| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find period from given information |
| Difficulty | Standard +0.3 This is a straightforward SHM problem requiring standard formulas (v = ω√(a²-x²), period T = 2π/ω) with clear given information. Students must find ω from the period, use initial conditions to find amplitude from the velocity equation, then apply standard results for maximum velocity and acceleration. It's slightly above average difficulty due to the multi-step nature and need to correctly apply SHM velocity formula, but requires no novel insight—just methodical application of standard M3 techniques. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{2\pi}{\omega} = 2.4 \Rightarrow \omega = \frac{5\pi}{6} (\approx 2.62)\) | M1 A1 | |
| \(x = 0, t = 0 \Rightarrow x = a\sin\omega t\); when \(t = 0.4\), \(x = a\sin\left(\frac{5\pi}{6} \times 0.4\right) = \frac{\sqrt{3}}{2}a\) | M1 | |
| \(v^2 = \omega^2(a^2 - x^2) \Rightarrow 16 = \frac{25\pi^2}{36}\left(a^2 - \frac{3a^2}{4}\right) \Rightarrow a = \frac{48}{5\pi}(\approx 3.06)\) | M1 A1 | |
| \(v_{\max} = a\omega = 8\) | M1 A1 | cao; awrt 8.0 if decimals used earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{2\pi}{\omega} = 2.4 \Rightarrow \omega = \frac{5\pi}{6}\); \(x = a\sin\omega t\), \(\dot{x} = a\omega\cos\omega t\) | M1 A1, M1 | |
| \(4 = a\omega\cos\left(\frac{5\pi}{6} \times 0.4\right)\) | M1 | |
| \(a = \frac{48}{5\pi}(\approx 3.06)\) or \(a\omega = 8\) | A1 | |
| \(v_{\max} = a\omega = 8\) | M1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\ddot{x}_{\max} = a\omega^2 = \frac{20\pi}{3}\) | M1 A1 | awrt 21; (2) |
| Total | [9] |
# Question 2:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{2\pi}{\omega} = 2.4 \Rightarrow \omega = \frac{5\pi}{6} (\approx 2.62)$ | M1 A1 | |
| $x = 0, t = 0 \Rightarrow x = a\sin\omega t$; when $t = 0.4$, $x = a\sin\left(\frac{5\pi}{6} \times 0.4\right) = \frac{\sqrt{3}}{2}a$ | M1 | |
| $v^2 = \omega^2(a^2 - x^2) \Rightarrow 16 = \frac{25\pi^2}{36}\left(a^2 - \frac{3a^2}{4}\right) \Rightarrow a = \frac{48}{5\pi}(\approx 3.06)$ | M1 A1 | |
| $v_{\max} = a\omega = 8$ | M1 A1 | cao; awrt 8.0 if decimals used earlier |
## Part (a) Alternative:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{2\pi}{\omega} = 2.4 \Rightarrow \omega = \frac{5\pi}{6}$; $x = a\sin\omega t$, $\dot{x} = a\omega\cos\omega t$ | M1 A1, M1 | |
| $4 = a\omega\cos\left(\frac{5\pi}{6} \times 0.4\right)$ | M1 | |
| $a = \frac{48}{5\pi}(\approx 3.06)$ or $a\omega = 8$ | A1 | |
| $v_{\max} = a\omega = 8$ | M1 A1 | (7) |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\ddot{x}_{\max} = a\omega^2 = \frac{20\pi}{3}$ | M1 A1 | awrt 21; (2) |
| **Total** | **[9]** | |
---2. A particle $P$ moves in a straight line with simple harmonic motion of period 2.4 s about a fixed origin $O$. At time $t$ seconds the speed of $P$ is $v \mathrm {~ms} ^ { - 1 }$. When $t = 0 , P$ is at $O$. When $t = 0.4 , v = 4$. Find
\begin{enumerate}[label=(\alph*)]
\item the greatest speed of $P$,
\item the magnitude of the greatest acceleration of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2010 Q2 [9]}}