| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2002 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Maximum speed in SHM |
| Difficulty | Standard +0.3 This is a standard M3 SHM question with elastic strings following a predictable structure: find equilibrium extension, show SHM using Hooke's law and F=ma, then use energy/amplitude relationships. Part (a) is routine Hooke's law, part (b) follows the standard 'show that' template for SHM problems, part (c) applies v_max = ωA formula, and part (d) requires recognizing when the string goes slack. While it requires multiple techniques and careful algebra, it's a textbook application of well-practiced methods with no novel insight required, making it slightly easier 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^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg = \frac{8mge}{4a}\) | M1 | |
| \(\frac{2}{3}a = AO\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg - \frac{8mg}{4a}(e + x) = m\ddot{x}\) | M1 M1 A1 | |
| \(\ddot{x} = -\frac{2g}{a}x\) | M1 A1 | |
| \(T = 2\pi\sqrt{\frac{a}{2g}} = \pi\sqrt{\frac{2a}{g}}\) | M1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = d\omega\) | M1 | |
| \(\frac{1}{2}\sqrt{ga} = d\sqrt{\frac{2g}{a}}\) | A1 ft on \(\omega\) | |
| \(d = \frac{a}{2\sqrt{2}} = a\frac{\sqrt{2}}{4} = 0.35a\) (awrt) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Partly under gravity, partly SHM | B1 B1 | (2) |
## Part (a)
$mg = \frac{8mge}{4a}$ | M1
$\frac{2}{3}a = AO$ | A1 | (2)
## Part (b)
$mg - \frac{8mg}{4a}(e + x) = m\ddot{x}$ | M1 M1 A1
$\ddot{x} = -\frac{2g}{a}x$ | M1 A1
$T = 2\pi\sqrt{\frac{a}{2g}} = \pi\sqrt{\frac{2a}{g}}$ | M1 A1 | (7)
## Part (c)
$v = d\omega$ | M1
$\frac{1}{2}\sqrt{ga} = d\sqrt{\frac{2g}{a}}$ | A1 ft on $\omega$
$d = \frac{a}{2\sqrt{2}} = a\frac{\sqrt{2}}{4} = 0.35a$ (awrt) | A1 | (3)
## Part (d)
Partly under gravity, partly SHM | B1 B1 | (2)
**Total: (14 marks)**
---A light elastic string, of natural length $4a$ and modulus of elasticity $8mg$, has one end attached to a fixed point $A$. A particle $P$ of mass $m$ is attached to the other end of the string and hangs in equilibrium at the point $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $AO$.
[2]
\end{enumerate}
The particle is now pulled down to a point $C$ vertically below $O$, where $OC = d$. It is released from rest. In the subsequent motion the string does not become slack.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ moves with simple harmonic motion of period $\pi\sqrt{\frac{2a}{g}}$.
[7]
\end{enumerate}
The greatest speed of $P$ during this motion is $\frac{1}{2}\sqrt{(ga)}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find $d$ in terms of $a$.
[3]
\end{enumerate}
Instead of being pulled down a distance $d$, the particle is pulled down a distance $a$. Without further calculation,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item describe briefly the subsequent motion of $P$.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2002 Q6 [14]}}