| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.3 This is a standard M3 elastic string SHM question with a slack phase, requiring multiple techniques (equilibrium analysis, SHM equations, energy conservation, projectile motion) across four parts. While it involves several steps and careful bookkeeping of phases, the methods are all routine for M3 students and follow predictable patterns without requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (Measuring \(x\) from \(E\)) \(\quad 2\ddot{x} = 2g - 98(x+0.2)\), and so \(\ddot{x} = -49x\) | M1 A1, A1 | |
| SHM period with \(\omega^2 = 49\) so \(T = \dfrac{2\pi}{7}\) | d M1 A1cso | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Max. acceleration \(= 49 \times\) max. \(x = 49 \times 0.4 = 19.6 \text{ m s}^{-2}\) | B1 | (1) No use of \(\ddot{x}\), just \(a\) is M1 A0, A0 then M1 A0 if otherwise correct. Quoted results are not acceptable. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| String slack when \(x = -0.2\): \(\quad v^2 = 49(0.4^2 - 0.2^2)\) | M1 A1 | M1 – Use correct formula with their \(\omega\), \(a\) and \(x\) but not \(x=0\). A1 Correct values but allow \(x = +0.2\) |
| \(\Rightarrow v \approx 2.42 \text{ m s}^{-1} = \dfrac{7\sqrt{3}}{5}\) | A1 | (3) Answer must be positive and evaluated for B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(x = a\cos\omega t\) or use \(x = a\sin\omega t\) but not with \(x=0\) or \(\pm a\) | M1 | |
| Attempt complete method for finding time when string goes slack \(-0.2 = 0.4\cos 7t \Rightarrow \cos 7t = -\dfrac{1}{2}\) | dM1 A1 | If they use \(x = a\sin\omega t\) with \(x = \pm 0.2\) and add \(\frac{\pi}{7}\) or \(\frac{\pi}{14}\) this is dM1, A1 if done correctly |
| \(t = \dfrac{2\pi}{21} \approx 0.299 \text{ s}\) | A1 | If they use \(x = a\cos\omega t\) with \(x = -0.2\) this is dM1, then A1 (as in scheme) |
| Time when string is slack \(= \dfrac{(2)\times 2.42}{g} = \dfrac{2\sqrt{3}}{7} \approx 0.495 \text{ s}\) (2 needed for A) | M1 A1ft | If they use \(x = a\cos\omega t\) with \(x = +0.2\) this needs *their* \(\dfrac{\pi}{7}\) minus answer to reach dM1, then A1 |
| Total time \(= 2 \times 0.299 + 0.495 \approx 1.09 \text{ s}\) | A1 | (7) |
| 16 |
| Answer | Marks |
|---|---|
| Case | Marks |
| \(2\ddot{x} = 2g - 98x\) is M1A1A0M0A0 | |
| \(2\ddot{x} = -98x\) is M0A0A0M0A0 |
# Question 7:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| (Measuring $x$ from $E$) $\quad 2\ddot{x} = 2g - 98(x+0.2)$, and so $\ddot{x} = -49x$ | M1 A1, A1 | |
| SHM period with $\omega^2 = 49$ so $T = \dfrac{2\pi}{7}$ | d M1 A1cso | (5) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Max. acceleration $= 49 \times$ max. $x = 49 \times 0.4 = 19.6 \text{ m s}^{-2}$ | B1 | (1) No use of $\ddot{x}$, just $a$ is M1 A0, A0 then M1 A0 if otherwise correct. Quoted results are not acceptable. |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| String slack when $x = -0.2$: $\quad v^2 = 49(0.4^2 - 0.2^2)$ | M1 A1 | M1 – Use correct formula with their $\omega$, $a$ and $x$ but **not** $x=0$. A1 Correct values but allow $x = +0.2$ |
| $\Rightarrow v \approx 2.42 \text{ m s}^{-1} = \dfrac{7\sqrt{3}}{5}$ | A1 | (3) Answer must be positive and evaluated for B1 |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $x = a\cos\omega t$ or use $x = a\sin\omega t$ but not with $x=0$ or $\pm a$ | M1 | |
| Attempt complete method for finding time when string goes slack $-0.2 = 0.4\cos 7t \Rightarrow \cos 7t = -\dfrac{1}{2}$ | dM1 A1 | If they use $x = a\sin\omega t$ with $x = \pm 0.2$ and add $\frac{\pi}{7}$ or $\frac{\pi}{14}$ this is dM1, A1 if done correctly |
| $t = \dfrac{2\pi}{21} \approx 0.299 \text{ s}$ | A1 | If they use $x = a\cos\omega t$ with $x = -0.2$ this is dM1, then A1 (as in scheme) |
| Time when string is slack $= \dfrac{(2)\times 2.42}{g} = \dfrac{2\sqrt{3}}{7} \approx 0.495 \text{ s}$ (2 needed for A) | M1 A1ft | If they use $x = a\cos\omega t$ with $x = +0.2$ this needs *their* $\dfrac{\pi}{7}$ minus answer to reach dM1, then A1 |
| Total time $= 2 \times 0.299 + 0.495 \approx 1.09 \text{ s}$ | A1 | (7) |
| | **16** | |
### Special Cases for (a):
| Case | Marks |
|---|---|
| $2\ddot{x} = 2g - 98x$ is M1A1A0M0A0 | |
| $2\ddot{x} = -98x$ is M0A0A0M0A0 | |
### Alternative for (d):
$$\tfrac{1}{2}mv^2 + mg \times 0.6 = \dfrac{\lambda \times 0.6^2}{2l} \quad \text{M1 A1}$$7. A particle $P$ of mass 2 kg is attached to one end of a light elastic string, of natural length 1 m and modulus of elasticity 98 N . The other end of the string is attached to a fixed point $A$. When $P$ hangs freely below $A$ in equilibrium, $P$ is at the point $E , 1.2 \mathrm {~m}$ below $A$. The particle is now pulled down to a point $B$ which is 0.4 m vertically below $E$ and released from rest.
\begin{enumerate}[label=(\alph*)]
\item Prove that, while the string is taut, $P$ moves with simple harmonic motion about $E$ with period $\frac { 2 \pi } { 7 } \mathrm {~s}$.
\item Find the greatest magnitude of the acceleration of $P$ while the string is taut.
\item Find the speed of $P$ when the string first becomes slack.
\item Find, to 3 significant figures, the time taken, from release, for $P$ to return to $B$ for the first time.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2008 Q7 [16]}}