| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring: (1) equilibrium analysis with two elastic strings using Hooke's law, (2) proving SHM by showing restoring force is proportional to displacement, (3) finding the period from the equation of motion, and (4) applying energy conservation or SHM amplitude formula. While the individual techniques are standard for FM students, the two-string setup requires careful bookkeeping of extensions and forces, and the full solution involves 4-5 distinct steps with algebraic manipulation. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(8mg\,\frac{(AP - 2a)}{2a}\) | M1 A1 | Find/verify \(AP\) by equating equilibrium tensions |
| \(= \frac{16mg(6a - AP)}{4a}\) | A1 | |
| \(AP = \frac{32a}{8} = 4a\) | A.G. A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m\,\frac{d^2x}{dt^2} = \frac{8mg(2a-x)}{2a} - \frac{16mg(2a+x)}{4a}\) | M1 A2 | Newton's law at general point (lose A1 for each incorrect term) |
| *Or* \(m\,\frac{d^2y}{dt^2} = -\frac{8mg(2a+y)}{2a} + \frac{16mg(2a-y)}{4a}\) | M1 A2 | |
| \(\frac{d^2x}{dt^2} = -\frac{8gx}{a}\) | A1 | Simplify to standard SHM equation |
| S.R.: B1 if no derivation (max 3/6) | ||
| \(T = \frac{2\pi}{\sqrt{(8g/a)}} = \pi\sqrt{\frac{a}{2g}}\) | M1 A1 | A.G. Find period \(T\) using SHM with \(\omega = \sqrt{(8g/a)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_{max} = \sqrt{(8g/a)} \times a = \sqrt{(8ag)}\) or \(2\sqrt{(2ag)}\) | M1 A1 | Find max speed using \(\omega A\) with \(A = a\) |
| Subtotal: 2 | Total: [11] |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8mg\,\frac{(AP - 2a)}{2a}$ | M1 A1 | Find/verify $AP$ by equating equilibrium tensions |
| $= \frac{16mg(6a - AP)}{4a}$ | A1 | |
| $AP = \frac{32a}{8} = 4a$ | **A.G.** A1 | |
**Subtotal: 3**
### Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m\,\frac{d^2x}{dt^2} = \frac{8mg(2a-x)}{2a} - \frac{16mg(2a+x)}{4a}$ | M1 A2 | Newton's law at general point (lose A1 for each incorrect term) |
| *Or* $m\,\frac{d^2y}{dt^2} = -\frac{8mg(2a+y)}{2a} + \frac{16mg(2a-y)}{4a}$ | M1 A2 | |
| $\frac{d^2x}{dt^2} = -\frac{8gx}{a}$ | A1 | Simplify to standard SHM equation |
| **S.R.:** B1 if no derivation (max 3/6) | | |
| $T = \frac{2\pi}{\sqrt{(8g/a)}} = \pi\sqrt{\frac{a}{2g}}$ | M1 A1 | **A.G.** Find period $T$ using SHM with $\omega = \sqrt{(8g/a)}$ |
**Subtotal: 6**
### Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_{max} = \sqrt{(8g/a)} \times a = \sqrt{(8ag)}$ or $2\sqrt{(2ag)}$ | M1 A1 | Find max speed using $\omega A$ with $A = a$ |
**Subtotal: 2 | Total: [11]**
---5 A particle $P$ of mass $m$ lies on a smooth horizontal surface. $A$ and $B$ are fixed points on the surface, where $A B = 10 a$. A light elastic string, of natural length $2 a$ and modulus of elasticity $8 m g$, joins $P$ to $A$. Another light elastic string, of natural length $4 a$ and modulus of elasticity $16 m g$, joins $P$ to $B$. Show that when $P$ is in equilibrium, $A P = 4 a$.
The particle is held at rest at the point $C$ between $A$ and $B$ on the line $A B$ where $A C = 3 a$. The particle is now released.\\
(i) Show that the subsequent motion of $P$ is simple harmonic with period $\pi \sqrt { } \left( \frac { a } { 2 g } \right)$.\\
(ii) Find the maximum speed of $P$.
\hfill \mbox{\textit{CAIE FP2 2012 Q5 [11]}}