| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Challenging +1.2 This is a standard M3 two-spring SHM problem requiring equilibrium analysis, verification of SHM conditions, and application of standard formulas. While it involves multiple parts and careful bookkeeping of extensions/compressions, the techniques are routine for Further Maths students: Hooke's law at equilibrium, showing restoring force proportional to displacement, using impulse to find initial velocity, and applying standard SHM equations. The multi-part structure and need for systematic working place it slightly above average difficulty, but no novel insight is required. |
| 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 |
**(a)**
M1A1
$T = \frac{182x}{20x}$
$A = 2.5$, $B = 1.5$
A1
$\frac{182x}{20x \cdot 2.5} = \frac{1.5}{12}$
A1ft (4)
$x = 1.2$
$AO = 3.7$ m
**(b)**
M1A1A1
$180.8 < y < 201.2$ with $y > 0.5$
A1cso (4)
$y = 1.5 + 2.5$
$40y = y$ shows SHM (or $y = 20/m$ where $y = 6$)
B1
$\text{Max speed} = 12 \text{ m s}^{-1}$
B1ft
$0.5 = 40 \times 2 = 10$
$12 = a \times 2\pi\omega = a\sqrt{10}$
M1
$a = \frac{12}{\sqrt{10}} = \frac{6\sqrt{10}}{5}$ or $1.897...$ (accept $1.9$, $1.90$ or better)
A1ft (4)
**(c)**
$1.2 = a\sin(\omega t)$
$1.2 = \frac{1.2\sqrt{10}}{t} \sin\left(\sqrt{10}t\right)$
M1 (must use radians)
$t = \sin^{-1}\left(\frac{1.2}{\sqrt{10}}\right) \div \sqrt{10}$
A1cso (3) [15]
$t = 0.1082...$ s (Accept $0.11$ or better)
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00388805-5d60-4327-a10e-c0df74a0cb75-11_186_1042_223_452}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two points $A$ and $B$ are 6 m apart on a smooth horizontal floor. A particle $P$ of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 2.5 m and modulus of elasticity 20 N . The other end of the spring is attached to $A$. A second light elastic spring, of natural length 1.5 m and modulus of elasticity 18 N , has one end attached to $P$ and the other end attached to $B$, as shown in Figure 3. Initially $P$ rests in equilibrium at the point $O$, where $A O B$ is a straight line.
\begin{enumerate}[label=(\alph*)]
\item Find the length of $A O$.
The particle $P$ now receives an impulse of magnitude 6 N s acting in the direction $O B$ and $P$ starts to move towards $B$.
\item Show that $P$ moves with simple harmonic motion about $O$.
\item Find the amplitude of the motion.
\item Find the time taken by $P$ to travel 1.2 m from $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q6 [15]}}