| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| 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 SHM question with elastic strings requiring systematic application of Hooke's law, equilibrium analysis, and SHM formulas. While it has multiple parts and requires careful bookkeeping of extensions and tensions, each step follows predictable methods without requiring novel insight. The 'show that' parts provide scaffolding, making it more accessible than open-ended SHM problems. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Total extension \(= 0.6\); \(T_b = \frac{\lambda \times \text{ext}}{l} = \frac{2(0.3-x)}{0.7} = \frac{2}{7}(3-10x)\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T_a = \frac{2(x+0.3)}{0.7}\left(=\frac{2}{7}(10x+3)\right)\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T_b - T_a = 0.5\ddot{x}\) leading to \(\frac{2}{7}(3-10x)-\frac{2}{7}(10x+3)=0.5\ddot{x}\) | M1 A1ft | |
| \(2\times\left(-\frac{20x}{7}\right) = 0.5\ddot{x}\) | ||
| \(\ddot{x} = -\frac{40}{7\times 0.5}x\) (\(\therefore\) S.H.M.) | M1 A1 | |
| Period \(= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{7\times 0.5}{40}} = 2\pi\sqrt{\frac{7}{80}}\) | M1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(v_{\max} = a\omega = 0.2\sqrt{\frac{80}{7}}\) o.e. or a.w.r.t. \(0.68\ \text{m s}^{-1}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(x = a\cos\omega t = 0.2\cos\left(\sqrt{\frac{80}{7}}t\right)\) | M1 | |
| \(x = -0.1\): \(-\frac{0.1}{0.2} = \cos\left(\sqrt{\frac{80}{7}}t\right)\) | A1 | |
| \(t = \sqrt{\frac{7}{80}}\cos^{-1}(-0.5)\) | ||
| \(t = \sqrt{\frac{7}{80}}\times\frac{2\pi}{3} = \frac{\pi}{3}\sqrt{\frac{7}{20}}\) o.e. (accept a.w.r.t. 0.62 s) | M1 A1 | (4) |
# Question 7:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| Total extension $= 0.6$; $T_b = \frac{\lambda \times \text{ext}}{l} = \frac{2(0.3-x)}{0.7} = \frac{2}{7}(3-10x)$ | M1 A1 | (2) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T_a = \frac{2(x+0.3)}{0.7}\left(=\frac{2}{7}(10x+3)\right)$ | B1 | (1) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T_b - T_a = 0.5\ddot{x}$ leading to $\frac{2}{7}(3-10x)-\frac{2}{7}(10x+3)=0.5\ddot{x}$ | M1 A1ft | |
| $2\times\left(-\frac{20x}{7}\right) = 0.5\ddot{x}$ | | |
| $\ddot{x} = -\frac{40}{7\times 0.5}x$ ($\therefore$ S.H.M.) | M1 A1 | |
| Period $= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{7\times 0.5}{40}} = 2\pi\sqrt{\frac{7}{80}}$ | M1 A1 | (6) |
## Part (d):
| Working/Answer | Marks | Notes |
|---|---|---|
| $v_{\max} = a\omega = 0.2\sqrt{\frac{80}{7}}$ o.e. or a.w.r.t. $0.68\ \text{m s}^{-1}$ | M1 A1 | (2) |
## Part (e):
| Working/Answer | Marks | Notes |
|---|---|---|
| $x = a\cos\omega t = 0.2\cos\left(\sqrt{\frac{80}{7}}t\right)$ | M1 | |
| $x = -0.1$: $-\frac{0.1}{0.2} = \cos\left(\sqrt{\frac{80}{7}}t\right)$ | A1 | |
| $t = \sqrt{\frac{7}{80}}\cos^{-1}(-0.5)$ | | |
| $t = \sqrt{\frac{7}{80}}\times\frac{2\pi}{3} = \frac{\pi}{3}\sqrt{\frac{7}{20}}$ o.e. (accept a.w.r.t. 0.62 s) | M1 A1 | (4) |
The image provided appears to be only the back/final page of an Edexcel mark scheme document (June 2011, Order Code UA028443). This page contains only publisher information, contact details, and logos — **no actual mark scheme content** (questions, answers, mark allocations, or guidance notes) is visible on this page.
To extract the mark scheme content you're looking for, please share the **earlier pages** of the document that contain the actual marking guidance.\begin{enumerate}
\item A particle $P$ of mass 0.5 kg is attached to the mid-point of a light elastic string of natural length 1.4 m and modulus of elasticity 2 N . The ends of the string are attached to the points $A$ and $B$ on a smooth horizontal table, where $A B = 2 \mathrm {~m}$. The mid-point of $A B$ is $O$ and the point $C$ is on the table between $O$ and $B$ where $O C = 0.2 \mathrm {~m}$. At time $t = 0$ the particle is released from rest at $C$. At time $t$ seconds the length of the string $A P$ is $( 1 + x ) \mathrm { m }$.\\
(a) Show that the tension in $B P$ is $\frac { 2 } { 7 } ( 3 - 10 x ) \mathrm { N }$.\\
(b) Find, in terms of $x$, the tension in $A P$.\\
(c) Show that $P$ performs simple harmonic motion with period $2 \pi \sqrt { } \left( \frac { 7 } { 80 } \right)$ s.\\
(d) Find the greatest speed of $P$ during the motion.
\end{enumerate}
The point $D$ lies between $O$ and $A$, where $O D = 0.1 \mathrm {~m}$.\\
(e) Find the time taken by $P$ to move directly from $C$ to $D$.\\
\hfill \mbox{\textit{Edexcel M3 2011 Q7 [15]}}