| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | Specimen |
| 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 Further Maths mechanics question requiring understanding of SHM with elastic strings, finding equilibrium, proving SHM conditions (F ∝ -x), and applying energy conservation. Part (iii) involves algebraic manipulation with the SHM energy equation. More demanding than standard A-level mechanics but routine for FM students who know the techniques. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find \(k\) by equating equilibrium tensions: \(mg(a/2)/a = 2mg(3a/2 - ka)/ka\) | M1 A1 | vertical motion can earn M1 only |
| \(\frac{1}{2} = 3/k - 2,\quad k = 6/5\) or \(1.2\) | A1 | |
| Apply Newton's law at general point: \(m\,d^2x/dt^2 = -mg(a/2+x)/a + 2mg(3a/2 - ka - x)/ka\) or \(m\,d^2y/dt^2 = +mg(a/2-y)/a - 2mg(3a/2 - ka + y)/ka\) | M1 A2 | lose A1 for each incorrect term |
| Simplify to give standard SHM equation: \(d^2x/dt^2 = -(1+2/k)gx/a = -8gx/3a\) | A1 | S.R.: B1 if no derivation (max 2/5) |
| State or find period using \(2\pi/\omega\) with \(\omega = \sqrt{8g/3a}\): \(T = 2\pi\sqrt{3a/8g}\) or \(\pi\sqrt{3a/2g}\) or \(3.85\sqrt{a/g}\) or \(1.22\sqrt{a}\) [s] | B1\(\checkmark\) | \(\checkmark\) on \(\omega\) |
| Substitute values in \(v^2 = \omega^2(x_0^2 - x^2)\): \(0.7^2 = (8g/3a)\{(0.2a)^2 - (0.05a)^2\}\) | M1 A1 | |
| Solve to find numerical value of \(a\): \(0.49 = (8g/3)\times 0.0375a,\quad a = 0.49\) | A1 |
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find $k$ by equating equilibrium tensions: $mg(a/2)/a = 2mg(3a/2 - ka)/ka$ | M1 A1 | vertical motion can earn M1 only |
| $\frac{1}{2} = 3/k - 2,\quad k = 6/5$ or $1.2$ | A1 | |
| Apply Newton's law at general point: $m\,d^2x/dt^2 = -mg(a/2+x)/a + 2mg(3a/2 - ka - x)/ka$ or $m\,d^2y/dt^2 = +mg(a/2-y)/a - 2mg(3a/2 - ka + y)/ka$ | M1 A2 | lose A1 for each incorrect term |
| Simplify to give standard SHM equation: $d^2x/dt^2 = -(1+2/k)gx/a = -8gx/3a$ | A1 | **S.R.**: B1 if no derivation (max 2/5) |
| State or find period using $2\pi/\omega$ with $\omega = \sqrt{8g/3a}$: $T = 2\pi\sqrt{3a/8g}$ or $\pi\sqrt{3a/2g}$ or $3.85\sqrt{a/g}$ or $1.22\sqrt{a}$ [s] | B1$\checkmark$ | $\checkmark$ on $\omega$ |
| Substitute values in $v^2 = \omega^2(x_0^2 - x^2)$: $0.7^2 = (8g/3a)\{(0.2a)^2 - (0.05a)^2\}$ | M1 A1 | |
| Solve to find numerical value of $a$: $0.49 = (8g/3)\times 0.0375a,\quad a = 0.49$ | A1 | |(i) Find the value of $k$.\\
(ii) The particle $P$ is released from rest at a point between $A$ and $B$ where both strings are taut. Show that $P$ performs simple harmonic motion and state the period of the motion.\\
(iii) In the case where $P$ is released from rest at a distance $0.2 a \mathrm {~m}$ from $M$, the speed of $P$ is $0.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $P$ is $0.05 a \mathrm {~m}$ from $M$. Find the value of $a$.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q3 [11]}}