| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question requiring proof of SHM condition, finding the period using Hooke's law and equilibrium, then applying energy methods and SHM equations. While it involves multiple steps and careful setup (finding spring constant from equilibrium, establishing centre of oscillation), these are well-practiced techniques for Further Maths students. The multi-part structure and need to track natural length vs extension elevates it slightly above average difficulty. |
| 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 | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(mg = \frac{\lambda \times 4l}{2l} \Rightarrow \lambda = \frac{1}{2}mg\) | M1 A1 | M1 Resolving vertically and using Hooke's Law; A1 cao |
| \(mg - \frac{\frac{1}{2}mg(x+4l)}{2l} = m\ddot{x}\) (or \(x\) replaced by \(-x\) on both sides) | M1 A2,1,0 | M1 equation of motion in general position with correct no. of terms, allow use of accln instead of derivative; A2 for correct equation, A1 for equation with at most one error |
| \(-\frac{g}{4l}x = \ddot{x}\), hence SHM with \(\omega = \sqrt{\frac{g}{4l}}\) | A1 | A1 for correct equation in correct form |
| \(T = \frac{2\pi}{\omega}\) | M1 | M1 Use of correct formula |
| \(2\pi\sqrt{\frac{4l}{g}} = 4\pi\sqrt{\frac{l}{g}}\) * | A1* | A1* Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Their \(\omega \times 2l\) | M1 | M1 for use of correct formula |
| \(\sqrt{gl}\) | A1 | A1 for correct speed |
| \(\frac{1}{2}mgl\) | A1 | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(-l = 2l\cos\omega t\) | M1 A1 | M1 for complete method to find \(t\); A1 for correct equation(s) (\(\omega\) does not need to be substituted for this mark) |
| \(t = \frac{2\pi}{3\omega}\) | M1 | M1 for solving for \(t\) |
| \(\frac{1}{3}T\) | A1 | A1 cso |
## Question 6:
### Part 6(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg = \frac{\lambda \times 4l}{2l} \Rightarrow \lambda = \frac{1}{2}mg$ | M1 A1 | M1 Resolving vertically and using Hooke's Law; A1 cao |
| $mg - \frac{\frac{1}{2}mg(x+4l)}{2l} = m\ddot{x}$ (or $x$ replaced by $-x$ on both sides) | M1 A2,1,0 | M1 equation of motion in general position with correct no. of terms, allow use of accln instead of derivative; A2 for correct equation, A1 for equation with at most one error |
| $-\frac{g}{4l}x = \ddot{x}$, hence SHM with $\omega = \sqrt{\frac{g}{4l}}$ | A1 | A1 for correct equation in correct form |
| $T = \frac{2\pi}{\omega}$ | M1 | M1 Use of correct formula |
| $2\pi\sqrt{\frac{4l}{g}} = 4\pi\sqrt{\frac{l}{g}}$ * | A1* | A1* Given answer correctly obtained |
**(8 marks)**
### Part 6(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Their $\omega \times 2l$ | M1 | M1 for use of correct formula |
| $\sqrt{gl}$ | A1 | A1 for correct speed |
| $\frac{1}{2}mgl$ | A1 | A1 cao |
**(3 marks)**
### Part 6(c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $-l = 2l\cos\omega t$ | M1 A1 | M1 for complete method to find $t$; A1 for correct equation(s) ($\omega$ does not need to be substituted for this mark) |
| $t = \frac{2\pi}{3\omega}$ | M1 | M1 for solving for $t$ |
| $\frac{1}{3}T$ | A1 | A1 cso |
**(4 marks)**
---\begin{enumerate}
\item A particle $P$ of mass $m$ is attached to one end of a light elastic spring of natural length 2l. The other end of the spring is attached to a fixed point $A$. The particle $P$ hangs in equilibrium vertically below $A$, at the point $E$ where $A E = 6 l$. The particle $P$ is then raised a vertical distance $2 l$ and released from rest.
\end{enumerate}
Air resistance is modelled as being negligible.\\
(a) Show that $P$ moves with simple harmonic motion of period $T$ where
$$T = 4 \pi \sqrt { \frac { l } { g } }$$
(b) Find, in terms of $m , l$ and $g$, the kinetic energy of $P$ as it passes through $E$\\
(c) Find, in terms of $T$, the exact time from the instant when $P$ is released to the instant when $P$ has moved a distance 31 .
\hfill \mbox{\textit{Edexcel M3 2022 Q6 [15]}}