| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 16 |
| 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 Further Maths SHM question requiring Hooke's law equilibrium analysis, proving SHM conditions (F ∝ -x), and applying standard SHM formulas for period and motion. Part (a) is routine equilibrium, (b) requires showing the restoring force condition but follows a well-practiced method, while (c)-(d) apply standard SHM amplitude and timing formulas. The multi-part structure and Further Maths context place it above average difficulty, but it's a textbook-style question without novel insight required. |
| Spec | 4.10f 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(mg = \frac{2mge}{l}\) | M1 | Use of Hooke's Law and \(T = mg\) |
| \(e = \frac{1}{2}l\) so \(AO = \frac{3l}{2}\) | A1* | Correct answer fully justified |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion vertically: \(mg - T = m\ddot{x}\) | M1 | All terms needed but allow \(a\) for the acceleration |
| \(mg - \frac{2mg(x+e)}{l} = m\ddot{x}\) | A1 | Correct unsimplified equation including \(\ddot{x}\) |
| \(-\frac{2g}{l}x = \ddot{x}\), so SHM with \(\omega^2 = \frac{2g}{l}\) | A1 | Must mention SHM |
| Use of \(\frac{2\pi}{\omega}\) | M1 | Correct method |
| \(2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}\) | A1* | Correct answer correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Complete method to find \(h\) | M1 | Correct no. of terms, dimensionally correct in energy equation OR use SHM to find \(v^2\) at unstretched position AND then use motion under gravity. Correct no. of terms, dimensionally correct in both equations |
| \(mgh = \frac{2mg\left(\frac{3l}{2}\right)^2}{2l}\) | A1 | Equation with at most one error |
| OR \(v^2 = \frac{2g}{l}\left(l^2 - \left(-\frac{1}{2}l\right)^2\right)\) and \(0 = \frac{3gl}{2} - 2gs\) | A1 | Correct equation OR correct equations |
| \(h = \frac{9l}{4}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(-\frac{1}{2}l = l\cos\omega t\) | M1 | Complete method to find time to reach unstretched position |
| \(t = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}\) | A1 | Correct time |
| \(v = \sqrt{\frac{2g}{l}\left(l^2 - \left(-\frac{1}{2}l\right)^2\right)}\) OR | M1 | Complete method to find speed at unstretched position |
| \(0 = \sqrt{\frac{3gl}{2}} - gt_1 \Rightarrow t_1 = \sqrt{\frac{3l}{2g}}\) | M1 | Complete method to find time to rest position |
| Total time \(= \frac{2\pi}{3}\sqrt{\frac{l}{2g}} + \sqrt{\frac{3l}{2g}}\) | A1 | cao |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $mg = \frac{2mge}{l}$ | M1 | Use of Hooke's Law and $T = mg$ |
| $e = \frac{1}{2}l$ so $AO = \frac{3l}{2}$ | A1* | Correct answer fully justified |
**Total: 2 marks**
---
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion vertically: $mg - T = m\ddot{x}$ | M1 | All terms needed but allow $a$ for the acceleration |
| $mg - \frac{2mg(x+e)}{l} = m\ddot{x}$ | A1 | Correct unsimplified equation including $\ddot{x}$ |
| $-\frac{2g}{l}x = \ddot{x}$, so SHM with $\omega^2 = \frac{2g}{l}$ | A1 | Must mention SHM |
| Use of $\frac{2\pi}{\omega}$ | M1 | Correct method |
| $2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}$ | A1* | Correct answer correctly shown |
**Total: 5 marks**
---
## Question 6(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Complete method to find $h$ | M1 | Correct no. of terms, dimensionally correct in energy equation OR use SHM to find $v^2$ at unstretched position AND then use motion under gravity. Correct no. of terms, dimensionally correct in both equations |
| $mgh = \frac{2mg\left(\frac{3l}{2}\right)^2}{2l}$ | A1 | Equation with at most one error |
| OR $v^2 = \frac{2g}{l}\left(l^2 - \left(-\frac{1}{2}l\right)^2\right)$ and $0 = \frac{3gl}{2} - 2gs$ | A1 | Correct equation OR correct equations |
| $h = \frac{9l}{4}$ | A1 | cao |
**Total: 4 marks**
---
## Question 6(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $-\frac{1}{2}l = l\cos\omega t$ | M1 | Complete method to find time to reach unstretched position |
| $t = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}$ | A1 | Correct time |
| $v = \sqrt{\frac{2g}{l}\left(l^2 - \left(-\frac{1}{2}l\right)^2\right)}$ OR | M1 | Complete method to find speed at unstretched position |
| $0 = \sqrt{\frac{3gl}{2}} - gt_1 \Rightarrow t_1 = \sqrt{\frac{3l}{2g}}$ | M1 | Complete method to find time to rest position |
| Total time $= \frac{2\pi}{3}\sqrt{\frac{l}{2g}} + \sqrt{\frac{3l}{2g}}$ | A1 | cao |
**Total: 5 marks**
---\begin{enumerate}
\item A light elastic string, of natural length $l$ and modulus of elasticity $2 m g$, has one end attached to a fixed point $A$ and the other end attached to a particle $P$ of mass $m$. The particle $P$ hangs in equilibrium at the point $O$.\\
(a) Show that $A O = \frac { 3 l } { 2 }$
\end{enumerate}
The particle $P$ is pulled down vertically from $O$ to the point $B$, where $O B = l$, and released from rest.
Air resistance is modelled as being negligible.\\
Using the model,\\
(b) prove that $P$ begins to move with simple harmonic motion about $O$ with period $\pi \sqrt { \frac { 2 l } { g } }$
The particle $P$ first comes to instantaneous rest at the point $C$.\\
Using the model,\\
(c) find the length $B C$ in terms of $l$,\\
(d) find, in terms of $l$ and $g$, the exact time it takes $P$ to move directly from $B$ to $C$.
\hfill \mbox{\textit{Edexcel FM2 2021 Q6 [16]}}