| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| Session | June |
| 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 Further Maths SHM question requiring proof of SHM from Hooke's law, finding equilibrium position, applying standard period formula, and using energy conservation. While it involves multiple steps and FM content (making it harder than average A-level), the techniques are routine for FM2 students with no 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 |
| Equation of motion about equilibrium position | M1 | 3.1a |
| \(\frac{4mg(x+e)}{l} - mg = -m\ddot{x}\) | A1 | 1.1b |
| Extension \(e\) at equilibrium: \(\frac{4mge}{l} = mg\), \(\left(e = \frac{l}{4}\right)\) | B1 | 1.1b |
| \(\Rightarrow \frac{4gx}{l} = -\ddot{x}\), \(\left(\ddot{x} = -\frac{4g}{l}x\right)\) | M1 | 3.1a |
| This is of the form \(\ddot{x} = -\omega^2 x\), so SHM * | A1* | 3.2a |
| Period \(= \frac{2\pi}{\omega}\) | M1 | 3.4 |
| \(= 2\pi\sqrt{\frac{l}{4g}} = \pi\sqrt{\frac{l}{g}}\) * | A1* | 2.2a |
| Total: (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion for extension \(x\): \(\frac{4mgx}{l} - mg = -m\ddot{x}\), \(\ddot{x} = -\frac{4g}{l}\left(x - \frac{l}{4}\right)\) | M1, A1 | 3.1a, 1.1b |
| Use substitution \(X = x - \frac{l}{4}\) | B1 | 1.1b |
| \(\Rightarrow \frac{4gX}{l} = -\ddot{X}\), \(\left(\ddot{X} = -\frac{4g}{l}X\right)\) | M1 | 3.1a |
| This is of the form \(\ddot{X} = -\omega^2 X\), so SHM * | A1* | 3.2a |
| Period \(= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{4g}} = \pi\sqrt{\frac{l}{g}}\) * | M1, A1* | 3.4, 2.2a |
| Total: (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Max speed \(= a\omega \left(= \frac{l}{2}\sqrt{\frac{4g}{l}}\right)\) | M1 | Use the model to find the max speed. Follow their \(\omega\) |
| Max KE \(= \frac{1}{2}m\left(\frac{l}{2}\sqrt{\frac{4g}{l}}\right)^2\) | M1 | Follow their \(a\), \(\omega\) |
| \(= \frac{1}{2}m\frac{l^2}{4}\times\frac{4g}{l} = \frac{1}{2}mlg\) | A1 | Correct simplified |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = a\cos\omega t = \frac{l}{2}\cos\sqrt{\frac{4g}{l}}\,t\) | B1ft | Or equivalent. Follow their \(a\), \(\omega\) |
| Length of spring \(< l \Rightarrow x = -\frac{l}{4}\), \(\quad -\frac{l}{4} = \frac{l}{2}\cos\sqrt{\frac{4g}{l}}\,t\) | M1 | Follow their \(e\) and solve for \(t\) |
| \(\Rightarrow \sqrt{\frac{4g}{l}}\,t = \frac{2\pi}{3}\) or \(\frac{4\pi}{3}\), \(\quad t = \frac{\pi}{3}\sqrt{\frac{l}{g}}\) or \(t = \frac{2\pi}{3}\sqrt{\frac{l}{g}}\) | A1 | One correct solution. Accept \(t = \frac{2\pi}{3\omega}\) or \(t = \frac{4\pi}{3\omega}\) |
| Correct strategy | M1 | Complete strategy to find the required interval: select formula for displacement as function of time and use symmetry of motion to find the time interval |
| Length of time \(= \frac{2\pi}{3}\sqrt{\frac{l}{g}} - \frac{\pi}{3}\sqrt{\frac{l}{g}} = \frac{\pi}{3}\sqrt{\frac{l}{g}}\) | A1 | Correct answer from correct working |
| (5) | ||
| (15 marks total) |
# Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion about equilibrium position | M1 | 3.1a |
| $\frac{4mg(x+e)}{l} - mg = -m\ddot{x}$ | A1 | 1.1b |
| Extension $e$ at equilibrium: $\frac{4mge}{l} = mg$, $\left(e = \frac{l}{4}\right)$ | B1 | 1.1b |
| $\Rightarrow \frac{4gx}{l} = -\ddot{x}$, $\left(\ddot{x} = -\frac{4g}{l}x\right)$ | M1 | 3.1a |
| This is of the form $\ddot{x} = -\omega^2 x$, so SHM * | A1* | 3.2a |
| Period $= \frac{2\pi}{\omega}$ | M1 | 3.4 |
| $= 2\pi\sqrt{\frac{l}{4g}} = \pi\sqrt{\frac{l}{g}}$ * | A1* | 2.2a |
| **Total: (7)** | | |
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# Question 7(a) alt:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion for extension $x$: $\frac{4mgx}{l} - mg = -m\ddot{x}$, $\ddot{x} = -\frac{4g}{l}\left(x - \frac{l}{4}\right)$ | M1, A1 | 3.1a, 1.1b |
| Use substitution $X = x - \frac{l}{4}$ | B1 | 1.1b |
| $\Rightarrow \frac{4gX}{l} = -\ddot{X}$, $\left(\ddot{X} = -\frac{4g}{l}X\right)$ | M1 | 3.1a |
| This is of the form $\ddot{X} = -\omega^2 X$, so SHM * | A1* | 3.2a |
| Period $= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{4g}} = \pi\sqrt{\frac{l}{g}}$ * | M1, A1* | 3.4, 2.2a |
| **Total: (7)** | | |
## Question (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Max speed $= a\omega \left(= \frac{l}{2}\sqrt{\frac{4g}{l}}\right)$ | M1 | Use the model to find the max speed. Follow their $\omega$ |
| Max KE $= \frac{1}{2}m\left(\frac{l}{2}\sqrt{\frac{4g}{l}}\right)^2$ | M1 | Follow their $a$, $\omega$ |
| $= \frac{1}{2}m\frac{l^2}{4}\times\frac{4g}{l} = \frac{1}{2}mlg$ | A1 | Correct simplified |
| | **(3)** | |
---
## Question (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = a\cos\omega t = \frac{l}{2}\cos\sqrt{\frac{4g}{l}}\,t$ | B1ft | Or equivalent. Follow their $a$, $\omega$ |
| Length of spring $< l \Rightarrow x = -\frac{l}{4}$, $\quad -\frac{l}{4} = \frac{l}{2}\cos\sqrt{\frac{4g}{l}}\,t$ | M1 | Follow their $e$ and solve for $t$ |
| $\Rightarrow \sqrt{\frac{4g}{l}}\,t = \frac{2\pi}{3}$ or $\frac{4\pi}{3}$, $\quad t = \frac{\pi}{3}\sqrt{\frac{l}{g}}$ or $t = \frac{2\pi}{3}\sqrt{\frac{l}{g}}$ | A1 | One correct solution. Accept $t = \frac{2\pi}{3\omega}$ or $t = \frac{4\pi}{3\omega}$ |
| Correct strategy | M1 | Complete strategy to find the required interval: select formula for displacement as function of time and use symmetry of motion to find the time interval |
| Length of time $= \frac{2\pi}{3}\sqrt{\frac{l}{g}} - \frac{\pi}{3}\sqrt{\frac{l}{g}} = \frac{\pi}{3}\sqrt{\frac{l}{g}}$ | A1 | Correct answer from correct working |
| | **(5)** | |
| | **(15 marks total)** | |\begin{enumerate}
\item A light elastic spring has natural length $l$ and modulus of elasticity $4 m g$. A particle $P$ of mass $m$ is attached to one end of the spring. The other end of the spring is attached to a fixed point $A$. The point $B$ is vertically below $A$ with $A B = \frac { 7 } { 4 } l$. The particle $P$ is released from rest at $B$.\\
(a) Show that $P$ moves with simple harmonic motion with period $\pi \sqrt { \frac { l } { g } }$\\
(b) Find, in terms of $m , l$ and $g$, the maximum kinetic energy of $P$ during the motion.\\
(c) Find the time within each complete oscillation for which the length of the spring is less than $l$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2020 Q7 [15]}}