| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Equilibrium position with elastic string/spring |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring equilibrium analysis, proving SHM from first principles using Hooke's law, finding amplitude from energy considerations, and determining when tension becomes zero. While systematic, it demands careful setup of equations of motion, integration of multiple mechanics concepts (elasticity, SHM, energy), and extended reasoning across 14 marks—significantly above average difficulty but still within standard M3/FM scope. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{4mge}{l} = mg\) | M1 | |
| \(e = \frac{1}{4}l\) | A1 | Correct answer |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg - T = m\ddot{x}\) | M1 A1 | |
| \(mg - \frac{4mg}{l}(x + \frac{1}{4}l) = m\ddot{x}\) | M1 | |
| \(-\frac{4g}{l}x = \ddot{x}\) | A1 | Fully correct, simplified equation |
| SHM, with \(\omega = \sqrt{\frac{4g}{l}}\) | A1 | Conclusion with all work correct |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{gl} = a\sqrt{\frac{4g}{l}}\) | M1 A1 | |
| \(a = \frac{1}{2}l\) | A1 | Correct amplitude |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{4g}{l}}t\) | M1 A1 | |
| \(t = \frac{7\pi}{12}\sqrt{\frac{l}{g}}\) | M1 A1 | Solving their equation must be in radians and must give a positive value |
| (4) | ||
| 14 |
## 6(a)
$\frac{4mge}{l} = mg$ | M1 |
$e = \frac{1}{4}l$ | A1 | Correct answer
| (2) |
## 6(b)
$mg - T = m\ddot{x}$ | M1 A1 |
$mg - \frac{4mg}{l}(x + \frac{1}{4}l) = m\ddot{x}$ | M1 |
$-\frac{4g}{l}x = \ddot{x}$ | A1 | Fully correct, simplified equation
SHM, with $\omega = \sqrt{\frac{4g}{l}}$ | A1 | Conclusion with all work correct
| (5) |
## 6(c)
$\sqrt{gl} = a\sqrt{\frac{4g}{l}}$ | M1 A1 |
$a = \frac{1}{2}l$ | A1 | Correct amplitude
| (3) |
## 6(d)
$-\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{4g}{l}}t$ | M1 A1 |
$t = \frac{7\pi}{12}\sqrt{\frac{l}{g}}$ | M1 A1 | Solving their equation must be in radians and must give a positive value
| (4) |
| 14 |A light elastic string, of natural length $l$ and modulus of elasticity $4mg$, has one end attached to a fixed point $A$. The other end is attached to a particle $P$ of mass $m$. The particle hangs freely at rest in equilibrium at the point $E$. The distance of $E$ below $A$ is $(l + e)$.
\begin{enumerate}[label=(\alph*)]
\item Find $e$ in terms of $l$. [2]
\end{enumerate}
At time $t = 0$, the particle is projected vertically downwards from $E$ with speed $\sqrt{gl}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Prove that, while the string is taut, $P$ moves with simple harmonic motion. [5]
\item Find the amplitude of the simple harmonic motion. [3]
\item Find the time at which the string first goes slack. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q6 [14]}}