| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| 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 formulas. Part (a) uses v_max = ωa with given frequency and amplitude. Part (b) requires using x = a cos(ωt) to find time for quarter-amplitude displacement. Both parts are direct applications of standard SHM results with minimal problem-solving, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2\pi}{\omega} = \frac{1}{2} \Rightarrow \omega = ...\) | M1 | Use period = 1/frequency to find \(\omega\). Must be correct way up. |
| \(\omega = 4\pi\) | A1 | Correct value for \(\omega\) |
| \(v = \omega \times 0.3\) | M1 | Use of \(v = a\omega\) or \(v^2 = \omega^2(a^2 - x^2)\) with \(x=0\) |
| \(v = 1.2\pi\), 3.8 or better (m s\(^{-1}\)) | A1 (4) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = a\sin\omega t \Rightarrow 0.15 = 0.3\sin 4\pi t \Rightarrow t = ...\) | M1 | Use \(0.15 = a\sin\omega t\) to obtain a value for \(t\). Use *their* \(a\) and \(\omega\). |
| \(t = \frac{1}{4\pi} \times \frac{\pi}{6} = \frac{1}{24}\) (s), \(0.04166... = 0.042\) or better | A1 (2) [6] | Correct value, 0.042 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = a\cos\omega t \Rightarrow 0.15 = 0.3\cos 4\pi t \Rightarrow t = ...\) | MI | Complete method using \(x = a\cos\omega t\) AND \(\frac{T}{4}\) to obtain value for \(t\) |
| \(\frac{T}{4} - t = \frac{0.5}{4} - t = ...\) | ||
| Correct value, 0.042 or better | A1 |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2\pi}{\omega} = \frac{1}{2} \Rightarrow \omega = ...$ | M1 | Use period = 1/frequency to find $\omega$. Must be correct way up. |
| $\omega = 4\pi$ | A1 | Correct value for $\omega$ |
| $v = \omega \times 0.3$ | M1 | Use of $v = a\omega$ or $v^2 = \omega^2(a^2 - x^2)$ with $x=0$ |
| $v = 1.2\pi$, 3.8 or better (m s$^{-1}$) | A1 (4) | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = a\sin\omega t \Rightarrow 0.15 = 0.3\sin 4\pi t \Rightarrow t = ...$ | M1 | Use $0.15 = a\sin\omega t$ to obtain a value for $t$. Use *their* $a$ and $\omega$. |
| $t = \frac{1}{4\pi} \times \frac{\pi}{6} = \frac{1}{24}$ (s), $0.04166... = 0.042$ or better | A1 (2) **[6]** | Correct value, 0.042 or better |
**ALT 1(b) Using cos:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = a\cos\omega t \Rightarrow 0.15 = 0.3\cos 4\pi t \Rightarrow t = ...$ | MI | Complete method using $x = a\cos\omega t$ AND $\frac{T}{4}$ to obtain value for $t$ |
| $\frac{T}{4} - t = \frac{0.5}{4} - t = ...$ | | |
| Correct value, 0.042 or better | A1 | |
---\begin{enumerate}
\item A particle $P$ moves in a straight line with simple harmonic motion between two fixed points $A$ and $B$. The particle performs 2 complete oscillations per second. The midpoint of $A B$ is $O$ and the midpoint of $O A$ is $C$
\end{enumerate}
The length of $A B$ is 0.6 m .\\
(a) Find the maximum speed of $P$\\
(b) Find the time taken by $P$ to move directly from $O$ to $C$
\hfill \mbox{\textit{Edexcel M3 2022 Q1 [6]}}