| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Challenging +1.2 This is a multi-part SHM question requiring equilibrium analysis with two springs, proving SHM conditions, and finding time proportions using inverse trig functions. While it involves several steps and careful bookkeeping with two springs, the techniques are standard M3 material: Hooke's law, F=ma leading to ẍ=-ω²x, and amplitude/period calculations. The most challenging aspect is part (c) requiring integration of the SHM equation to find time intervals, which is beyond routine but well within expected M3 problem-solving. |
| 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 |
|---|---|
| Hooke's law: Equilibrium \(\Rightarrow \frac{16(d-2)}{2} = \frac{12(4-d)}{1}\) | M1A1A1 |
| \(\Rightarrow d = 3.2\) so extensions are 1.2m and 0.8m. | A1 |
| Answer | Marks |
|---|---|
| If the particle is displaced distance \(x\) towards B then \(-m\ddot{x} = \frac{16(1.2 + x)}{2} - \frac{12(0.8 - x)}{1}(= 20x)\) | M1A1ft, A1ft |
| \(\Rightarrow \ddot{x} = -40x\) or \(\ddot{x} = -\frac{20}{m}\) (\(\Rightarrow\) SHM) | A1 |
| Answer | Marks |
|---|---|
| \(T = \frac{2\pi}{\sqrt{40}}\) | B1ft |
| \(a = \frac{\sqrt{10}}{?}\) their \(\omega\) | B1ft |
| \(x = a\sin\omega t\) their \(a\), their \(\omega\) | M1 |
| \(\frac{1}{4} = \frac{1}{2}\sin\sqrt{40}t\) | A1 |
| \(\sqrt{40}t = \frac{\pi}{6}\) (\(\Rightarrow t = \frac{\pi}{6\sqrt{40}}\)) | M1 |
| Proportion \(\frac{4t}{T} = \frac{4\pi}{6\sqrt{40}} \times \frac{\sqrt{40}}{2\pi} = \frac{1}{3}\) | M1A1 |
## Part (a)
Hooke's law: Equilibrium $\Rightarrow \frac{16(d-2)}{2} = \frac{12(4-d)}{1}$ | M1A1A1 |
$\Rightarrow d = 3.2$ so extensions are 1.2m and 0.8m. | A1 |
## Part (b)
If the particle is displaced distance $x$ towards B then $-m\ddot{x} = \frac{16(1.2 + x)}{2} - \frac{12(0.8 - x)}{1}(= 20x)$ | M1A1ft, A1ft |
$\Rightarrow \ddot{x} = -40x$ or $\ddot{x} = -\frac{20}{m}$ ($\Rightarrow$ SHM) | A1 |
## Part (c)
$T = \frac{2\pi}{\sqrt{40}}$ | B1ft |
$a = \frac{\sqrt{10}}{?}$ their $\omega$ | B1ft |
$x = a\sin\omega t$ their $a$, their $\omega$ | M1 |
$\frac{1}{4} = \frac{1}{2}\sin\sqrt{40}t$ | A1 |
$\sqrt{40}t = \frac{\pi}{6}$ ($\Rightarrow t = \frac{\pi}{6\sqrt{40}}$) | M1 |
Proportion $\frac{4t}{T} = \frac{4\pi}{6\sqrt{40}} \times \frac{\sqrt{40}}{2\pi} = \frac{1}{3}$ | M1A1 |\includegraphics{figure_4}
$A$ and $B$ are two points on a smooth horizontal floor, where $AB = 5$ m.
A particle $P$ has mass $0.5$ kg. One end of a light elastic spring, of natural length $2$ m and modulus of elasticity $16$ N, is attached to $P$ and the other end is attached to $A$. The ends of another light elastic spring, of natural length $1$ m and modulus of elasticity $12$ N, are attached to $P$ and $B$, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Find the extensions in the two springs when the particle is at rest in equilibrium. [5]
\end{enumerate}
Initially $P$ is at rest in equilibrium. It is then set in motion and starts to move towards $B$. In the subsequent motion $P$ does not reach $A$ or $B$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ oscillates with simple harmonic motion about the equilibrium position. [4]
\item Given that the initial speed of $P$ is $\sqrt{10}$ m s$^{-1}$, find the proportion of time in each complete oscillation for which $P$ stays within $0.25$ m of the equilibrium position. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q7 [16]}}