| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find amplitude from speed conditions |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring application of the velocity formula v² = ω²(a² - x²) with two simultaneous equations. While it involves algebraic manipulation and solving a system, the approach is routine for students who have practiced SHM problems. The 'show that' in part (b) provides the answer, reducing difficulty slightly. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks |
|---|---|
| (a) \(64 = n^2(a^2-1)\), \(16 = n^2(a^2-4)\) | M1 A1 |
| Divide: \(4 = \frac{a^2-1}{a^2-4}\) | M1 A1 |
| \(3a^2 = 15\) | M1 A1 |
| \(a = \sqrt{5}\) m (= 2.24 m) | |
| (b) \(n^2 = \frac{64}{a^2-1} = 16\) | M1 A1 A1 |
| \(n = 4\) | |
| \(T = \frac{2\pi}{4} = \frac{\pi}{2}\) s | 7 marks total |
**(a)** $64 = n^2(a^2-1)$, $16 = n^2(a^2-4)$ | M1 A1
Divide: $4 = \frac{a^2-1}{a^2-4}$ | M1 A1
$3a^2 = 15$ | M1 A1
$a = \sqrt{5}$ m (= 2.24 m) |
**(b)** $n^2 = \frac{64}{a^2-1} = 16$ | M1 A1 A1
$n = 4$ |
$T = \frac{2\pi}{4} = \frac{\pi}{2}$ s | 7 marks totalA particle $P$ moves with simple harmonic motion in a straight line. The centre of oscillation is $O$. When $P$ is at a distance 1 m from $O$, its speed is 8 ms$^{-1}$. When it is at a distance 2 m from $O$, its speed is 4 ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the amplitude of the motion. [4 marks]
\item Show that the period of motion is $\frac{\pi}{2}$ s. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [7]}}