| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Tidal/harbour water level SHM |
| Difficulty | Standard +0.3 This is a standard SHM modelling question requiring setup of amplitude/centre/period from given data, then application of velocity formula and solving a trigonometric inequality. While it involves multiple steps (14 marks), the techniques are routine for M3: finding SHM parameters, using v = ±ω√(a²-x²), and solving cos/sin equations. No novel insight required, just careful application of standard methods. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = 3\), \(T = 12\) (or \(\frac{1}{2}T = 6\)) | B1, B1 | |
| \(T = \frac{2\pi}{\omega} = 12 \Rightarrow \omega = \frac{\pi}{6}\) (\(\approx 0.52\)) | M1 A1 | |
| Taking \(x = a\) when \(t=0\): \(x = a\cos\omega t\) | M1 | |
| \(\dot{x} = -a\omega\sin\omega t\) | M1 A1 | |
| When \(t=5\): \(\dot{x} = -3 \times \frac{\pi}{6}\sin\frac{5\pi}{6}\) | M1 | |
| \( | \dot{x} | = \frac{\pi}{4}\) (\(\text{m h}^{-1}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(t=5\): \(x = 3\cos\frac{5\pi}{6} = -\frac{3\sqrt{3}}{2}\) (\(\approx -2.60\)) | M1 | |
| \(v^2 = \omega^2(a^2 - x^2)\) | M1 | |
| \(= \frac{\pi^2}{6^2}\left(9 - \frac{9\times 3}{4}\right) = \frac{\pi^2}{16}\) | M1 A1 | |
| \( | v | = \frac{\pi}{4}\) (\(\text{m h}^{-1}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = a\sin\omega t\), \(\dot{x} = a\omega\cos\omega t\) | M1 A1 | |
| When \(t=2\): \(\dot{x} = 3 \times \frac{\pi}{6}\cos\frac{2\pi}{6}\) | M1 | \(t=2\) oe is essential for this M |
| \(= \frac{\pi}{4}\) (\(\text{m h}^{-1}\)) | A1 | (9 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Depth of 5.5 m \(\Rightarrow x = -1.5\) | ||
| \(-1.5 = a\cos\omega t\) | M1 | |
| \(\cos\omega t = -\frac{1}{2}\) | A1ft | |
| \(\frac{\pi}{6}t = \frac{2\pi}{3}, \frac{4\pi}{3}\) | M1 | |
| \(t = 4, 8\) | A1 | |
| Required time: \(t_2 - t_1 = 8 - 4 = 4\) (h) | A1 | (5 marks) [14 total] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.5 = a\cos\omega t\) | M1 | |
| \(\cos\omega t = \frac{1}{2}\) | A1ft | |
| \(\frac{\pi}{6}t = \frac{\pi}{3}\), \(t=2\) | M1, A1 | |
| Required time: \(2t = 4\) (h) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.5 = 3\sin\omega t\) | M1 | |
| \(\sin\omega t = \frac{1}{2}\) | A1ft | |
| \(\frac{\pi}{6}t = \frac{\pi}{6}, \frac{5\pi}{6}\) | M1 | |
| \(t = 1, 5\) | A1 | |
| Required time: \(t_2 - t_1 = 5 - 1 = 4\) (h) | A1 | (5 marks) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 3$, $T = 12$ (or $\frac{1}{2}T = 6$) | B1, B1 | |
| $T = \frac{2\pi}{\omega} = 12 \Rightarrow \omega = \frac{\pi}{6}$ ($\approx 0.52$) | M1 A1 | |
| Taking $x = a$ when $t=0$: $x = a\cos\omega t$ | M1 | |
| $\dot{x} = -a\omega\sin\omega t$ | M1 A1 | |
| When $t=5$: $\dot{x} = -3 \times \frac{\pi}{6}\sin\frac{5\pi}{6}$ | M1 | |
| $|\dot{x}| = \frac{\pi}{4}$ ($\text{m h}^{-1}$) | A1 | awrt 0.79 (9 marks) |
**Alternative to 6(a) (last 5 marks):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $t=5$: $x = 3\cos\frac{5\pi}{6} = -\frac{3\sqrt{3}}{2}$ ($\approx -2.60$) | M1 | |
| $v^2 = \omega^2(a^2 - x^2)$ | M1 | |
| $= \frac{\pi^2}{6^2}\left(9 - \frac{9\times 3}{4}\right) = \frac{\pi^2}{16}$ | M1 A1 | |
| $|v| = \frac{\pi}{4}$ ($\text{m h}^{-1}$) | A1 | awrt 0.79 |
**Alternative measuring $x$ from centre (using 1400 as $t=0$):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = a\sin\omega t$, $\dot{x} = a\omega\cos\omega t$ | M1 A1 | |
| When $t=2$: $\dot{x} = 3 \times \frac{\pi}{6}\cos\frac{2\pi}{6}$ | M1 | $t=2$ oe is essential for this M |
| $= \frac{\pi}{4}$ ($\text{m h}^{-1}$) | A1 | (9 marks) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Depth of 5.5 m $\Rightarrow x = -1.5$ | | |
| $-1.5 = a\cos\omega t$ | M1 | |
| $\cos\omega t = -\frac{1}{2}$ | A1ft | |
| $\frac{\pi}{6}t = \frac{2\pi}{3}, \frac{4\pi}{3}$ | M1 | |
| $t = 4, 8$ | A1 | |
| Required time: $t_2 - t_1 = 8 - 4 = 4$ (h) | A1 | (5 marks) [14 total] |
**Alternative for 6(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.5 = a\cos\omega t$ | M1 | |
| $\cos\omega t = \frac{1}{2}$ | A1ft | |
| $\frac{\pi}{6}t = \frac{\pi}{3}$, $t=2$ | M1, A1 | |
| Required time: $2t = 4$ (h) | A1 | (5 marks) |
**Alternative 6(b) using $x = a\sin\omega t$:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.5 = 3\sin\omega t$ | M1 | |
| $\sin\omega t = \frac{1}{2}$ | A1ft | |
| $\frac{\pi}{6}t = \frac{\pi}{6}, \frac{5\pi}{6}$ | M1 | |
| $t = 1, 5$ | A1 | |
| Required time: $t_2 - t_1 = 5 - 1 = 4$ (h) | A1 | (5 marks) |6. The rise and fall of the water level in a harbour is modelled as simple harmonic motion. On a particular day the maximum and minimum depths of water in the harbour are 10 m and 4 m and these occur at 1100 hours and 1700 hours respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the speed, in $\mathrm { m } \mathrm { h } ^ { - 1 }$, at which the water level in the harbour is falling at 1600 hours on this particular day.
\item Find the total time, between 1100 hours and 2300 hours on this particular day, for which the depth of water in the harbour is less than 5.5 m .\\
(Total 14 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q6 [14]}}