| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Simple Harmonic Motion |
| Type | String becomes slack timing |
| Difficulty | Challenging +1.8 This is a substantial SHM problem requiring multiple connected steps: energy conservation to find λ, deriving the SHM equation from Hooke's law, and finding the time between taut/slack transitions using phase analysis. The final part demands careful identification of when the string becomes slack (at natural length, not equilibrium) and working with inverse trig functions in the phase equation. While the techniques are standard for Further Maths mechanics, the multi-stage reasoning and phase calculation with the specific answer form elevate this above routine exercises. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^2 |
**(i)** Conservation of energy: $2mgl = \frac{1}{2}\frac{\lambda}{l}l^2$ — M1
$\lambda = 4mg$ (AG) — A1 [2]
**(ii)** $\frac{\lambda d}{l} = mg$ where $d$ is extension at equilibrium position — M1
Newton II: $mg - \frac{\lambda(d+x)}{l} = m\ddot{x}$ when $x$ from equilibrium — M1A1
$\ddot{x} = -\frac{\lambda x}{ml} = -\frac{4gx}{l}$ (AG) — A1 [4]
**(iii)** $d = \frac{mgl}{\lambda} \Rightarrow d = \frac{l}{4}$ and $a = \frac{3l}{4}$; where $a$ is the amplitude. — M1A1
Using $x = a\sin\omega t \Rightarrow \frac{l}{4} = \frac{3l}{4}\sin\omega t$ (or $x = a\cos\omega t \Rightarrow -\frac{1}{3} = \cos\omega t$ etc.) — M1
$\Rightarrow t = \frac{1}{\omega}\sin^{-1}\left(\frac{1}{3}\right) = \frac{1}{2}\sqrt{\frac{l}{g}}\sin^{-1}\left(\frac{1}{3}\right)$ — A1
Required time $= \frac{1}{2}\text{Period} + 2t = \left(\frac{\pi}{2} + \sin^{-1}\left(\frac{1}{3}\right)\right)\sqrt{\frac{l}{g}}$ (AG) — A1 [5]
**[Total: 11]**5 A particle of mass $m$ is attached by a light elastic string of natural length $l$ and modulus of elasticity $\lambda$ to a fixed point $A$, from which it is allowed to fall freely. The particle first comes to rest, instantaneously, at $B$, where $A B = 2 l$. Prove that\\
(i) $\lambda = 4 m g$,\\
(ii) while the string is taut, $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - \frac { 4 g } { l } x$, where $x$ is the displacement from the equilibrium position at time $t$,\\
(iii) the time taken between the first occasion when the string becomes taut and the next occasion when it becomes slack is
$$\left[ \frac { 1 } { 2 } \pi + \sin ^ { - 1 } \left( \frac { 1 } { 3 } \right) \right] \sqrt { \frac { l } { g } }$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q5 [11]}}