| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| 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 compound pendulum problem requires calculating elastic and gravitational PE with non-trivial geometry (horizontal strings to vertical wires), finding equilibrium via energy methods, and deriving the SHM period. The geometric setup is more complex than standard pendulum questions, requiring careful coordinate work and understanding of elastic PE. However, it's a structured multi-part question with clear guidance, and the techniques (energy methods, small angle approximations) are standard for M4 level. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(= mga(1 + \cos\theta + \sin^2\theta)\) | B1 | [5] |
| Answer | Marks |
|---|---|
| \(\ddot{\theta} = \left(\frac{3g}{4a}\right)\sin\theta(2\cos\theta - 1)\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ddot{\theta} \approx \left(\frac{3g}{4a}\right)\theta\) | A1 | |
| Approximate SHM with period: \(T = 2\pi\sqrt{\frac{4a}{3g}} = 4\pi\sqrt{\frac{a}{3g}}\) | A1 | [8] |
**Potential Energy Function**
$V = \text{G.P.E. of rod} + \text{E.P.E. of } R_1B + \text{E.P.E. of } R_2B$
$= mg(a\cos\theta) + \frac{1}{2} \cdot \frac{4ka}{8}\left[(2a + 2a\sin\theta)^2 + (2a - 2a\sin\theta)^2\right]$
$= mga\cos\theta + \frac{1}{2}mga(2 + 2\sin^2\theta)$
$= mga(1 + \cos\theta + \sin^2\theta)$ | B1 | [5]
**Small Oscillations**
Conservation of mechanical energy:
$mga(1 + \cos\theta + \sin^2\theta) + \frac{1}{2}\left(\frac{1}{3}ma^2\right)\dot{\theta}^2 = \text{constant}$
$\frac{d}{dt}$: $mga\dot{\theta}(2\sin\theta\cos\theta - \sin\theta) + \frac{1}{3}mga\dot{\theta}\ddot{\theta} = 0$
$\ddot{\theta} = \left(\frac{3g}{4a}\right)\sin\theta(2\cos\theta - 1)$ | M1 |
For small oscillations: $\sin\theta \approx \theta$ and $\cos\theta \approx 1$
$\ddot{\theta} \approx \left(\frac{3g}{4a}\right)\theta$ | A1 |
Approximate SHM with period: $T = 2\pi\sqrt{\frac{4a}{3g}} = 4\pi\sqrt{\frac{a}{3g}}$ | A1 | [8]7\\
\includegraphics[max width=\textwidth, alt={}, center]{de53978b-aa96-4fa2-a928-81a16450154e-4_557_1036_278_553}
A uniform rod $A B$, of mass $m$ and length $2 a$, is pivoted to a fixed point at $A$ and is free to rotate in a vertical plane. Two fixed vertical wires in this plane are a distance $6 a$ apart and the point $A$ is half-way between the two wires. Light smooth rings $R _ { 1 }$ and $R _ { 2 }$ slide on the wires and are connected to $B$ by light elastic strings, each of natural length $a$ and modulus of elasticity $\frac { 1 } { 4 } m g$. The strings $B R _ { 1 }$ and $B R _ { 2 }$ are always horizontal and the angle between $A B$ and the upward vertical is $\theta$, where $- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$ (see diagram).\\
(i) Taking $A$ as the reference level for gravitational potential energy, show that the total potential energy of the system is
$$m g a \left( 1 + \cos \theta + \sin ^ { 2 } \theta \right) .$$
(ii) Given that $\theta = 0$ is a position of stable equilibrium, find the approximate period of small oscillations about this position.
\hfill \mbox{\textit{OCR M4 2003 Q7 [13]}}