| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.2 This is a standard compound pendulum problem from Further Maths M4 requiring energy methods and small angle approximations. While it involves multiple steps (stability analysis, kinetic energy derivation, and period calculation), these are well-rehearsed techniques for this specification. The question provides clear scaffolding through parts (i)-(iii), making it more accessible than unstructured SHM problems, though still above average difficulty due to the Further Maths content. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{d\theta} = mga(3\cos\theta + 4\sin\theta - 3)\) | B1 | |
| When \(\theta = 0\), \(\frac{dV}{d\theta} = mga(3+0-3) = 0\) | M1 | Considering \(\frac{dV}{d\theta} = 0\) |
| so \(\theta = 0\) is a position of equilibrium | A1 ag | Correctly shown |
| \(\frac{d^2V}{d\theta^2} = mga(-3\sin\theta + 4\cos\theta)\) | ||
| When \(\theta = 0\), \(\frac{d^2V}{d\theta^2} = 4mga > 0\) | M1 | Considering \(\frac{d^2V}{d\theta^2}\) (or other method); \(V'' = 4mga \Rightarrow\) Stable M1A0; \(V'' = 4mga \Rightarrow\) Minimum \(\Rightarrow\) Stable M1A1 |
| hence the equilibrium is stable | A1 ag [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Speed of \(P\) and \(Q\) is \(a\dot\theta\) | M1 | Or moment of inertia of \(P\) is \(5ma^2\) |
| KE is \(\frac{1}{2}(5m)(a\dot\theta)^2 + \frac{1}{2}(3m)(a\dot\theta)^2\) or \(\frac{1}{2}(8m)(a\dot\theta)^2\) | \(\frac{5}{2}ma^2\dot\theta^2 + \frac{3}{2}ma^2\dot\theta^2\) M1A1; \(\frac{1}{2}(5ma^2)\dot\theta^2 + \frac{1}{3}(3ma^2)\dot\theta^2\) M1A0; \(\frac{1}{2}(8ma^2)\dot\theta^2\) M1A0 | |
| \(= \frac{5}{2}ma^2\dot\theta^2 + \frac{3}{2}ma^2\dot\theta^2 = 4ma^2\dot\theta^2\) | A1 ag [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V + 4ma^2\dot\theta^2 = K\) | ||
| \(\frac{dV}{d\theta}\dot\theta + 8ma^2\dot\theta\ddot\theta = 0\) | M1 | |
| \(mga(3\cos\theta + 4\sin\theta - 3)\dot\theta + 8ma^2\dot\theta\ddot\theta = 0\) | A1 | \(= 0\) is required for A1 (may be implied by later work) |
| For small \(\theta\): \(\sin\theta \approx \theta\), \(\cos\theta \approx 1\) | M1 | |
| \(mga(3 + 4\theta - 3) + 8ma^2\ddot\theta \approx 0\) | A1 ft | Linear approximation (ft dep on M1M1) |
| \(\ddot\theta \approx -\dfrac{g}{2a}\theta\) | ||
| Approximate period is \(2\pi\sqrt{\dfrac{2a}{g}}\) | A1 [5] |
# Question 6:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{d\theta} = mga(3\cos\theta + 4\sin\theta - 3)$ | B1 | |
| When $\theta = 0$, $\frac{dV}{d\theta} = mga(3+0-3) = 0$ | M1 | Considering $\frac{dV}{d\theta} = 0$ |
| so $\theta = 0$ is a position of equilibrium | A1 ag | Correctly shown |
| $\frac{d^2V}{d\theta^2} = mga(-3\sin\theta + 4\cos\theta)$ | | |
| When $\theta = 0$, $\frac{d^2V}{d\theta^2} = 4mga > 0$ | M1 | Considering $\frac{d^2V}{d\theta^2}$ (or other method); $V'' = 4mga \Rightarrow$ Stable M1A0; $V'' = 4mga \Rightarrow$ Minimum $\Rightarrow$ Stable M1A1 |
| hence the equilibrium is stable | A1 ag [5] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed of $P$ and $Q$ is $a\dot\theta$ | M1 | Or moment of inertia of $P$ is $5ma^2$ |
| KE is $\frac{1}{2}(5m)(a\dot\theta)^2 + \frac{1}{2}(3m)(a\dot\theta)^2$ or $\frac{1}{2}(8m)(a\dot\theta)^2$ | | $\frac{5}{2}ma^2\dot\theta^2 + \frac{3}{2}ma^2\dot\theta^2$ M1A1; $\frac{1}{2}(5ma^2)\dot\theta^2 + \frac{1}{3}(3ma^2)\dot\theta^2$ M1A0; $\frac{1}{2}(8ma^2)\dot\theta^2$ M1A0 |
| $= \frac{5}{2}ma^2\dot\theta^2 + \frac{3}{2}ma^2\dot\theta^2 = 4ma^2\dot\theta^2$ | A1 ag [2] | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V + 4ma^2\dot\theta^2 = K$ | | |
| $\frac{dV}{d\theta}\dot\theta + 8ma^2\dot\theta\ddot\theta = 0$ | M1 | |
| $mga(3\cos\theta + 4\sin\theta - 3)\dot\theta + 8ma^2\dot\theta\ddot\theta = 0$ | A1 | $= 0$ is required for A1 (may be implied by later work) |
| For small $\theta$: $\sin\theta \approx \theta$, $\cos\theta \approx 1$ | M1 | |
| $mga(3 + 4\theta - 3) + 8ma^2\ddot\theta \approx 0$ | A1 ft | Linear approximation (ft dep on M1M1) |
| $\ddot\theta \approx -\dfrac{g}{2a}\theta$ | | |
| Approximate period is $2\pi\sqrt{\dfrac{2a}{g}}$ | A1 [5] | |(i) Show that $\theta = 0$ is a position of stable equilibrium.\\
(ii) Show that the kinetic energy of the system is $4 m a ^ { 2 } \dot { \theta } ^ { 2 }$.\\
(iii) By differentiating the energy equation, then making suitable approximations for $\sin \theta$ and $\cos \theta$, find the approximate period of small oscillations about the equilibrium position $\theta = 0$.
\section*{[Question 7 is printed overleaf.]}
\hfill \mbox{\textit{OCR M4 2010 Q6 [12]}}