| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | January |
| 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 | Standard +0.3 This is a standard M3 SHM question requiring equilibrium analysis, proving SHM by showing acceleration proportional to displacement, and applying standard SHM formulas. While it involves multiple parts and careful setup with elastic forces, the techniques are routine for Further Maths Mechanics 3 students with no novel problem-solving required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{8mge}{l} = mg\) | M1 | For Hooke's law and equating tension to weight |
| \(e = \frac{1}{8}l\) | A1 (2) | A1cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-mg - T = m\ddot{x}\) | M1 A1 | M1 for NL2 vertically, weight and tension needed, \(\ddot{x}\) or \(a\) for acceleration; A1 for correct equation |
| \(-mg - \dfrac{8mg}{l}(x - \tfrac{1}{8}l) = m\ddot{x}\) | M1dep A1 | M1dep for using Hooke's law to replace tension with expression in \(x\); A1 for correct equation |
| \(-\dfrac{8g}{l}x = \ddot{x}\) | A1 | For rearranging correctly |
| SHM, period \(2\pi\sqrt{\dfrac{l}{8g}}\) | A1cso (6) | A1cso for conclusion SHM and correct period |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = \frac{1}{2}l - \frac{1}{8}l = \frac{3}{8}l\) | B1 | For using information in question to obtain amplitude \(= \frac{3}{8}l\) |
| \(u^2 = \dfrac{8g}{l}\left((\tfrac{3}{8}l)^2 - (-\tfrac{1}{8}l)^2\right)\) | M1 A1 | M1 for using \(v^2 = \omega^2(a^2 - x^2)\) with their \(\omega\) and \(a\); A1 for correct unsimplified expression |
| \(u = \sqrt{gl}\) | A1 (4) | A1cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -a\cos\omega t\), \(\dot{x} = a\omega\sin\omega t\) | ||
| \(\sqrt{\dfrac{9gl}{32}} = \dfrac{3l}{8}\sqrt{\dfrac{8g}{l}}\sin\sqrt{\dfrac{8g}{l}}t\) | M1 A1 | M1 for using \(\dot{x} = a\omega\sin\omega t\) with their \(a\) and \(\omega\) and given speed; A1 for fully correct equation |
| \(\dfrac{1}{2} = \sin\sqrt{\dfrac{8g}{l}}t\) | ||
| \(t = \dfrac{\pi}{6}\sqrt{\dfrac{l}{8g}}\) | M1dep A1 (4) | M1dep for solving their equation (must use radians); A1cao for \(t = \dfrac{\pi}{6}\sqrt{\dfrac{l}{8g}}\) or \(0.5235\ldots\sqrt{\dfrac{l}{8g}}\) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{8mge}{l} = mg$ | M1 | For Hooke's law and equating tension to weight |
| $e = \frac{1}{8}l$ | A1 (2) | A1cao |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-mg - T = m\ddot{x}$ | M1 A1 | M1 for NL2 vertically, weight and tension needed, $\ddot{x}$ or $a$ for acceleration; A1 for correct equation |
| $-mg - \dfrac{8mg}{l}(x - \tfrac{1}{8}l) = m\ddot{x}$ | M1dep A1 | M1dep for using Hooke's law to replace tension with expression in $x$; A1 for correct equation |
| $-\dfrac{8g}{l}x = \ddot{x}$ | A1 | For rearranging correctly |
| SHM, period $2\pi\sqrt{\dfrac{l}{8g}}$ | A1cso (6) | A1cso for conclusion SHM and correct period |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = \frac{1}{2}l - \frac{1}{8}l = \frac{3}{8}l$ | B1 | For using information in question to obtain amplitude $= \frac{3}{8}l$ |
| $u^2 = \dfrac{8g}{l}\left((\tfrac{3}{8}l)^2 - (-\tfrac{1}{8}l)^2\right)$ | M1 A1 | M1 for using $v^2 = \omega^2(a^2 - x^2)$ with their $\omega$ and $a$; A1 for correct unsimplified expression |
| $u = \sqrt{gl}$ | A1 (4) | A1cao |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -a\cos\omega t$, $\dot{x} = a\omega\sin\omega t$ | | |
| $\sqrt{\dfrac{9gl}{32}} = \dfrac{3l}{8}\sqrt{\dfrac{8g}{l}}\sin\sqrt{\dfrac{8g}{l}}t$ | M1 A1 | M1 for using $\dot{x} = a\omega\sin\omega t$ with their $a$ and $\omega$ and given speed; A1 for fully correct equation |
| $\dfrac{1}{2} = \sin\sqrt{\dfrac{8g}{l}}t$ | | |
| $t = \dfrac{\pi}{6}\sqrt{\dfrac{l}{8g}}$ | M1dep A1 (4) | M1dep for solving their equation (must use radians); A1cao for $t = \dfrac{\pi}{6}\sqrt{\dfrac{l}{8g}}$ or $0.5235\ldots\sqrt{\dfrac{l}{8g}}$ |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2c0bb9ea-31a6-42f1-9e2e-d792eee8fd10-11_517_254_278_845}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A smooth hollow narrow tube of length $l$ has one open end and one closed end. The tube is fixed in a vertical position with the closed end at the bottom. A light elastic spring has natural length $l$ and modulus of elasticity $8 m g$. The spring is inside the tube and has one end attached to a fixed point $O$ on the closed end of the tube. The other end of the spring is attached to a particle $P$ of mass $m$. The particle rests in equilibrium at a distance $e$ below the top of the tube, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Find $e$ in terms of $l$.
The particle $P$ is now held inside the tube at a distance $\frac { 1 } { 2 } l$ below the top of the tube and released from rest at time $t = 0$
\item Prove that $P$ moves with simple harmonic motion of period $2 \pi \sqrt { \left( \frac { l } { 8 g } \right) }$.
The particle $P$ passes through the open top of the tube with speed $u$.
\item Find $u$ in terms of $g$ and $l$.
\item Find the time taken for $P$ to first attain a speed of $\sqrt { \left( \frac { 9 g l } { 32 } \right) }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q7 [16]}}