| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question with two springs in equilibrium. Part (a) requires setting up tension equations using Hooke's law and solving simultaneously—straightforward but requires care. Parts (b)-(d) follow standard SHM procedures: showing restoring force proportional to displacement, using energy conservation, and applying SHM equations. While it involves multiple steps and careful bookkeeping of extensions, it's a textbook application of well-practiced techniques without requiring novel insight. Slightly above average difficulty due to the two-spring setup and multi-part nature, but well within expected M3 standard. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tensions equal when \(P\) in equilibrium: \(\dfrac{15\times x}{0.8} = \dfrac{10\times(2.4-x)}{0.8}\) | M1A2 | |
| \(25x = 24\), \(\quad x = \dfrac{24}{25} = 0.96\) | ||
| \(AC = 1.76\text{ (m)}\) | A1 | *AG* (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(P\) is distance \(x\) from \(C\), restoring force: \(\dfrac{15\times(0.96+x)}{0.8} - \dfrac{10\times(1.44-x)}{0.8} = \dfrac{25}{0.8}x = -m\ddot{x} = -0.2\ddot{x}\) | M1A2 | |
| \(\ddot{x} = -156.25x\left(= -12.5^2\,x\right) \implies\) SHM | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Speed at \(C\) = max speed \(= a\omega = 0.4\times 12.5 = 5\ (\text{m s}^{-1})\) | M1A1ft | \(0.4\times\) their \(\omega\) (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = a\cos\omega t\) | B1ft | their \(\omega\) |
| \(\dot{x} = -a\omega\sin\omega t\) | M1 | their \(\omega\) |
| \((-)2 = (-)5\sin 12.5t\) | A1ft | their \(\omega\) |
| \(12.5t = 0.4115\ldots\), \(\quad t = 0.0329\ldots \approx 0.033\text{ (s)}\) | A1 | (4) [14] |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tensions equal when $P$ in equilibrium: $\dfrac{15\times x}{0.8} = \dfrac{10\times(2.4-x)}{0.8}$ | M1A2 | |
| $25x = 24$, $\quad x = \dfrac{24}{25} = 0.96$ | | |
| $AC = 1.76\text{ (m)}$ | A1 | *AG* **(4)** |
---
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $P$ is distance $x$ from $C$, restoring force: $\dfrac{15\times(0.96+x)}{0.8} - \dfrac{10\times(1.44-x)}{0.8} = \dfrac{25}{0.8}x = -m\ddot{x} = -0.2\ddot{x}$ | M1A2 | |
| $\ddot{x} = -156.25x\left(= -12.5^2\,x\right) \implies$ SHM | A1 | **(4)** |
---
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed at $C$ = max speed $= a\omega = 0.4\times 12.5 = 5\ (\text{m s}^{-1})$ | M1A1ft | $0.4\times$ their $\omega$ **(2)** |
---
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = a\cos\omega t$ | B1ft | their $\omega$ |
| $\dot{x} = -a\omega\sin\omega t$ | M1 | their $\omega$ |
| $(-)2 = (-)5\sin 12.5t$ | A1ft | their $\omega$ |
| $12.5t = 0.4115\ldots$, $\quad t = 0.0329\ldots \approx 0.033\text{ (s)}$ | A1 | **(4) [14]** |\begin{enumerate}
\item Two points $A$ and $B$ are 4 m apart on a smooth horizontal surface. A light elastic string, of natural length 0.8 m and modulus of elasticity 15 N , has one end attached to the point A. A light elastic string, of natural length 0.8 m and modulus of elasticity 10 N , has one end attached to the point $B$. A particle $P$ of mass 0.2 kg is attached to the free end of each string. The particle rests in equilibrium on the surface at the point $C$ on the straight line between $A$ and $B$.\\
(a) Show that the length of $A C$ is 1.76 m .
\end{enumerate}
The particle $P$ is now held at the point $D$ on the line $A B$ such that $A D = 2.16 \mathrm {~m}$. The particle is then released from rest and in the subsequent motion both strings remain taut.\\
(b) Show that $P$ moves with simple harmonic motion.\\
(c) Find the speed of $P$ as it passes through the point $C$.\\
(d) Find the time from the instant when $P$ is released from $D$ until the instant when $P$ is first moving with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
\hfill \mbox{\textit{Edexcel M3 2013 Q7 [14]}}