| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 2 |
| 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, resolution of forces in 2D, binomial approximation for small oscillations, and multi-step application of SHM theory. While the structure is guided through parts (a)-(d), it demands strong spatial reasoning, careful algebraic manipulation with surds, and understanding of approximation techniques—significantly harder than routine SHM questions but follows a standard Pre-U/Further Maths framework. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
**Question 13(a)**
Indication (e.g. from diagram) that each string has extension $e$ given by $e = \sqrt{(3l)^2 + x^2} - l$ [B1]
Use of Hooke's Law, $T = \dfrac{\lambda e}{l}$, to get $T = \dfrac{6mg\left(\sqrt{9l^2 + x^2} - l\right)}{l}$ **AG** [B1]
**Total: 2 marks**
**Question 13(b)**
Let $\theta$ be the angle between each string and line of motion of particle. [M1]
$m\ddot{x} = -2T\cos\theta = -\dfrac{12mg}{l}\left(\sqrt{9l^2 + x^2} - l\right) \times \dfrac{x}{\sqrt{9l^2 + x^2}}$ [A1A1]
$\Rightarrow \ddot{x} = -\dfrac{12gx}{l}\left(1 - \dfrac{l}{\sqrt{9l^2 + x^2}}\right)$ **AG** [A1]
**Total: 4 marks**
**Question 13(c)**
$\therefore \ddot{x} \approx (-12g + 4g)\dfrac{x}{l} = -\dfrac{8g}{l}x$ **AG** [M1A1]
**Total: 2 marks**
**Question 13(d)**
$v_{\max} = \omega a \Rightarrow \dfrac{gl}{200} = \dfrac{8g}{l}a^2 \Rightarrow a^2 = \dfrac{l^2}{1600} \Rightarrow a = \dfrac{1}{40}l$ [M1A1]
Hence time $t$ given by $\dfrac{1}{80}l = a\sin\sqrt{\dfrac{8g}{l}}\,t \Rightarrow t = \dfrac{\pi}{6}\sqrt{\dfrac{l}{8g}}$ o.e. [M1A1]
**Total: 4 marks**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 $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in each string is $\frac { 6 m g } { l } \left( \sqrt { 9 l ^ { 2 } + x ^ { 2 } } - l \right)$.
\item Show that
$$\ddot { x } = - \frac { 12 g x } { l } \left( 1 - \frac { l } { \sqrt { 9 l ^ { 2 } + x ^ { 2 } } } \right)$$
\item Given that throughout the motion $\frac { x ^ { 2 } } { l ^ { 2 } }$ is small enough to be negligible, show that the equation of motion is approximately
$$\ddot { x } = - \frac { 8 g x } { l } .$$
\item 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 $\frac { 1 } { 80 } l$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q13 [2]}}