| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: horizontal or non-standard geometry |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring routine application of formulas (proving SHM from Hooke's law, finding amplitude and period, using standard SHM equations for velocity at a given time). Part (a) follows a textbook template, parts (b-c) involve straightforward substitution into SHM formulas, and part (d) requires simple reasoning about the motion's direction. While multi-part with 13 marks total, each step is procedural without requiring novel insight or complex problem-solving. |
| 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 |
|---|---|---|
| Hooke's Law: \(T = \frac{12x}{0.6} = [20x]\) | M1 | |
| Equation of motion: \((--)T = 0.8\ddot{x}\) | M1 | |
| \(-\frac{12x}{0.6} = 0.8\ddot{x}\) → \(\ddot{x} = -25x\) | A1 | |
| Finding \(\omega\) from derived equation of form \(\ddot{x} = -\omega^2 x\) | M1 | |
| Period \(= \frac{2\pi}{\omega} = \frac{2\pi}{5}\) (*) | (no incorrect working seen) | A1 |
| Answer | Marks |
|---|---|
| Substituting (candidate's) \(\omega\) and \(a\) in \(\omega^2 a\): \(= 25 \times 0.25 = 6.25\) (m s\(^{-2}\)) | M1; A1 |
| (or finding \(T_{\max} = 0.8a \Rightarrow a = 5/0.8 = 6.25\)) | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Complete method for \(x\): \(x = 0.25\cos 10° (-0.2098)\) | M1 A1 | |
| Using \(v^2 = \omega^2(a^2 - x^2) \Rightarrow v = (\pm)5\sqrt{(0.25)^2 - (0.25\cos 10°)^2}\) | M1 A1 ft | |
| \(v = (\pm)0.68\) (m s\(^{-1}\)) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Direction \(\overrightarrow{OB}\) or equivalent | B1 | (1 mark) |
## (a)
Hooke's Law: $T = \frac{12x}{0.6} = [20x]$ | M1 |
Equation of motion: $(--)T = 0.8\ddot{x}$ | M1 |
$-\frac{12x}{0.6} = 0.8\ddot{x}$ → $\ddot{x} = -25x$ | A1 |
Finding $\omega$ from derived equation of form $\ddot{x} = -\omega^2 x$ | M1 |
Period $= \frac{2\pi}{\omega} = \frac{2\pi}{5}$ (*) | (no incorrect working seen) | A1 | (5 marks)
## (b)
Substituting (candidate's) $\omega$ and $a$ in $\omega^2 a$: $= 25 \times 0.25 = 6.25$ (m s$^{-2}$) | M1; A1 |
(or finding $T_{\max} = 0.8a \Rightarrow a = 5/0.8 = 6.25$) | (2 marks)
## (c)
Complete method for $x$: $x = 0.25\cos 10° (-0.2098)$ | M1 A1 |
Using $v^2 = \omega^2(a^2 - x^2) \Rightarrow v = (\pm)5\sqrt{(0.25)^2 - (0.25\cos 10°)^2}$ | M1 A1 ft |
$v = (\pm)0.68$ (m s$^{-1}$) | A1 | (5 marks)
## (d)
Direction $\overrightarrow{OB}$ or equivalent | B1 | (1 mark)
**(13 marks total)**
---A particle $P$ of mass $0.8$ kg is attached to one end $A$ of a light elastic spring $OA$, of natural length $60$ cm and modulus of elasticity $12$ N. The spring is placed on a smooth horizontal table and the end $O$ is fixed. The particle $P$ is pulled away from $O$ to a point $B$, where $OB = 85$ cm, and is released from rest.
\begin{enumerate}[label=(\alph*)]
\item Prove that the motion of $P$ is simple harmonic with period $\frac{2\pi}{5}$ s. [5]
\item Find the greatest magnitude of the acceleration of $P$ during the motion. [2]
\end{enumerate}
Two seconds after being released from rest, $P$ passes through the point $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, to 2 significant figures, the speed of $P$ as it passes through $C$. [5]
\item State the direction in which $P$ is moving 2 s after being released. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q5 [13]}}