| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with elastic strings/springs |
| Difficulty | Challenging +1.2 This is a standard small oscillations problem requiring resolution of forces, use of Hooke's law, and binomial approximation for small angles. While it involves multiple steps and some geometric reasoning (finding string extensions), the techniques are routine for Further Maths mechanics students. The SHM identification and amplitude calculation in parts (ii)-(iii) are straightforward applications of standard formulas once the equation is established. |
| Spec | 3.02h Motion under gravity: vector form6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
**Question 11(i)**
$T = \frac{6mg(\sqrt{9a^2 + x^2} - a)}{a}$ **M1A1**
Let $\theta$ be the angle between each string and line of motion of particle. **M1**
$m\ddot{x} = -2T\cos\theta = -\frac{12mg}{a}\left(\sqrt{9a^2 + x^2} - a\right) \times \frac{x}{\sqrt{9a^2 + x^2}}$ **A1A1**
$\Rightarrow \ddot{x} = -\frac{12gx}{a}\left(1 - \frac{a}{\sqrt{9a^2 + x^2}}\right)$ **A1** [6]
**Question 11(ii)**
$\therefore \ddot{x} \approx (-12g + 4g)\frac{x}{a} = -\frac{8g}{a}x$ **M1A1**
Which is SHM of period $2\pi\sqrt{\frac{a}{8g}}$ or $\pi\sqrt{\frac{a}{2g}}$. **A1** [3]
**Question 11(iii)**
$v_{\max} = \varpi a \Rightarrow \frac{ga}{200} = \frac{8g}{a}A^2 \Rightarrow A^2 = \frac{a^2}{1600} \Rightarrow A = \frac{a}{40}$
where $A$ is the amplitude. **M1A1** [2] **[11]**11 Two light strings, each of natural length $a$ 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 $6 a$ 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
$$\ddot { x } = - \frac { 12 g x } { a } \left( 1 - \frac { a } { \sqrt { 9 a ^ { 2 } + x ^ { 2 } } } \right) .$$
(ii) Given that $\frac { x } { a }$ is small throughout the motion, show that the equation of motion is approximately
$$\ddot { x } = - \frac { 8 g x } { a }$$
and state the period of the simple harmonic motion that this equation represents.\\
(iii) Given that the initial speed of $P$ is $\sqrt { \frac { g a } { 200 } }$, show that the amplitude of the simple harmonic motion is $\frac { 1 } { 40 } a$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2012 Q11 [11]}}