| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Challenging +1.2 This is a standard two-spring SHM problem requiring equilibrium analysis using Hooke's law, verification of SHM conditions (F ∝ -x), and application of energy conservation. While it involves multiple steps and careful bookkeeping of extensions, the techniques are routine for Further Maths students and follow predictable patterns without requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{6mge_A}{3a} = \frac{mg(3a-e_A)}{2a}\) | M1 A1 | Find extension of either string by equating equilibrium tensions |
| \(e_A = \frac{3a}{5}\) or \(e_B = \frac{12a}{5}\) | A1 | |
| \(AO = 3a + e_A\) or \(6a - e_B = 3.6a\) | B1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m\frac{d^2x}{dt^2} = \frac{mg(3a-e_A-x)}{2a} - \frac{6mg(e_A+x)}{3a}\) | M1 A2 | Apply Newton's law at general point; lose A1 for each incorrect term |
| \(\frac{d^2x}{dt^2} = -\frac{5gx}{2a}\) | A1 | Simplify to standard SHM equation; S.R.: B1 if no derivation (max 3/6) |
| \(T = 2\pi\sqrt{\frac{2a}{5g}}\) | M1 A1 | Find period using SHM with \(\omega = \sqrt{\frac{5g}{2a}}\); A.E.F. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_{max} = \sqrt{\frac{5g}{2a}} \times 0.2a = \sqrt{\frac{ag}{10}}\) or \(\sqrt{a}\) | M1 A1 | Find max speed using \(\omega A\) with \(A = 0.2a\); A.E.F. |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{6mge_A}{3a} = \frac{mg(3a-e_A)}{2a}$ | M1 A1 | Find extension of either string by equating equilibrium tensions |
| $e_A = \frac{3a}{5}$ or $e_B = \frac{12a}{5}$ | A1 | |
| $AO = 3a + e_A$ or $6a - e_B = 3.6a$ | B1 | A.G. |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m\frac{d^2x}{dt^2} = \frac{mg(3a-e_A-x)}{2a} - \frac{6mg(e_A+x)}{3a}$ | M1 A2 | Apply Newton's law at general point; lose A1 for each incorrect term |
| $\frac{d^2x}{dt^2} = -\frac{5gx}{2a}$ | A1 | Simplify to standard SHM equation; S.R.: B1 if no derivation (max 3/6) |
| $T = 2\pi\sqrt{\frac{2a}{5g}}$ | M1 A1 | Find period using SHM with $\omega = \sqrt{\frac{5g}{2a}}$; A.E.F. |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_{max} = \sqrt{\frac{5g}{2a}} \times 0.2a = \sqrt{\frac{ag}{10}}$ or $\sqrt{a}$ | M1 A1 | Find max speed using $\omega A$ with $A = 0.2a$; A.E.F. |
---5 The points $A$ and $B$ are on a smooth horizontal table at a distance $8 a$ apart. A particle $P$ of mass $m$ lies on the table on the line $A B$, between $A$ and $B$. The particle is attached to $A$ by a light elastic string of natural length $3 a$ and modulus of elasticity 6 mg , and to $B$ by a light elastic string of natural length $2 a$ and modulus of elasticity $m g$. In equilibrium, $P$ is at the point $O$ on $A B$.\\
(i) Show that $A O = 3.6 a$.
The particle is released from rest at the point $C$ on $A B$, between $A$ and $B$, where $A C = 3.4 a$.\\
(ii) Show that $P$ moves in simple harmonic motion and state the period.\\
(iii) Find the greatest speed of $P$.
\hfill \mbox{\textit{CAIE FP2 2014 Q5 [12]}}