| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 2 |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with elastic strings/springs |
| Difficulty | Challenging +1.2 This is a structured multi-part SHM question with clear scaffolding through parts (a)-(d). While it requires understanding of elastic strings, geometry, and small-angle approximations, each step is guided and uses standard techniques. The binomial approximation and SHM solution are routine for Further Maths students, making this moderately above average difficulty but not requiring novel insight. |
| 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
$$\ddot { x } = - \frac { 12 g x } { l } \left( 1 - \frac { 1 } { \sqrt { 9 l ^ { 2 } + x ^ { 2 } } } \right)$$
\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 2020 Q13 [2]}}