| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Standard +0.3 This is a standard SHM question requiring routine application of Hooke's law, proving SHM from F=-kx, and using standard SHM formulas (period, amplitude, velocity). While it's a multi-part question worth several marks, each step follows a predictable template with no novel insight required. The calculations are straightforward for Further Maths students familiar with SHM, making it slightly easier than average. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| At equilibrium: \(12g = \frac{\lambda \times 0.05}{0.75}\) | M1 | AO3, use of Hooke's Law |
| \(\lambda = \underline{1764}\) (N) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Consider displacement \(x\) from the equilibrium position. Apply N2L: \(12g - T = 12\ddot{x}\) | M1 | AO3 |
| \(12g - \frac{\lambda(0.05+x)}{0.75} = 12\ddot{x}\) | A1 | AO3, ft \(\lambda\) |
| \(\ddot{x} = -(14)^2 x\), therefore is SHM (with \(\omega = 14\)) | A1 | AO2 |
| Amplitude \(= \underline{0.05}\) (m) | B1 | AO1 |
| Period \(= \frac{2\pi}{\omega} = \frac{\pi}{7}\) s | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Maximum speed \(= a\omega = 0.05 \times 14 = \underline{0.7}\) (ms\(^{-1}\)) | M1, A1 | AO3, AO1, ft \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Use of \(v^2 = \omega^2(a^2 - x^2)\) with \(\omega = 14,\ a = 0.05,\ x = 0.03\) | M1, A1 | AO3, AO2, ft \(a\) |
| \(v^2 = 14^2(0.05^2 - 0.03^2) = 14^2 \times 0.04^2\) | ||
| \(v = \underline{0.56}\) (ms\(^{-1}\)) | A1 | AO1, cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Displacement from origin \(= x = 0.05\cos(14t)\) | M1 | AO3, (Accept \(\pm\)) |
| When \(t = 1.6\): \(x = 0.05\cos(14 \times 1.6)\) | A1 | AO2, ft \(a\) (Accept \(\pm\)) |
| \(x = (-)\underline{0.046}\) (m) | A1 | AO1, cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| The seat is modelled as a particle. | B1 | AO3 |
| The spring is assumed to be light. | B1 | AO3 |
| Total | [17] |
# Question 6:
## Part (a)
| Answer | Mark | Notes |
|--------|------|-------|
| At equilibrium: $12g = \frac{\lambda \times 0.05}{0.75}$ | M1 | AO3, use of Hooke's Law |
| $\lambda = \underline{1764}$ (N) | A1 | AO1 |
## Part (b)
| Answer | Mark | Notes |
|--------|------|-------|
| Consider displacement $x$ from the equilibrium position. Apply N2L: $12g - T = 12\ddot{x}$ | M1 | AO3 |
| $12g - \frac{\lambda(0.05+x)}{0.75} = 12\ddot{x}$ | A1 | AO3, ft $\lambda$ |
| $\ddot{x} = -(14)^2 x$, therefore is SHM (with $\omega = 14$) | A1 | AO2 |
| Amplitude $= \underline{0.05}$ (m) | B1 | AO1 |
| Period $= \frac{2\pi}{\omega} = \frac{\pi}{7}$ s | B1 | AO1 |
## Part (c)
| Answer | Mark | Notes |
|--------|------|-------|
| Maximum speed $= a\omega = 0.05 \times 14 = \underline{0.7}$ (ms$^{-1}$) | M1, A1 | AO3, AO1, ft $a$ |
## Part (d)
| Answer | Mark | Notes |
|--------|------|-------|
| Use of $v^2 = \omega^2(a^2 - x^2)$ with $\omega = 14,\ a = 0.05,\ x = 0.03$ | M1, A1 | AO3, AO2, ft $a$ |
| $v^2 = 14^2(0.05^2 - 0.03^2) = 14^2 \times 0.04^2$ | | |
| $v = \underline{0.56}$ (ms$^{-1}$) | A1 | AO1, cao |
## Part (e)
| Answer | Mark | Notes |
|--------|------|-------|
| Displacement from origin $= x = 0.05\cos(14t)$ | M1 | AO3, (Accept $\pm$) |
| When $t = 1.6$: $x = 0.05\cos(14 \times 1.6)$ | A1 | AO2, ft $a$ (Accept $\pm$) |
| $x = (-)\underline{0.046}$ (m) | A1 | AO1, cao |
## Part (f)
| Answer | Mark | Notes |
|--------|------|-------|
| The seat is modelled as a particle. | B1 | AO3 |
| The spring is assumed to be light. | B1 | AO3 |
| **Total** | **[17]** | |6. The diagram shows a playground ride consisting of a seat $P$, of mass 12 kg , attached to a vertical spring, which is fixed to a horizontal board. When the ride is at rest with nobody on it, the compression of the spring is 0.05 m .\\
\includegraphics[max width=\textwidth, alt={}, center]{3efc4ef6-8a80-4267-8e95-733200e875c5-4_305_654_1032_667}
The spring is of natural length 0.75 m and modulus of elasticity $\lambda$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\lambda$.
The seat $P$ is now pushed vertically downwards a further 0.05 m and is then released from rest.
\item Show that $P$ makes Simple Harmonic oscillations of period $\frac { \pi } { 7 }$ and write down the amplitude of the motion.
\item Find the maximum speed of $P$.
\item Calculate the speed of $P$ when it is at a distance 0.03 m from the equilibrium position.
\item Find the distance of $P$ from the equilibrium position 1.6 s after it is released.[3]
\item State one modelling assumption you have made about the seat and one modelling assumption you have made about the spring.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 Q6 [17]}}