| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Standard +0.3 This is a standard M3 SHM question involving springs with familiar setup (equilibrium extension, frequency relationship). Part (a) requires routine application of Hooke's law and SHM frequency formula ω²=λ/(ml), part (b) uses equilibrium condition mg=λe/l, and part (c) applies Hooke's law directly. All techniques are textbook exercises with no novel insight required, though the multi-step nature and algebraic manipulation place it slightly above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks |
|---|---|
| (a) \(mg = \lambda e\) | M1 A1 |
| \(\lambda = \frac{mgl}{e}\) | |
| SHM: \(m\ddot{x} = mg - \frac{mgl}{e}(e + x)\) | M1 A1 |
| \(\ddot{x} = -\omega^2 x = -\frac{g}{e}x\) | M1 A1 M1 |
| \(\omega^2 = \frac{g}{e}\) | |
| 5 osc. per second, so \(\omega = 100\pi\) | M1 A1 M1 |
| Thus \(\frac{g}{e} = 100\pi^2\) | A1 |
| \(\frac{\lambda}{l} = 100\pi^2 m\) | |
| (b) \(mg = \lambda e = 100\pi^2 me\), so \(e = \frac{g}{100\pi^2}\) | M1 A1 |
| (c) \(T_1 = \frac{\lambda}{l} = \lambda = 100\pi^2 ml\) N | M1 A1 |
(a) $mg = \lambda e$ | M1 A1 |
$\lambda = \frac{mgl}{e}$ | |
SHM: $m\ddot{x} = mg - \frac{mgl}{e}(e + x)$ | M1 A1 |
$\ddot{x} = -\omega^2 x = -\frac{g}{e}x$ | M1 A1 M1 |
$\omega^2 = \frac{g}{e}$ | |
5 osc. per second, so $\omega = 100\pi$ | M1 A1 M1 |
Thus $\frac{g}{e} = 100\pi^2$ | A1 |
$\frac{\lambda}{l} = 100\pi^2 m$ | |
(b) $mg = \lambda e = 100\pi^2 me$, so $e = \frac{g}{100\pi^2}$ | M1 A1 |
(c) $T_1 = \frac{\lambda}{l} = \lambda = 100\pi^2 ml$ N | M1 A1 |
**Total: 12 marks**
---A particle $P$ of mass $m$ kg hangs in equilibrium at one end of a light spring, of natural length $l$ m and modulus of elasticity $\lambda$ N, whose other end is fixed at a point vertically above $P$. In this position the length of the spring is $(l + e)$ m. When $P$ is displaced vertically through a small distance and released, it performs simple harmonic motion with 5 oscillations per second.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{\lambda}{l} = 100n^2m$. [8 marks]
\item Express $e$ in terms of $g$. [2 marks]
\item Determine, in terms of $m$ and $l$, the magnitude of the tension in the spring when it is stretched to twice its natural length. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [12]}}