2}{a}\right)$ *or* $\frac{1}{2}T - \omega^{-1}\cos^{-1}\left(\frac{1.6}{a}\right) - \omega^{-1}\cos^{-1}\left(\frac{1.2}{a}\right)$ | 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 |
| $= \omega^{-1}(0.9273 + 0.6435)$ *or* $\omega^{-1}(2.498 - 0.9273)$ *or* $6 - \omega^{-1}(0.6435 + 0.927)$ (AEF throughout) | A1 | |
| $= 1.771 + 1.229$ *or* $4.771 - 1.771$ *or* $6 - 1.229 - 1.771$ | A1 | *OR* use geometry of circular motion with radius $a$ and angular velocity $\omega$ |
| | **3** | |
---2 A particle $P$ moves on a straight line in simple harmonic motion. The centre of the motion is $O$. The points $A$ and $B$ are on the line, on opposite sides of $O$, with $O A = 1.6 \mathrm {~m}$ and $O B = 1.2 \mathrm {~m}$. The ratio of the speed of $P$ at $A$ to its speed at $B$ is $3 : 4$.\\
(i) Find the amplitude of the motion.\\
The maximum speed of $P$ during its motion is $\frac { 1 } { 3 } \pi \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the period of the motion.\\
(iii) Find the time taken for $P$ to travel directly from $A$ to $B$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q2 [9]}}