| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.8 This is a challenging Pre-U Further Maths mechanics problem requiring analysis of elastic strings with slack/taut transitions, SHM identification, energy methods, and integration to find time periods. Part (iii) demands careful handling of two distinct motion phases and produces a non-trivial exact answer involving inverse trig functions, requiring strong technical facility beyond standard A-level. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{\lambda x}{a} = m\ddot{x} \Rightarrow \ddot{x} = -\frac{\lambda}{ma}x\) (where \(x\) is the extension of the taut part) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: \(v^2 = \frac{\lambda}{ma}(4a^2 - a^2) \Rightarrow v = \sqrt{\frac{3\lambda a}{m}}\) using \(v^2 = \omega^2(a^2-x^2)\) | M1, A1 | |
| OR: \(\frac{\lambda(2a)^2}{2a} - \frac{\lambda a^2}{2a} - \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{3\lambda a}{m}}\) using conservation of energy | M1, A1 | 2 |
| Answer | Marks |
|---|---|
| \(a = 2a\cos\omega t \Rightarrow t = \frac{1}{\omega}\cos^{-1}\left(\frac{1}{2}\right) \Rightarrow t = \sqrt{\frac{ma\pi}{\lambda 3}}\) when one side is slack | M1, A1 |
| \(\frac{\lambda\left(\frac{a}{2}-y\right) - \lambda\left(\frac{a}{2}+y\right)}{a} = m\ddot{y} \Rightarrow \ddot{y} = -\frac{2\lambda}{ma}y\) when both sides are taut | M1, A1 |
| Answer | Marks |
|---|---|
| \(\frac{3\lambda a}{m} - \frac{2\lambda}{ma}\left(A^2 - \frac{a^2}{4}\right) \Rightarrow A = \frac{\sqrt{7}}{2}a\) using \(v^2 = \omega^2(a^2-x^2)\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a}{2} = \frac{\sqrt{7}}{2}a\sin\omega t \Rightarrow t = \sqrt{\frac{ma}{2\lambda}}\sin^{-1}\left(\frac{1}{\sqrt{7}}\right)\) | M1 | |
| \(\therefore\) Total time to centre is \(\sqrt{\frac{ma}{\lambda}}\left(\frac{\pi}{3} + \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{1}{\sqrt{7}}\right)\right)\) (AG) | A1 | 8 |
### (i)
$-\frac{\lambda x}{a} = m\ddot{x} \Rightarrow \ddot{x} = -\frac{\lambda}{ma}x$ (where $x$ is the extension of the taut part) | M1, A1 | **2**
### (ii)
EITHER: $v^2 = \frac{\lambda}{ma}(4a^2 - a^2) \Rightarrow v = \sqrt{\frac{3\lambda a}{m}}$ using $v^2 = \omega^2(a^2-x^2)$ | M1, A1 |
OR: $\frac{\lambda(2a)^2}{2a} - \frac{\lambda a^2}{2a} - \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{3\lambda a}{m}}$ using conservation of energy | M1, A1 | **2**
### (iii)
$a = 2a\cos\omega t \Rightarrow t = \frac{1}{\omega}\cos^{-1}\left(\frac{1}{2}\right) \Rightarrow t = \sqrt{\frac{ma\pi}{\lambda 3}}$ when one side is slack | M1, A1 |
$\frac{\lambda\left(\frac{a}{2}-y\right) - \lambda\left(\frac{a}{2}+y\right)}{a} = m\ddot{y} \Rightarrow \ddot{y} = -\frac{2\lambda}{ma}y$ when both sides are taut | M1, A1 |
(where $y$ is the displacement from the equilibrium position)
$\frac{3\lambda a}{m} - \frac{2\lambda}{ma}\left(A^2 - \frac{a^2}{4}\right) \Rightarrow A = \frac{\sqrt{7}}{2}a$ using $v^2 = \omega^2(a^2-x^2)$ | M1, A1 |
(where $A$ is the amplitude of the motion)
$\frac{a}{2} = \frac{\sqrt{7}}{2}a\sin\omega t \Rightarrow t = \sqrt{\frac{ma}{2\lambda}}\sin^{-1}\left(\frac{1}{\sqrt{7}}\right)$ | M1 |
$\therefore$ Total time to centre is $\sqrt{\frac{ma}{\lambda}}\left(\frac{\pi}{3} + \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{1}{\sqrt{7}}\right)\right)$ (AG) | A1 | **8**A light elastic string of natural length $2a$ and modulus of elasticity $\lambda$ is stretched between two points $A$ and $B$, which are $3a$ apart on a smooth horizontal table. A particle of mass $m$ is attached to the mid-point of the string, pulled aside to $A$ and released.
\begin{enumerate}[label=(\roman*)]
\item Prove that, while one part of the string is taut and the other part is slack, the particle is describing simple harmonic motion. [2]
\item Find the speed of the particle when the slack part of the string becomes taut. [2]
\item Prove that the total time for the particle to reach the mid-point of the string for the first time is
$$\sqrt{\frac{ma}{\lambda}} \left( \frac{\pi}{3} + \frac{1}{\sqrt{2}} \sin^{-1} \frac{1}{\sqrt{7}} \right).$$ [8]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q6 [12]}}