| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find period from given information |
| Difficulty | Challenging +1.2 This is a multi-part SHM question requiring understanding of amplitude, period relationships, and standard formulas. Part (a) involves geometric reasoning about positions and timing to find the period (a 'show that' proof), while parts (b-d) apply standard SHM formulas. The geometric setup with midpoints adds moderate complexity beyond routine textbook exercises, but the techniques are all standard M3 material with no novel insights required. |
| Spec | 3.02h Motion under gravity: vector form4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = a\sin\omega t\); \(0.125 = 0.25\sin 0.1\omega\) | M1A1 | |
| \(\sin 0.1\omega = \frac{1}{2}\), so \(0.1\omega = \frac{\pi}{6}\) | ||
| \(\omega = \frac{\pi}{0.6} = \frac{10\pi}{6}\) | M1depA1 | |
| Period \(= \frac{2\pi}{\omega} = \frac{6}{5}\ (=1.2)\) | A1 | B1 on e-pen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = 0.25\sin\frac{5}{3}\pi t\) | ||
| \(t=2\): \(x = 0.25\sin\left(2\times\frac{5}{3}\pi\right)\) | M1 | |
| \(x = -0.2165...\) | A1 | |
| Dist from \(B = 0.25 + x = 0.033\) m | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Max accel \(= a\omega^2 = 0.25\times\left(\frac{5\pi}{3}\right)^2 = 6.853...= 6.85\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Max speed \(= a\omega = 0.25\times\left(\frac{5\pi}{3}\right) = 1.308...= 1.31\) | M1A1 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = a\sin\omega t$; $0.125 = 0.25\sin 0.1\omega$ | M1A1 | |
| $\sin 0.1\omega = \frac{1}{2}$, so $0.1\omega = \frac{\pi}{6}$ | | |
| $\omega = \frac{\pi}{0.6} = \frac{10\pi}{6}$ | M1depA1 | |
| Period $= \frac{2\pi}{\omega} = \frac{6}{5}\ (=1.2)$ | A1 | B1 on e-pen |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = 0.25\sin\frac{5}{3}\pi t$ | | |
| $t=2$: $x = 0.25\sin\left(2\times\frac{5}{3}\pi\right)$ | M1 | |
| $x = -0.2165...$ | A1 | |
| Dist from $B = 0.25 + x = 0.033$ m | A1ft | |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| Max accel $= a\omega^2 = 0.25\times\left(\frac{5\pi}{3}\right)^2 = 6.853...= 6.85$ | M1A1 | |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| Max speed $= a\omega = 0.25\times\left(\frac{5\pi}{3}\right) = 1.308...= 1.31$ | M1A1 | |
---5. A particle $P$ is moving in a straight line with simple harmonic motion on a smooth horizontal floor. The particle comes to instantaneous rest at points $A$ and $B$ where $A B$ is 0.5 m . The mid-point of $A B$ is $O$. The mid-point of $O A$ is $C$. The mid-point of $O B$ is $D$. The particle takes 0.2 s to travel directly from $C$ to $D$. At time $t = 0 , P$ is moving through $O$ towards $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the period of the motion is $\frac { 6 } { 5 } \mathrm {~s}$.
\item Find the distance of $P$ from $B$ when $t = 2 \mathrm {~s}$.
\item Find the maximum magnitude of the acceleration of $P$.
\item Find the maximum speed of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q5 [12]}}