| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with elastic strings/springs |
| Difficulty | Challenging +1.8 This is a challenging M4 compound pendulum problem requiring cosine rule application, equilibrium analysis with elastic strings, energy methods or moment equations, and small angle approximation for SHM. It demands multiple sophisticated techniques and careful geometric reasoning, placing it well above average difficulty but within reach of strong Further Maths students. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(RB^2 = a^2 + (2a)^2 - 2(a)(2a)\cos\left(\theta+\frac{1}{4}\pi\right)\) | M1 | |
| \(= 5a^2 - 4a^2\left(\cos\theta\cos\frac{1}{4}\pi - \sin\theta\sin\frac{1}{4}\pi\right)\) | ||
| \(= a^2(5 - 2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta)\) | A1 (ag) | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = -mg(2a\sin\theta) + \dfrac{mg\sqrt{2}}{2a}\times RB^2\) | M1 | Considering PE and EE |
| \(= \frac{5}{2}\sqrt{2}mga - 2mga\cos\theta\) | A1 | |
| \(\dfrac{dV}{d\theta} = 2mga\sin\theta\) | M1 | |
| When \(\theta=0\), \(\dfrac{dV}{d\theta}=0\), hence equilibrium | A1 | Correctly shown |
| \(\dfrac{d^2V}{d\theta^2} = 2mga\cos\theta\) | M1 | Or other method for max/min |
| When \(\theta=0\), \(\dfrac{d^2V}{d\theta^2} = 2mga > 0\), hence stable | A1 | Correctly shown |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| KE is \(\frac{1}{2}m(2a\dot{\theta})^2\) | B1 | |
| \(\frac{5}{2}\sqrt{2}mga - 2mga\cos\theta + 2ma^2\dot{\theta}^2 = E\) | M1 | |
| Differentiating w.r.t. \(t\): \(2mga\sin\theta\,\dot{\theta} + 4ma^2\dot{\theta}\ddot{\theta} = 0\) | M1 | |
| \(\ddot{\theta} = -\dfrac{g}{2a}\sin\theta\) | A1 | Requires fully correct working |
| OR \((mg\cos\theta - T\sin\phi)(2a) = I\ddot{\theta}\), where \(T = \dfrac{mg\sqrt{2}(RB)}{a}\) and \(\dfrac{\sin\phi}{a} = \dfrac{\sin(\theta+\frac{1}{4}\pi)}{RB}\) | M2 | or \(mg\cos\theta - T\sin\phi = m(2a\ddot{\theta})\) |
| \(\ddot{\theta} = -\dfrac{g}{2a}\sin\theta\) | A2 | Give A1 if just one minor error |
| Period is \(2\pi\sqrt{\dfrac{2a}{g}}\) | B1 ft | ft provided that \(k\) is in terms of \(a\) and \(g\) only |
| Total: 5 |
# Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $RB^2 = a^2 + (2a)^2 - 2(a)(2a)\cos\left(\theta+\frac{1}{4}\pi\right)$ | M1 | |
| $= 5a^2 - 4a^2\left(\cos\theta\cos\frac{1}{4}\pi - \sin\theta\sin\frac{1}{4}\pi\right)$ | | |
| $= a^2(5 - 2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta)$ | A1 (ag) | |
| **Total: 2** | | |
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# Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = -mg(2a\sin\theta) + \dfrac{mg\sqrt{2}}{2a}\times RB^2$ | M1 | Considering PE and EE |
| $= \frac{5}{2}\sqrt{2}mga - 2mga\cos\theta$ | A1 | |
| $\dfrac{dV}{d\theta} = 2mga\sin\theta$ | M1 | |
| When $\theta=0$, $\dfrac{dV}{d\theta}=0$, hence equilibrium | A1 | Correctly shown |
| $\dfrac{d^2V}{d\theta^2} = 2mga\cos\theta$ | M1 | Or other method for max/min |
| When $\theta=0$, $\dfrac{d^2V}{d\theta^2} = 2mga > 0$, hence stable | A1 | Correctly shown |
| **Total: 6** | | |
---
# Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| KE is $\frac{1}{2}m(2a\dot{\theta})^2$ | B1 | |
| $\frac{5}{2}\sqrt{2}mga - 2mga\cos\theta + 2ma^2\dot{\theta}^2 = E$ | M1 | |
| Differentiating w.r.t. $t$: $2mga\sin\theta\,\dot{\theta} + 4ma^2\dot{\theta}\ddot{\theta} = 0$ | M1 | |
| $\ddot{\theta} = -\dfrac{g}{2a}\sin\theta$ | A1 | Requires fully correct working |
| **OR** $(mg\cos\theta - T\sin\phi)(2a) = I\ddot{\theta}$, where $T = \dfrac{mg\sqrt{2}(RB)}{a}$ and $\dfrac{\sin\phi}{a} = \dfrac{\sin(\theta+\frac{1}{4}\pi)}{RB}$ | M2 | or $mg\cos\theta - T\sin\phi = m(2a\ddot{\theta})$ |
| $\ddot{\theta} = -\dfrac{g}{2a}\sin\theta$ | A2 | Give A1 if just one minor error |
| Period is $2\pi\sqrt{\dfrac{2a}{g}}$ | B1 ft | ft provided that $k$ is in terms of $a$ and $g$ only |
| **Total: 5** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{b86c4b97-13a9-4aaf-8c95-20fe043b4532-3_585_801_991_647}
A light rod $A B$ of length $2 a$ can rotate freely in a vertical plane about a fixed horizontal axis through $A$. A particle of mass $m$ is attached to the rod at $B$. A fixed smooth ring $R$ lies in the same vertical plane as the rod, where $A R = a$ and $A R$ makes an angle $\frac { 1 } { 4 } \pi$ above the horizontal. A light elastic string, of natural length $a$ and modulus of elasticity $m g \sqrt { } 2$, passes through the ring $R$; one end is fixed to $A$ and the other end is fixed to $B$. The rod makes an angle $\theta$ below the horizontal, where $- \frac { 1 } { 4 } \pi < \theta < \frac { 3 } { 4 } \pi$ (see diagram).\\
(i) Use the cosine rule to show that $R B ^ { 2 } = a ^ { 2 } ( 5 - ( 2 \sqrt { } 2 ) \cos \theta + ( 2 \sqrt { } 2 ) \sin \theta )$.\\
(ii) Show that $\theta = 0$ is a position of stable equilibrium.\\
(iii) Show that $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - k \sin \theta$, expressing the constant $k$ in terms of $a$ and $g$, and hence write down the approximate period of small oscillations about the equilibrium position $\theta = 0$.
\hfill \mbox{\textit{OCR M4 2005 Q7 [13]}}