| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Topic | Simple Harmonic Motion |
| Type | Find amplitude from speed conditions |
| Difficulty | Challenging +1.2 This is a standard SHM problem requiring application of the energy equation v² = ω²(a² - x²) at two positions to form simultaneous equations. While it involves algebraic manipulation with surds and requires careful setup, it follows a well-established method taught in Further Maths mechanics. The second part adds a timing calculation using SHM phase equations, making it slightly above average difficulty but still routine for Pre-U Further Maths students. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
(i) $80 = \omega^2(a^2 - 4)$ **B1**
$64 = \omega^2(a^2 - 5)$ **B1**
Dividing: $\frac{5}{4} = \frac{a^2 - 4}{a^2 - 5} \Rightarrow a^2 = 9 \Rightarrow a = 3$ **M1A1**
$\omega = 4$, $T = \frac{2\pi}{4} = \frac{\pi}{2}$ seconds. **A1A1** [6]
(ii) $t_1 - t_2 = \frac{1}{4}\sin^{-1}\left(\frac{2}{3}\right) - \frac{1}{4}\sin^{-1}\left(\frac{-\sqrt{5}}{3}\right)$ $(= 0.1824... + 0.2102...)$ **M1A1**
$\Rightarrow t_1 - t_2 = 0.393$, or $\frac{1}{8}\pi$, seconds (3sf) **A1** [3]8 A particle, $P$, is moving in a straight line with simple harmonic motion about a centre $O$. When $P$ is at the point $A , 2 \mathrm {~m}$ from $O$, it has speed $4 \sqrt { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When $P$ is at the point $B , \sqrt { 5 } \mathrm {~m}$ from $O$, it has speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the amplitude and period of the motion.\\
(ii) Given that $A$ and $B$ are on opposite sides of $O$, find the time taken for $P$ to travel directly from $A$ to $B$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q8 [9]}}