| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 11 |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with elastic strings/springs |
| Difficulty | Challenging +1.8 This is a sophisticated SHM problem requiring geometric analysis of elastic strings, binomial approximation for small oscillations, and multi-step reasoning through four parts. While the techniques (Hooke's law, Newton's second law, binomial expansion) are standard Further Maths content, the geometric setup and the need to derive the approximation make it significantly harder than routine SHM exercises. However, the question provides clear guidance through each step, preventing it from reaching the highest difficulty levels. |
| 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 |
From Question 11 in the mark scheme:
**(i)** $T = \frac{6mg(\sqrt{9a^2+x^2}-a)}{a}$ **M1 A1**
Let $\theta$ be the angle between each string and line of motion of particle. **M1**
$m\ddot{x} = -2T\cos\theta = -\frac{12mg}{a}(\sqrt{9a^2+x^2}-a)\times\frac{x}{\sqrt{9a^2+x^2}}$ **A1 A1**
$\Rightarrow \ddot{x} = -\frac{12gx}{a}\left(1 - \frac{a}{\sqrt{9a^2+x^2}}\right)$ **A1**
**(ii)** $\therefore \ddot{x} \approx (-12g + 4g)\frac{x}{a} = -\frac{8g}{a}x$ **M1 A1**
Which is simple harmonic motion of period $2\pi\sqrt{\frac{a}{8g}}$ or $\pi\sqrt{\frac{a}{2g}}$. **A1**
**(iii)** $v_{\max} = \varomega a \Rightarrow \frac{ga}{200} = \frac{8g}{a}A^2 \Rightarrow A^2 = \frac{a^2}{1600} \Rightarrow A = \frac{a}{40}$
where $A$ is the amplitude. **M1 A1**13 Two light strings, each of natural length $l$ and modulus of elasticity $6 m g$, are attached at their ends to a particle $P$ of mass $m$. The other ends of the strings are attached to two fixed points $A$ and $B$, which are at a distance $6 l$ apart on a smooth horizontal table. Initially $P$ is at rest at the mid-point of $A B$. The particle is now given a horizontal impulse in the direction perpendicular to $A B$. At time $t$ the displacement of $P$ from the line $A B$ is $x$.\\
(i) Show that the tension in each string is $\frac { 6 m g } { l } \left( \sqrt { 9 l ^ { 2 } + x ^ { 2 } } - l \right)$.\\
(ii) Show that
$$\ddot { x } = - \frac { 12 g x } { l } \left( 1 - \frac { l } { \sqrt { 9 l ^ { 2 } + x ^ { 2 } } } \right) .$$
(iii) Given that throughout the motion $\frac { x ^ { 2 } } { l ^ { 2 } }$ is small enough to be negligible, show that the equation of motion is approximately
$$\ddot { x } = - \frac { 8 g x } { l } .$$
(iv) Given that the initial speed of $P$ is $\sqrt { \frac { g l } { 200 } }$, find the time taken for the particle to travel a distance of $\frac { 1 } { 80 } l$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q13 [11]}}