| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Maximum speed in SHM |
| Difficulty | Standard +0.8 This is a substantial M3 question requiring multiple techniques: Hooke's law equilibrium, deriving SHM from first principles, applying SHM formulas for period and speed, and solving a non-trivial timing problem involving inverse trigonometric functions. While the individual components are standard M3 material, the multi-part structure, the need to carefully track the equilibrium position as the centre of oscillation, and the final timing calculation (part e) requiring precise SHM phase analysis make this moderately challenging. It's harder than average A-level but standard for M3 students who have practiced SHM with elastic strings. |
| 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^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| (a) \(mg = \frac{mg}{2}e\) | M1 A1 |
| \(e : l = 2 : 1\) | |
| (b) \(mx = mg - \frac{mg}{2l}(2l + x)\) | M1 A1 M1 A1 |
| \(\ddot{x} = -\frac{g}{2}x\), so S.H.M. | |
| (c) \(\omega^2 = \frac{g}{l}\) | M1 A1 |
| Period = \(\frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2l}{g}} \, \text{s}\) | |
| (d) At \(E\), \(v = a\omega = \frac{3l}{2}\sqrt{\frac{2l}{g}} = \sqrt{\frac{8gl}{8}} \, \text{ms}^{-1}\) | M1 A1 |
| (e) \(x = \frac{3l}{4} \cos \omega t\) | M1 A1 I1 |
| \(x = -\frac{3l}{4}\) when \(\frac{3l}{4} \cos \omega t = \frac{3l}{4}\) | |
| \(\cos \omega t = -\frac{1}{2}\) | |
| \(\omega t = \frac{2\pi}{3}\) | M1 A1 |
| \(t = \frac{2\pi}{3\omega} = \frac{2\pi}{3}\sqrt{\frac{2l}{g}} \, \text{s}\) | |
| Total: 15 marks |
**(a)** $mg = \frac{mg}{2}e$ | M1 A1 |
$e : l = 2 : 1$ | |
**(b)** $mx = mg - \frac{mg}{2l}(2l + x)$ | M1 A1 M1 A1 |
$\ddot{x} = -\frac{g}{2}x$, so S.H.M. | |
**(c)** $\omega^2 = \frac{g}{l}$ | M1 A1 |
Period = $\frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2l}{g}} \, \text{s}$ | |
**(d)** At $E$, $v = a\omega = \frac{3l}{2}\sqrt{\frac{2l}{g}} = \sqrt{\frac{8gl}{8}} \, \text{ms}^{-1}$ | M1 A1 |
**(e)** $x = \frac{3l}{4} \cos \omega t$ | M1 A1 I1 |
$x = -\frac{3l}{4}$ when $\frac{3l}{4} \cos \omega t = \frac{3l}{4}$ | |
$\cos \omega t = -\frac{1}{2}$ | |
$\omega t = \frac{2\pi}{3}$ | M1 A1 |
$t = \frac{2\pi}{3\omega} = \frac{2\pi}{3}\sqrt{\frac{2l}{g}} \, \text{s}$ | |
| | **Total: 15 marks** |A light elastic string, of natural length $l$ m and modulus of elasticity $\frac{mg}{2}$ newtons, has one end fastened to a fixed point $O$. A particle $P$, of mass $m$ kg, is attached to the other end of the string. $P$ hangs in equilibrium at the point $E$, vertically below $O$, where $OE = (l + e)$ m
\begin{enumerate}[label=(\alph*)]
\item Find the numerical value of the ratio $e : l$. [2 marks]
\end{enumerate}
$P$ is now pulled down a further distance $\frac{3l}{2}$ m from $E$ and is released from rest.
In the subsequent motion, the string remains taut. At time $t$ s after being released, $P$ is at a distance $x$ m below $E$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Write down a differential equation for the motion of $P$ and show that the motion is simple harmonic. [4 marks]
\item Write down the period of the motion. [2 marks]
\item Find the speed with which $P$ first passes through $E$ again. [2 marks]
\item Show that the time taken by $P$ after it is released to reach the point $A$ above $E$, where $AE = \frac{3l}{4}$ m, is $\frac{2\pi}{3}\sqrt{\frac{2l}{g}}$ s. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [15]}}