| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | String becomes slack timing |
| Difficulty | Challenging +1.2 This is a standard M3 SHM question with elastic strings requiring multiple techniques (equilibrium, differential equation formation, energy/amplitude calculations, and timing). While it involves several steps and careful bookkeeping of extensions, the methods are entirely routine for this module with no novel insights required. The 'show that' format provides targets to work towards, reducing problem-solving demand. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| In equilibrium: \(\dfrac{6mge}{a} = mg \Rightarrow e = \dfrac{a}{6}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(m\ddot{x} = -\dfrac{6mg(e+x)}{a} + mg\) | M1 A1 A1 | |
| \(\ddot{x} = -\dfrac{6g}{a}x \Rightarrow\) SHM | M1 A1 | |
| Period \(= \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{a}{6g}}\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Greatest speed \(= a\omega = \dfrac{a}{3}\sqrt{\dfrac{6g}{a}} = \dfrac{1}{3}\sqrt{6ga}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(x = \dfrac{a}{3}\cos\omega t\) | M1 | |
| String slack \(\Rightarrow x = -e \Rightarrow -\dfrac{a}{6} = \dfrac{a}{3}\cos\omega t\) | M1 A1 | |
| \(\omega t = \dfrac{2\pi}{3},\quad t = \dfrac{2\pi}{3}\sqrt{\dfrac{a}{6g}}\) | M1 A1 ft | (5) |
# Question 7:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| In equilibrium: $\dfrac{6mge}{a} = mg \Rightarrow e = \dfrac{a}{6}$ | M1 A1 | (2) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $m\ddot{x} = -\dfrac{6mg(e+x)}{a} + mg$ | M1 A1 A1 | |
| $\ddot{x} = -\dfrac{6g}{a}x \Rightarrow$ SHM | M1 A1 | |
| Period $= \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{a}{6g}}$ | A1 | (6) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| Greatest speed $= a\omega = \dfrac{a}{3}\sqrt{\dfrac{6g}{a}} = \dfrac{1}{3}\sqrt{6ga}$ | M1 A1 | (2) |
## Part (d):
| Working/Answer | Marks | Notes |
|---|---|---|
| $x = \dfrac{a}{3}\cos\omega t$ | M1 | |
| String slack $\Rightarrow x = -e \Rightarrow -\dfrac{a}{6} = \dfrac{a}{3}\cos\omega t$ | M1 A1 | |
| $\omega t = \dfrac{2\pi}{3},\quad t = \dfrac{2\pi}{3}\sqrt{\dfrac{a}{6g}}$ | M1 A1 ft | (5) |7. A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $a$ and modulus of elasticity 6 mg . The other end of the string is attached to a fixed point $O$. When the particle hangs in equilibrium with the string vertical, the extension of the string is $e$.
\begin{enumerate}[label=(\alph*)]
\item Find $e$.\\
(2)
The particle is now pulled down a vertical distance $\frac { 1 } { 3 } a$ below its equilibrium position and released from rest. At time $t$ after being released, during the time when the string remains taut, the extension of the string is $e + x$. By forming a differential equation for the motion of $P$ while the string remains taut,
\item show that during this time $P$ moves with simple harmonic motion of period $2 \pi \sqrt { \frac { a } { 6 g } }$.\\
(6)
\item Show that, while the string remains taut, the greatest speed of $P$ is $\frac { 1 } { 3 } \sqrt { } ( 6 g a )$.
\item Find $t$ when the string becomes slack for the first time.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [15]}}