| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Difficulty | Standard +0.3 This is a standard SHM problem requiring application of well-known formulas (a = -ω²x, v² = ω²(a² - x²)) and basic trigonometric manipulation. Part (c) requires setting up and solving a simple equation involving arccos, which is routine for M3. All steps follow predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\ddot{x} = -\omega^2 x \Rightarrow 1 = \omega^2 \times 0.04\) \((\Rightarrow \omega = 5)\) | M1 A1 | |
| \(T = \frac{2\pi}{5}\) awrt 1.3 | A1 | 3 marks |
| (b) \(v^2 = \omega^2(a^2 - x^2) \Rightarrow 0.2^2 = 5^2(a^2 - 0.04^2)\) | M1 A1 ft | |
| \(a = \frac{\sqrt{2}}{25}\) accept exact equivalents or awrt 0.057 | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}a = a\cos\omega t\) ft their \(\omega\) | M1 A1 ft | |
| \(5t = \frac{\pi}{3}\) | ||
| \(t = \frac{\pi}{15}\) | A1 | |
| \(T' = 4t = \frac{4\pi}{15}\) awrt 0.84 | M1 A1 | 5 marks [11] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}a = a\sin\omega t\) ft their \(\omega\) | M1 A1 ft | |
| \(5t = \frac{\pi}{6}\) | ||
| \(t = \frac{\pi}{30}\) | A1 | |
| \(T' = T - 4t = \frac{4\pi}{15}\) awrt 0.84 | M1 A1 | 5 marks |
**(a)** $\ddot{x} = -\omega^2 x \Rightarrow 1 = \omega^2 \times 0.04$ $(\Rightarrow \omega = 5)$ | M1 A1 |
$T = \frac{2\pi}{5}$ awrt 1.3 | A1 | 3 marks
**(b)** $v^2 = \omega^2(a^2 - x^2) \Rightarrow 0.2^2 = 5^2(a^2 - 0.04^2)$ | M1 A1 ft |
$a = \frac{\sqrt{2}}{25}$ accept exact equivalents or awrt 0.057 | A1 | 3 marks
**(c)** Using $x = a\cos\omega t$:
$\frac{1}{2}a = a\cos\omega t$ ft their $\omega$ | M1 A1 ft |
$5t = \frac{\pi}{3}$ | |
$t = \frac{\pi}{15}$ | A1 |
$T' = 4t = \frac{4\pi}{15}$ awrt 0.84 | M1 A1 | 5 marks [11]
**Alternative to (c) Using $x = a\sin\omega t$:**
$\frac{1}{2}a = a\sin\omega t$ ft their $\omega$ | M1 A1 ft |
$5t = \frac{\pi}{6}$ | |
$t = \frac{\pi}{30}$ | A1 |
$T' = T - 4t = \frac{4\pi}{15}$ awrt 0.84 | M1 A1 | 5 marks
---A particle $P$ moves on the $x$-axis with simple harmonic motion about the origin $O$ as centre. When $P$ is at a distance $0.04$ m from $O$, its speed is $0.2$ m s$^{-1}$ and the magnitude of its acceleration is $1$ m s$^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Find the period of the motion. [3]
\end{enumerate}
The amplitude of the motion is $a$ metres.
Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the value of $a$, [3]
\item the total time, within one complete oscillation, for which the distance $OP$ is greater than $\frac{3}{4}a$ metres. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q5 [11]}}