| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Challenging +1.3 This is a multi-step SHM problem requiring application of standard formulas (v² = ω²(a² - x²)) and integration for time calculation. While it involves algebraic manipulation to find the amplitude and period from given conditions, the techniques are standard for Further Maths students. The conceptual demand is moderate—recognizing the relationship between speeds at different positions and using maximum acceleration to find ω—but follows established SHM methods without requiring novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_B = 2v_A\), \(v_B^2 = 4v_A^2\), \(\omega^2(a^2 - 1^2) = 4\omega^2(a^2 - 3.5^2)\) | M1 | Find amplitude \(a\): allow M1 for \(v_A = 2v_B\) |
| \(3a^2 = 48\), \(a = [\pm]\ 4\) [m] | A1 | |
| \(1 = a\omega^2\), \(\omega = \frac{1}{\sqrt{a}} = \frac{1}{2}\) | B1 | Find \(\omega\) from given maximum acceleration |
| \(v_O = a\omega = \sqrt{a} = 2\) [m s\(^{-1}\)] | B1 | Find speed \(v_O\) at \(O\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \sin^{-1}(0.875) + \sin^{-1}(0.25)\) | M1 A1 | Find equation (AEF) for \(\omega t_{AB}\), combining \(t_A\) and \(t_B\), using \(x = a\sin\omega t\); allow sign errors for M1 |
| \(= 1.065 + 0.253\) (or \(t_{AB} = 2.131 + 0.505\)) | A1 | or \(x = a\cos\omega t\) |
| \(t_{AB} = 2 \times 1.318 = 2.64\) [s] | A1 | Hence find \(t_{AB}\) |
| Answer | Marks | Guidance |
|---|---|---|
| or \(\pi - \cos^{-1}(0.25) - \cos^{-1}(0.875)\) (AEF) | M1 A1 | or \(x = a\cos\omega t\) |
| \(= 1.823 - 0.505\) (or \(t_{AB} = 3.647 - 1.011\)) | A1 | |
| \(t_{AB} = 2 \times 1.318 = 2.64\) [s] | A1 | Hence find \(t_{AB}\) |
**Question 2:**
**Part (i):**
$v_B = 2v_A$, $v_B^2 = 4v_A^2$, $\omega^2(a^2 - 1^2) = 4\omega^2(a^2 - 3.5^2)$ | M1 | Find amplitude $a$: allow M1 for $v_A = 2v_B$
$3a^2 = 48$, $a = [\pm]\ 4$ [m] | A1 |
$1 = a\omega^2$, $\omega = \frac{1}{\sqrt{a}} = \frac{1}{2}$ | B1 | Find $\omega$ from given maximum acceleration
$v_O = a\omega = \sqrt{a} = 2$ [m s$^{-1}$] | B1 | Find speed $v_O$ at $O$
**Part (ii):**
$\omega t_{AB} = \sin^{-1}(\frac{3.5}{a}) + \sin^{-1}(\frac{1}{a})$
$= \sin^{-1}(0.875) + \sin^{-1}(0.25)$ | M1 A1 | Find equation (AEF) for $\omega t_{AB}$, combining $t_A$ and $t_B$, using $x = a\sin\omega t$; allow sign errors for M1
$= 1.065 + 0.253$ (or $t_{AB} = 2.131 + 0.505$) | A1 | or $x = a\cos\omega t$
$t_{AB} = 2 \times 1.318 = 2.64$ [s] | A1 | Hence find $t_{AB}$
**Alternative method for Part (ii):**
$\omega t_{AB} = \cos^{-1}(\frac{-1}{a}) - \cos^{-1}(\frac{3.5}{a})$
$= \cos^{-1}(-0.25) - \cos^{-1}(0.875)$
or $\pi - \cos^{-1}(0.25) - \cos^{-1}(0.875)$ (AEF) | M1 A1 | or $x = a\cos\omega t$
$= 1.823 - 0.505$ (or $t_{AB} = 3.647 - 1.011$) | A1 |
$t_{AB} = 2 \times 1.318 = 2.64$ [s] | A1 | Hence find $t_{AB}$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 $O$ such that $O A = 3.5 \mathrm {~m}$ and $O B = 1 \mathrm {~m}$. The speed of $P$ when it is at $B$ is twice its speed when it is at $A$. The maximum acceleration of $P$ is $1 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the speed of $P$ when it is at $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{2aaf3493-6509-4668-91a2-9f4708bbbb58-04_64_1566_492_328}\\
(ii) Find the time taken by $P$ to travel directly from $A$ to $B$.\\
\hfill \mbox{\textit{CAIE FP2 2019 Q2 [8]}}