| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| 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 energy methods, stability analysis, and SHM approximation. Part (i) demands showing equilibrium and deriving a stability condition from energy considerations. Part (ii) requires differentiating total energy to find the equation of motion, then applying small angle approximations. While the elastic energy formula is given (removing one difficult step), students must still handle rotational dynamics, energy differentiation, and multi-step reasoning across both parts. This exceeds standard M4 difficulty but remains within the module's scope. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force4.10f Simple harmonic motion: x'' = -omega^2 x6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total PE is \(V = \lambda a(1+\cos\theta) - mg(4a\cos\theta)\) | M1, A1 | Using EPE and GPE |
| \(\frac{dV}{d\theta} = -\lambda a\sin\theta + 4mga\sin\theta\) | M1 | Obtaining \(\frac{dV}{d\theta}\) |
| When \(\theta=0\), \(\frac{dV}{d\theta}=0\), so it is in equilibrium | B1 | AG Correctly shown |
| \(\frac{d^2V}{d\theta^2} = -\lambda a\cos\theta + 4mga\cos\theta\) | M1 | Obtaining \(\frac{d^2V}{d\theta^2}\) |
| When \(\theta=0\), \(\frac{d^2V}{d\theta^2} = -\lambda a + 4mga = a(4mg-\lambda)\) | ||
| If \(\lambda < 4mg\), \(\frac{d^2V}{d\theta^2} > 0\), so equilibrium is stable | A1 [6] | AG Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE is \(\frac{1}{2}m(4a\dot{\theta})^2\) | M1 | Using KE |
| Total energy is \(3mga - mga\cos\theta + 8ma^2\dot{\theta}^2 = K\) | A1 | FT — May still involve \(\lambda\) |
| \(mga\sin\theta\dot{\theta} + 16ma^2\dot{\theta}\ddot{\theta} = 0\) | M1 | Differentiating the energy equation |
| \(\ddot{\theta} = -\frac{1}{16}\frac{g}{a}\sin\theta\) \(\left(k=\frac{1}{16}\right)\) | A1 | |
| \(\ddot{\theta} \approx -\frac{g}{16a}\theta\), so approximate SHM | M1 | Implied by \(\frac{2\pi}{\omega}\) with appropriate \(\omega\) |
| Approximate period is \(8\pi\sqrt{\frac{a}{g}}\) | A1 [6] | FT \(2\pi\sqrt{\frac{a}{kg}}\) |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total PE is $V = \lambda a(1+\cos\theta) - mg(4a\cos\theta)$ | M1, A1 | Using EPE and GPE |
| $\frac{dV}{d\theta} = -\lambda a\sin\theta + 4mga\sin\theta$ | M1 | Obtaining $\frac{dV}{d\theta}$ |
| When $\theta=0$, $\frac{dV}{d\theta}=0$, so it is in equilibrium | B1 | AG Correctly shown | B1 for alternative proof of equilibrium (e.g. symmetry) |
| $\frac{d^2V}{d\theta^2} = -\lambda a\cos\theta + 4mga\cos\theta$ | M1 | Obtaining $\frac{d^2V}{d\theta^2}$ |
| When $\theta=0$, $\frac{d^2V}{d\theta^2} = -\lambda a + 4mga = a(4mg-\lambda)$ | | |
| If $\lambda < 4mg$, $\frac{d^2V}{d\theta^2} > 0$, so equilibrium is stable | A1 [6] | AG Correctly shown |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE is $\frac{1}{2}m(4a\dot{\theta})^2$ | M1 | Using KE | $\frac{1}{2}m\dot{\theta}^2$ is M0; $\frac{1}{2}I\dot{\theta}^2$ also needs attempt at $I=m(4a)^2$ for M1 |
| Total energy is $3mga - mga\cos\theta + 8ma^2\dot{\theta}^2 = K$ | A1 | FT — May still involve $\lambda$ |
| $mga\sin\theta\dot{\theta} + 16ma^2\dot{\theta}\ddot{\theta} = 0$ | M1 | Differentiating the energy equation | If wrt $\theta$, $\frac{d}{d\theta}(\dot{\theta}^2)=2\ddot{\theta}$ must be derived or clearly implied |
| $\ddot{\theta} = -\frac{1}{16}\frac{g}{a}\sin\theta$ $\left(k=\frac{1}{16}\right)$ | A1 | |
| $\ddot{\theta} \approx -\frac{g}{16a}\theta$, so approximate SHM | M1 | Implied by $\frac{2\pi}{\omega}$ with appropriate $\omega$ |
| Approximate period is $8\pi\sqrt{\frac{a}{g}}$ | A1 [6] | FT $2\pi\sqrt{\frac{a}{kg}}$ |6\\
\includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-3_716_483_890_790}
Two small smooth pegs $P$ and $Q$ are fixed at a distance $2 a$ apart on the same horizontal level, and $A$ is the mid-point of $P Q$. A light rod $A B$ of length $4 a$ is freely pivoted at $A$ and can rotate in the vertical plane containing $P Q$, with $B$ below the level of $P Q$. A particle of mass $m$ is attached to the rod at $B$. A light elastic string, of natural length $2 a$ and modulus of elasticity $\lambda$, passes round the pegs $P$ and $Q$ and its two ends are attached to the rod at the point $X$, where $A X = a$. The angle between the rod and the downward vertical is $\theta$, where $- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$ (see diagram). You are given that the elastic energy stored in the string is $\lambda a ( 1 + \cos \theta )$.\\
(i) Show that $\theta = 0$ is a position of equilibrium, and show that the equilibrium is stable if $\lambda < 4 m g$.\\
(ii) Given that $\lambda = 3 m g$, show that $\ddot { \theta } = - k \frac { g } { a } \sin \theta$, stating the value of the constant $k$. Hence find the approximate period of small oscillations of the system about the equilibrium position $\theta = 0$.
\hfill \mbox{\textit{OCR M4 2012 Q6 [12]}}