| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2019 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find period from given information |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring systematic application of well-known formulas (v² = ω²(a² - x²), max acceleration = ω²a, period = 2π/ω). While it has multiple parts and requires careful setup of the geometry (identifying amplitude and displacement), each step follows directly from standard SHM theory without requiring novel insight or complex problem-solving. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks |
|---|---|
| M1 | |
| A1 | |
| A1ft |
| Answer | Marks |
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| M1A1ft | |
| (2) |
| Answer | Marks |
|---|---|
| M1A1 | |
| M1A1 | |
| A1ft | |
| (3) |
| Answer | Marks |
|---|---|
| M1 | |
| A1 | |
| M1 | |
| A1 | |
| A1ft | |
| (5) |
**(a)**
$v^2 = \omega^2\left(a^2 - \frac{a^2}{4}\right) = \frac{3a^2\omega^2}{4}$
$\frac{27a^2}{4} = \frac{3a^2\omega^2}{4}$
$\omega = 3$
Period $= \frac{2\pi}{3}$
Max magnitude of acceleration $= a\omega^2$
$45 = 9a$, $a = 5$
| M1 |
| A1 |
| A1ft |
**(b)**
$x = a\sin\omega t$, $\dot{x} = a\omega\cos\omega t$ (or $x = a\cos\omega t$, $\dot{x} = -a\omega\sin\omega t$)
$\dot{x}_{\text{max}} = a\omega = 5 \times 3 = 15$ (ms$^{-1}$)
OR $v_{\text{max}} = a\omega = 5 \times 3 = 15$ (ms$^{-1}$)
| M1A1ft |
| (2) |
**(c)**
Time A to C: $a = a\cos\omega t \Rightarrow \frac{1}{2} = \cos 3t$
$t_{AC} = \frac{1}{3}\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{9}$ (0.3490...)
Time A to D: $\frac{\pi}{9} + \frac{2\pi}{12} = \frac{5\pi}{18}$
$x_D = 5\cos\left(3 \times \frac{5\pi}{18}\right) = -\frac{5\sqrt{3}}{2}$ (= −4.330...)
Distance CD: $\frac{5}{2} + \frac{5\sqrt{3}}{2} = \frac{5(1 + \sqrt{3})}{2}$ or 6.8 (m) (or better)
| M1A1 |
| M1A1 |
| A1ft |
| (3) |
**(d)**
Use of $v^2 = \omega^2(a^2 - x^2)$ with $v = \frac{3a}{2}$, $x = \frac{a}{2}$, amplitude $= a$ (or any other complete method)
Correct value for $\omega$
Correct period, follow through their $\omega$
Use max magnitude of acceleration $= a\omega^2$ with their $\omega$
$a = 5$
Use either method shown with their values of $a$ and $\omega$ to obtain a value for the max speed
$v_{\text{max}} = 15$ (ms$^{-1}$)
Attempt time A to C with their value of $\omega$ and $x = \frac{a}{2}$ or half their amplitude. Must reach a value for $t$ using radians
Correct time, exact or minimum 3 sf (no penalty for using an incorrect amplitude)
The above 2 marks can be awarded for a time even if no indication of which time they are finding (i.e., not stated to be time from end to C or centre to C). Following marks can only be awarded if work is consistent with their work for these 2 marks.
Add $\frac{1}{4}$ period and use this time to obtain a value for $x$ at D using their value for $a$ or just $a$
Correct value of $x$ exact or minimum 3 sf or a multiple of $a$
Correct distance CD, follow through their $x_D$. Must be positive. Minimum 2 sf for decimal.
| M1 |
| A1 |
| M1 |
| A1 |
| A1ft |
| (5) |
**Alternative for (d):**
Time C to centre O: $a = a\sin\omega t \Rightarrow \frac{1}{2} = \sin 3t$
$t_{CO} = \frac{1}{3}\sin^{-1}\left(\frac3. A particle $P$ is moving in a straight line with simple harmonic motion between two points $A$ and $B$, where $A B$ is $2 a$ metres. The point $C$ lies on the line $A B$ and $A C = \frac { 1 } { 2 } a$ metres. The particle passes through $C$ with speed $\frac { 3 a \sqrt { 3 } } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the period of the motion.
The maximum magnitude of the acceleration of $P$ is $45 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find
\item the value of $a$,
\item the maximum speed of $P$.
The point $D$ lies on $A B$ and $P$ takes a quarter of one period to travel directly from $C$ to $D$.
\item Find the distance CD.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2019 Q3 [12]}}