| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Standard +0.8 This is a standard two-spring SHM problem requiring equilibrium analysis using Hooke's law, verification of SHM conditions by showing restoring force proportional to displacement, and finding when tension becomes zero. While it involves multiple steps and careful bookkeeping of extensions/compressions, the techniques are routine for Further Maths mechanics students with no novel insights required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
6. The diagram shows a particle $P$, of mass 4 kg , lying on a smooth horizontal surface. It is attached by two light springs to fixed points $A$ and $B$, where $A B = 2.8 \mathrm {~m}$.\\
Spring $A P$ has natural length 0.8 m and modulus of elasticity 60 N .\\
Spring $P B$ has natural length 1.2 m and modulus of elasticity 30 N .\\
\includegraphics[max width=\textwidth, alt={}, center]{b9c63cb4-d446-4548-be42-e30b10cb4b99-5_231_1253_612_404}
When $P$ is in equilibrium, it is at the point $C$.
\begin{enumerate}[label=(\alph*)]
\item Show that $A C = 1 \mathrm {~m}$.
\item The particle $P$ is pulled horizontally and is initially held at rest at the midpoint of $A B$. The system is then released.
\begin{enumerate}[label=(\roman*)]
\item Show that $P$ performs Simple Harmonic Motion about centre $C$ and find the period of its motion.
\item Determine the shortest time taken for $P$ to reach a position where there is no tension in the spring $A P$.
\section*{END OF PAPER}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 2022 Q6 [14]}}