| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find amplitude from speed conditions |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring application of v² = ω²(a² - x²) with simultaneous equations to find amplitude and ω, followed by period calculation, energy considerations, and time integration. While the techniques are standard M3 content, the question requires careful algebraic manipulation, understanding of SHM energy relationships, and geometric reasoning about positions, making it moderately challenging but within expected M3 scope. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v^2 = \dot{\omega}(a^2-x^2)\) | B1 | |
| \(x = \pm 3, v = \pm 6 \therefore 36 = \dot{\omega}(a^2-9)\) giving \(\dot{\omega} = \frac{36}{a^2-9}\) | M1 A1 | |
| \(x = \pm 2.25, v = \pm 8 \therefore 64 = \dot{\omega}(a^2-\frac{81}{16})\) | M1 | |
| combining, \(64 = \frac{36}{a^2-9}(a^2-\frac{81}{16})\) | M1 | |
| giving \(16(a^2-9) = 9(a^2-\frac{81}{16})\) | A1 | |
| \(256a^2 - 2304 = 144a^2 - 729\) | ||
| so \(a^2 = \frac{1575}{112} = \frac{225}{16}, a = \frac{15}{4} \therefore AB = 7.5 \text{ m}\) | M1 A1 | |
| (b) \(a^2 = \frac{225}{16} \therefore 64 = \dot{\omega}(\frac{225}{16}-\frac{81}{16})\) so \(\dot{\omega} = \frac{9}{4}, \omega = \frac{8}{5}\) | M1 A1 | |
| period = \(\frac{2\pi}{\omega} = 2\pi \times \frac{5}{8} = \frac{5}{4}\pi \text{ s}\) | M1 A1 | |
| (c) KE \(= \frac{1}{2}m\dot{v}^2 = \frac{1}{2} \times 2.5 \times (\frac{64}{9}(\frac{225}{16}-1.05^2)) = 115.2 \text{ J}\) | M1 A2 | |
| (d) \(x = a\cos\omega t \therefore -\frac{3.75}{2} = 3.75\cos\omega t\) | M1 A1 | |
| \(\cos\omega t = -\frac{1}{2} \therefore \omega t = \frac{2\pi}{3}\) | M1 | |
| \(\therefore t = \frac{2\pi}{3} \times \frac{5}{8} = \frac{5\pi}{12} \text{ s}\) | A1 | (19) |
**(a)** $v^2 = \dot{\omega}(a^2-x^2)$ | B1 |
$x = \pm 3, v = \pm 6 \therefore 36 = \dot{\omega}(a^2-9)$ giving $\dot{\omega} = \frac{36}{a^2-9}$ | M1 A1 |
$x = \pm 2.25, v = \pm 8 \therefore 64 = \dot{\omega}(a^2-\frac{81}{16})$ | M1 |
combining, $64 = \frac{36}{a^2-9}(a^2-\frac{81}{16})$ | M1 |
giving $16(a^2-9) = 9(a^2-\frac{81}{16})$ | A1 |
$256a^2 - 2304 = 144a^2 - 729$ | |
so $a^2 = \frac{1575}{112} = \frac{225}{16}, a = \frac{15}{4} \therefore AB = 7.5 \text{ m}$ | M1 A1 |
**(b)** $a^2 = \frac{225}{16} \therefore 64 = \dot{\omega}(\frac{225}{16}-\frac{81}{16})$ so $\dot{\omega} = \frac{9}{4}, \omega = \frac{8}{5}$ | M1 A1 |
period = $\frac{2\pi}{\omega} = 2\pi \times \frac{5}{8} = \frac{5}{4}\pi \text{ s}$ | M1 A1 |
**(c)** KE $= \frac{1}{2}m\dot{v}^2 = \frac{1}{2} \times 2.5 \times (\frac{64}{9}(\frac{225}{16}-1.05^2)) = 115.2 \text{ J}$ | M1 A2 |
**(d)** $x = a\cos\omega t \therefore -\frac{3.75}{2} = 3.75\cos\omega t$ | M1 A1 |
$\cos\omega t = -\frac{1}{2} \therefore \omega t = \frac{2\pi}{3}$ | M1 |
$\therefore t = \frac{2\pi}{3} \times \frac{5}{8} = \frac{5\pi}{12} \text{ s}$ | A1 | **(19)**
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**Total: (75)**6. A particle $P$ of mass 2.5 kg is moving with simple harmonic motion in a straight line between two points $A$ and $B$ on a smooth horizontal table. When $P$ is 3 m from $O$, the centre of the oscillations, its speed is $6 \mathrm {~ms} ^ { - 1 }$. When $P$ is 2.25 m from $O$, its speed is $8 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $A B = 7.5 \mathrm {~m}$.
\item Find the period of the motion.
\item Find the kinetic energy of $P$ when it is 2.7 m from $A$.
\item Show that the time taken by $P$ to travel directly from $A$ to the midpoint of $O B$ is $\frac { \pi } { 4 }$ seconds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [19]}}