| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Standard +0.3 This is a straightforward SHM question requiring standard techniques: differentiation to verify SHM, reading amplitude/period from the equation, finding maximum speed using v_max = ωa, and solving a trigonometric equation for time between two positions. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks |
|---|---|
| \(x = 5\sin(\frac{\pi t}{3})\) | |
| \(\dot{x} = 5 \times \frac{\pi}{3}\cos(\frac{\pi t}{3})\) | |
| \(\ddot{x} = -5 \times (\frac{\pi}{3})^2 \sin(\frac{\pi t}{3})\) | M1A1 |
| \(\ddot{x} = -\frac{\pi^2}{9}x\) \((\therefore \text{S.H.M.})\) | A1 |
| Answer | Marks |
|---|---|
| \(\text{period} = \frac{2\pi}{\pi/3} = 6\) | B1 |
| \(\text{amplitude} = 5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ldots = 5 \times \frac{\pi}{3}\cos(\frac{\pi t}{3})\) or \( | v_{\max} | = a\omega\) |
| \(\max . v = \frac{5\pi}{3}\) | A1 |
| Answer | Marks |
|---|---|
| At \(A, x = 2\): \(2 = 5\sin(\frac{\pi t}{3})\) | M1 |
| \(\sin\frac{\pi}{3}t = 0.4\) | |
| \(t_A = \frac{3}{\pi} \times \sin^{-1}0.4\) | A1 |
| At \(B, x = 3\): \(t_B = \frac{3}{\pi} \times \sin^{-1}0.6\) | |
| \(\text{time } A \to B = \frac{3}{\pi} \times \sin^{-1}0.6 - \frac{3}{\pi} \times \sin^{-1}0.4\) | A1 |
| \(= 0.2215... = 0.22\) s accept awrt \(0.22\) | A1 |
## Part (a):
$x = 5\sin(\frac{\pi t}{3})$ | |
$\dot{x} = 5 \times \frac{\pi}{3}\cos(\frac{\pi t}{3})$ | |
$\ddot{x} = -5 \times (\frac{\pi}{3})^2 \sin(\frac{\pi t}{3})$ | M1A1 |
$\ddot{x} = -\frac{\pi^2}{9}x$ $(\therefore \text{S.H.M.})$ | A1 |
## Part (b):
$\text{period} = \frac{2\pi}{\pi/3} = 6$ | B1 |
$\text{amplitude} = 5$ | B1 |
## Part (c):
$\ldots = 5 \times \frac{\pi}{3}\cos(\frac{\pi t}{3})$ or $|v_{\max}| = a\omega$ | M1 |
$\max . v = \frac{5\pi}{3}$ | A1 |
## Part (d):
At $A, x = 2$: $2 = 5\sin(\frac{\pi t}{3})$ | M1 |
$\sin\frac{\pi}{3}t = 0.4$ | |
$t_A = \frac{3}{\pi} \times \sin^{-1}0.4$ | A1 |
At $B, x = 3$: $t_B = \frac{3}{\pi} \times \sin^{-1}0.6$ | |
$\text{time } A \to B = \frac{3}{\pi} \times \sin^{-1}0.6 - \frac{3}{\pi} \times \sin^{-1}0.4$ | A1 |
$= 0.2215... = 0.22$ s accept awrt $0.22$ | A1 |A particle $P$ moves along the $x$-axis. At time $t$ seconds its displacement, $x$ metres, from the origin $O$ is given by $x = 5 \sin (\frac{1}{4}\pi t)$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $P$ is moving with simple harmonic motion.
[3]
\item Find the period and the amplitude of the motion.
[2]
\item Find the maximum speed of $P$.
[2]
\end{enumerate}
The points $A$ and $B$ on the positive $x$-axis are such that $OA = 2$ m and $OB = 3$ m.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the time taken by $P$ to travel directly from $A$ to $B$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q4 [11]}}