| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.8 This is a challenging M3 question requiring multiple sophisticated techniques: proving SHM from elastic forces, finding when the string becomes slack, applying energy conservation during projectile motion, and calculating time periods across different motion phases. It demands strong conceptual understanding of the transition between SHM and free fall, careful coordinate system management, and extended multi-step reasoning across all four parts. While the individual techniques are A-level standard, the integration and length make this significantly harder than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) In eq. position, \(\frac{\lambda}{4}l = mg\) so \(\lambda = 3mg\) | M1 A1 | |
| At depth \(x\) below eq. position, \(mg - T = mx\) | M1 | |
| \(mg - \frac{3mg}{3l}(l + x) = mx\) so \(x = -\frac{4}{3}x\) | A1 A1 | |
| SHM with \(\omega^2 = \frac{8}{l}\) | ||
| (b) \(x = 2l \cos \omega t\); When \(x = -l\), \(\omega t = \frac{2\pi}{3}\) so \(t = \frac{2\pi}{3\omega} - \frac{2\pi}{3}\sqrt{\frac{l}{g}}\) | B1 M1 A1 A1 | |
| (c) When released, E.P.E. \(= \frac{3mg(9l)}{2(3l)} = \frac{9mgl}{2}\) | M1 A1 | |
| At max. height \(H\), P.E. \(= mgl + \frac{9mgl}{2}\) | M1 A1 | |
| \(H = 2l\) | ||
| (d) When slack, \(v^2 = 3gl\) so \(0 = \sqrt{(3gl)} - gt_h\) giving \(t_h = \sqrt{\frac{l}{g}}\) and \(T_b = \frac{2\pi}{3}\sqrt{\frac{l}{g}} + \sqrt{3}\sqrt{\frac{l}{g}}\) | M1 A1 M1 A1 | 17 marks |
**(a)** In eq. position, $\frac{\lambda}{4}l = mg$ so $\lambda = 3mg$ | M1 A1 |
At depth $x$ below eq. position, $mg - T = mx$ | M1 |
$mg - \frac{3mg}{3l}(l + x) = mx$ so $x = -\frac{4}{3}x$ | A1 A1 |
SHM with $\omega^2 = \frac{8}{l}$ | |
**(b)** $x = 2l \cos \omega t$; When $x = -l$, $\omega t = \frac{2\pi}{3}$ so $t = \frac{2\pi}{3\omega} - \frac{2\pi}{3}\sqrt{\frac{l}{g}}$ | B1 M1 A1 A1 |
**(c)** When released, E.P.E. $= \frac{3mg(9l)}{2(3l)} = \frac{9mgl}{2}$ | M1 A1 |
At max. height $H$, P.E. $= mgl + \frac{9mgl}{2}$ | M1 A1 |
$H = 2l$ | |
**(d)** When slack, $v^2 = 3gl$ so $0 = \sqrt{(3gl)} - gt_h$ giving $t_h = \sqrt{\frac{l}{g}}$ and $T_b = \frac{2\pi}{3}\sqrt{\frac{l}{g}} + \sqrt{3}\sqrt{\frac{l}{g}}$ | M1 A1 M1 A1 | 17 marksOne end of a light elastic string, of natural length $3l$ m, is attached to a fixed point $O$. A particle of mass $m$ kg is attached to the other end of the string. When the particle hangs freely in equilibrium, the string is extended by a length of $l$ m. The particle is then pulled down through a further distance $2l$ m and released from rest.
\begin{enumerate}[label=(\alph*)]
\item Prove that as long as the string is taut, the particle performs simple harmonic motion about its equilibrium position. [5 marks]
\item Show that the time between the release of the particle and the instant when the string becomes slack is $\frac{2\pi}{3}\sqrt{\frac{l}{g}}$ s. [4 marks]
\item Find the greatest height reached by the particle above its point of release. [4 marks]
\item Show that the time $T$ s taken to reach this greatest height from the moment of release is given by $T = \left(\frac{2\pi}{3} + \sqrt{3}\right)\sqrt{\frac{l}{g}}$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [17]}}