| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: horizontal or non-standard geometry |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring equilibrium analysis with two springs, deriving the SHM equation from Hooke's law, and using the period formula. While systematic, it demands careful bookkeeping of forces, algebraic manipulation to show the restoring force is proportional to displacement, and understanding of the SHM period relationship—more demanding than standard single-spring problems but follows established Further Maths techniques. |
| 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 |
|---|---|---|
| 5(i) | λ (AE – 1.25) / 1.25 = | M1A1 |
| 0.6λ (2.6 – AE – 1.0) / 1.0 (AEF) | A1 | SC: vertical motion can earn M1 only |
| AE – 1.25 = 1.2 – 0.75 AE , AE = 1.4 AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | [±] m d2x/dt2 = – λ (0.15 + x) / 1.25 + 0.6λ (0.2 – x) / 1.0 | |
| or + λ (0.15 – x) / 1.25 – 0.6λ (0.2 + x) / 1.0 | M1A2 | Apply Newton’s law at 1.4 + x or 1.4 – x from A |
| Answer | Marks | Guidance |
|---|---|---|
| d2x/dt2 = – (1.4 λ / 0.4) x = – 3.5 λx | M1A1 | Simplify to give SHM eqn. in standard form |
| Answer | Marks | Guidance |
|---|---|---|
| 5(iii) | ω = 2π / (π/7) [= 14] = √(3.5 λ) | M1A1 |
| λ = 142 / 3.5 = 56 | A1 | Solve for λ |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | λ (AE – 1.25) / 1.25 = | M1A1 | Verify AE by equating equilibrium tensions
0.6λ (2.6 – AE – 1.0) / 1.0 (AEF) | A1 | SC: vertical motion can earn M1 only
AE – 1.25 = 1.2 – 0.75 AE , AE = 1.4 AG | A1
4
--- 5(ii) ---
5(ii) | [±] m d2x/dt2 = – λ (0.15 + x) / 1.25 + 0.6λ (0.2 – x) / 1.0
or + λ (0.15 – x) / 1.25 – 0.6λ (0.2 + x) / 1.0 | M1A2 | Apply Newton’s law at 1.4 + x or 1.4 – x from A
(lose A1 for one incorrect tension term)
d2x/dt2 = – (1.4 λ / 0.4) x = – 3.5 λx | M1A1 | Simplify to give SHM eqn. in standard form
(A0 if no minus sign, or dirn. of acceln. is undefined)
SC: B2 if result stated without derivation (max 2/5)
5
--- 5(iii) ---
5(iii) | ω = 2π / (π/7) [= 14] = √(3.5 λ) | M1A1 | Find eqn. for λ using T = 2π/ω with FT on ω from SHM eqn.
λ = 142 / 3.5 = 56 | A1 | Solve for λ
3
Question | Answer | Marks | GuidanceThe fixed points $A$ and $B$ are on a smooth horizontal surface with $AB = 2.6$ m. One end of a light elastic spring, of natural length 1.25 m and modulus of elasticity $0.6$ N, is attached to $A$. The other end is attached to a particle $P$ of mass 0.4 kg. One end of a second light elastic spring, of natural length 1.0 m and modulus of elasticity $0.62$ N, is attached to $B$; its other end is attached to $P$. The system is in equilibrium with $P$ on the surface at the point $E$.
\begin{enumerate}[label=(\roman*)]
\item Show that $AE = 1.4$ m. [4]
\end{enumerate}
The particle $P$ is now displaced slightly from $E$, along the line $AB$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that, in the subsequent motion, $P$ performs simple harmonic motion. [5]
\item Given that the period of the motion is $\frac{4}{\pi}$ s, find the value of $\lambda$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q5 [12]}}