| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Challenging +1.2 This is a standard SHM derivation question requiring application of Newton's second law, chain rule manipulation (v dv/dx), and integration using a standard trigonometric substitution. While it involves multiple steps and requires careful algebraic manipulation, the techniques are well-practiced in M4 and follow a predictable pattern. The question guides students through each stage, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08k Separable differential equations: dy/dx = f(x)g(y)6.02e Calculate KE and PE: using formulae6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
2 A light elastic string AB has stiffness $k$. The end A is attached to a fixed point and a particle of mass $m$ is attached at the end B . With the string vertical, the particle is released from rest from a point at a distance $a$ below its equilibrium position. At time $t$, the displacement of the particle below the equilibrium position is $x$ and the velocity of the particle is $v$.\\
(i) Show that
$$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - k x$$
(ii) Show that, while the particle is moving upwards and the string is taut,
$$v = - \sqrt { \frac { k } { m } \left( a ^ { 2 } - x ^ { 2 } \right) }$$
(iii) Hence use integration to find an expression for $x$ at time $t$ while the particle is moving upwards and the string is taut.
\hfill \mbox{\textit{OCR MEI M4 2012 Q2 [13]}}