| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| 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 set up the model (amplitude = 3m, centre = 7m, period from timing), differentiate to find velocity at a given time, and solve an equation for when depth reaches 8.5m. While it involves multiple steps and careful time conversions, it follows a completely routine template for SHM problems with no novel insight required—slightly easier than average due to its predictable structure. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(a = 3\) (m) | B1 | \(a=3\) seen or implied |
| \(\frac{38}{3} = \frac{2\pi}{\omega} \Rightarrow \omega = \frac{3\pi}{19}\) | M1A1 | Use of \(T = \frac{2\pi}{\omega}\) where \(T = 2 \times \frac{19}{3}\) (double the time between 12:00 and 18:20). Condone use of \(T = 760\) min or \(45600\) s. A1: correct equation using hrs, min or seconds |
| \(x = -a\cos\omega t \Rightarrow v = a\omega\sin\omega t\) or similar | M1 | Form relevant equation in \(v\) and \(t\) using their \(a\) and \(\omega\) |
| \(v = 3 \times \frac{3\pi}{19}\sin\left(\frac{3\pi}{19} \times \frac{95}{60}\right)\) | M1 | Use correct equation with appropriate value of \(t\) |
| \(= \frac{9\pi\sqrt{2}}{38}\), \(1.1\), \(1.05\), \(1.052\),... (m h⁻¹) | A1 | (6) Correct answer, must be positive and in metres per hour |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(-1.5 = 3\cos\frac{3\pi t}{19}\) | M1A1ft | Complete method to find required time. Eg: use of \(-1.5 = a\cos\omega t\) to find required time is \(\frac{1}{\omega}\cos^{-1}\!\left(\frac{x}{a}\right)\); or use of \(1.5 = a\sin\omega t\) giving \(\frac{1}{4}\text{Period} + \frac{1}{\omega}\sin^{-1}\!\left(\frac{x}{a}\right)\); or use of \(1.5 = a\cos\omega t\) giving \(\frac{1}{\omega}\cos^{-1}\!\left(\frac{x}{a}\right)\) subtracted from 18:20. A1ft on their \(a\) and \(\omega\): \(\frac{1}{4}\!\left(\frac{38}{3}\right) + \frac{1}{\omega}\sin^{-1}\!\left(\frac{x}{a}\right)\) |
| \(t = \frac{38}{9}\) (h) | A1 | Correct \(t\) value in hours, minutes or seconds: \(t = \frac{38}{9}\) h, \(t = \frac{760}{3}\) min, \(t = 15200\) s |
| Time is 16:13 or 16:14 | A1 | (4) Accept 4.13 pm or 4.14 pm or 16:13 or 16:14 |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $a = 3$ (m) | B1 | $a=3$ seen or implied |
| $\frac{38}{3} = \frac{2\pi}{\omega} \Rightarrow \omega = \frac{3\pi}{19}$ | M1A1 | Use of $T = \frac{2\pi}{\omega}$ where $T = 2 \times \frac{19}{3}$ (double the time between 12:00 and 18:20). Condone use of $T = 760$ min or $45600$ s. A1: correct equation using hrs, min or seconds |
| $x = -a\cos\omega t \Rightarrow v = a\omega\sin\omega t$ or similar | M1 | Form relevant equation in $v$ and $t$ using their $a$ and $\omega$ |
| $v = 3 \times \frac{3\pi}{19}\sin\left(\frac{3\pi}{19} \times \frac{95}{60}\right)$ | M1 | Use correct equation with appropriate value of $t$ |
| $= \frac{9\pi\sqrt{2}}{38}$, $1.1$, $1.05$, $1.052$,... (m h⁻¹) | A1 | (6) Correct answer, must be positive and in metres per hour |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $-1.5 = 3\cos\frac{3\pi t}{19}$ | M1A1ft | Complete method to find required time. Eg: use of $-1.5 = a\cos\omega t$ to find required time is $\frac{1}{\omega}\cos^{-1}\!\left(\frac{x}{a}\right)$; or use of $1.5 = a\sin\omega t$ giving $\frac{1}{4}\text{Period} + \frac{1}{\omega}\sin^{-1}\!\left(\frac{x}{a}\right)$; or use of $1.5 = a\cos\omega t$ giving $\frac{1}{\omega}\cos^{-1}\!\left(\frac{x}{a}\right)$ subtracted from 18:20. A1ft on their $a$ and $\omega$: $\frac{1}{4}\!\left(\frac{38}{3}\right) + \frac{1}{\omega}\sin^{-1}\!\left(\frac{x}{a}\right)$ |
| $t = \frac{38}{9}$ (h) | A1 | Correct $t$ value in hours, minutes or seconds: $t = \frac{38}{9}$ h, $t = \frac{760}{3}$ min, $t = 15200$ s |
| Time is 16:13 or 16:14 | A1 | (4) Accept 4.13 pm or 4.14 pm or 16:13 or 16:14 |\begin{enumerate}
\item In a harbour, the water level rises and falls with the tides with simple harmonic motion.
\end{enumerate}
On a particular day, the depths of water in the harbour at low and high tide are 4 m and 10 m respectively.
Low tide occurs at 12:00 and high tide occurs at 18:20\\
(a) Find, in $\mathrm { mh } ^ { - 1 }$, the speed at which the water level is rising on this particular day at 13:35
A ship can only safely enter the harbour when the depth of water is at least 8.5 m .\\
(b) Find the earliest time after 12:00 on this particular day at which it is safe for the ship to enter the harbour, giving your answer to the nearest minute.
\hfill \mbox{\textit{Edexcel M3 2024 Q4 [10]}}