| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 12 |
| 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 application question requiring students to identify amplitude/period from context, then solve trigonometric equations for specific depth values. While it involves multiple parts and careful time arithmetic, the mathematical techniques are routine for M3 (setting up x = a cos(ωt) + centre, solving for t). The 'show that' in part (b) guides students to the answer, reducing problem-solving demand. Slightly easier than average due to its structured, textbook-style format. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| amplitude \(= \frac{1}{2}(14 - 6) = 4\) m | B1 | |
| period \(= 2 \times 6\frac{1}{4} = 12\frac{1}{2}\) hours | B1 | |
| \(x = a\cos\omega t\) \(\therefore 1 = 4\cos\omega t\) | M1 A1 | |
| \(\cos\omega t = -\frac{1}{4} \therefore \omega t = 1.8235, \ldots\) | A1 | |
| period \(= \frac{2\pi}{\omega} = 12\frac{1}{2} \therefore \omega = \frac{4\pi}{25}\) | M1 | |
| giving \(t = 1.8235 + \frac{4\pi}{25} = 3.6277\) hours | A1 | |
| depth 9 m at 11 am + 3.6277 hours = 2.38 pm (nearest min.) | A1 | |
| depth 9 m again when \(\omega t = 2\pi - 1.8235 = 4.4597\) | M1 | |
| \(t = 4.4597 + \frac{4\pi}{25} = 8.8723\) | A1 | |
| wait until 11 am + 8.8723 hours = 7.52 pm | M1 | |
| \(\therefore\) 2 hours 52 min. wait (nearest min.) | A1 | (12) |
| amplitude $= \frac{1}{2}(14 - 6) = 4$ m | B1 | |
| period $= 2 \times 6\frac{1}{4} = 12\frac{1}{2}$ hours | B1 | |
| $x = a\cos\omega t$ $\therefore 1 = 4\cos\omega t$ | M1 A1 | |
| $\cos\omega t = -\frac{1}{4} \therefore \omega t = 1.8235, \ldots$ | A1 | |
| period $= \frac{2\pi}{\omega} = 12\frac{1}{2} \therefore \omega = \frac{4\pi}{25}$ | M1 | |
| giving $t = 1.8235 + \frac{4\pi}{25} = 3.6277$ hours | A1 | |
| depth 9 m at 11 am + 3.6277 hours = 2.38 pm (nearest min.) | A1 | |
| depth 9 m again when $\omega t = 2\pi - 1.8235 = 4.4597$ | M1 | |
| $t = 4.4597 + \frac{4\pi}{25} = 8.8723$ | A1 | |
| wait until 11 am + 8.8723 hours = 7.52 pm | M1 | |
| $\therefore$ 2 hours 52 min. wait (nearest min.) | A1 | (12) |4. On a particular day, high tide at the entrance to a harbour occurs at 11 a.m. and the water depth is 14 m . Low tide occurs $6 \frac { 1 } { 4 }$ hours later at which time the water depth is 6 m .
In a model of the situation, the water level is assumed to perform simple harmonic motion.\\
Using this model,
\begin{enumerate}[label=(\alph*)]
\item write down the amplitude and period of the motion.
A ship needs a depth of 9 m before it can enter or leave the harbour.
\item Show that on this day a ship must enter the harbour by 2.38 p.m., correct to the nearest minute, or wait for low tide to pass.\\
(6 marks)\\
Given that a ship is not ready to enter the harbour until 5 p.m.,
\item find, to the nearest minute, how long the ship must wait before it can enter the harbour.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [12]}}