| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Standard +0.3 This is a standard SHM question requiring knowledge of period/amplitude relationships and the position-time equation. Part (c) involves solving x = a cos(ωt) for t, which is routine A-level mechanics. The multi-step nature and need to set up the equation correctly places it slightly above average, but all techniques are standard M3 material with no novel insight required. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) period = 2 × 3 = 6 s | B1 | |
| (ii) amplitude = \(\frac{1}{4} \times 4 = 2\) m | B1 | |
| (b) period = 6 = \(\frac{2\pi}{\omega}\) \(\therefore \omega = \frac{\pi}{3}\) | M1 | |
| \(v_{\text{max}} = a\omega = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}\) | A1 | |
| \(KE_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2} \times \frac{1}{2} \times \left(\frac{2\pi}{3}\right)^2 = \frac{1}{9}\pi^2 \text{ J}\) | M1 A1 | |
| (c) \(x = a \cos \omega t\) \(\therefore 0.8 = 2 \cos \omega t\) cos \(\omega t = 0.4\) \(\therefore \omega t = 1.9823\) \(\therefore t = 1.9823 \div \frac{\pi}{3} = 1.89 \text{ s}\) (2dp) | M1 A1; M1 A1 | (10) |
(a) (i) period = 2 × 3 = 6 s | B1 |
(ii) amplitude = $\frac{1}{4} \times 4 = 2$ m | B1 |
(b) period = 6 = $\frac{2\pi}{\omega}$ $\therefore \omega = \frac{\pi}{3}$ | M1 |
$v_{\text{max}} = a\omega = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$ | A1 |
$KE_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2} \times \frac{1}{2} \times \left(\frac{2\pi}{3}\right)^2 = \frac{1}{9}\pi^2 \text{ J}$ | M1 A1 |
(c) $x = a \cos \omega t$ $\therefore 0.8 = 2 \cos \omega t$ cos $\omega t = 0.4$ $\therefore \omega t = 1.9823$ $\therefore t = 1.9823 \div \frac{\pi}{3} = 1.89 \text{ s}$ (2dp) | M1 A1; M1 A1 | (10)
---4. A particle of mass 0.5 kg is moving on a straight line with simple harmonic motion.
At time $t = 0$ the particle is instantaneously at rest at the point $A$. It next comes instantaneously to rest 3 seconds later at the point $B$ where $A B = 4 \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item For the motion of the particle write down
\begin{enumerate}[label=(\roman*)]
\item the period,
\item the amplitude.
\end{enumerate}\item Find the maximum kinetic energy of the particle in terms of $\pi$.
The point $C$ lies on $A B$ at a distance of 1.2 m from $B$.
\item Find the time it takes the particle to travel directly from $A$ to $C$, giving your answer in seconds correct to 2 decimal places.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [10]}}