| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove motion is SHM from equation |
| Difficulty | Moderate -0.3 Part (i) is a straightforward differentiation exercise showing SHM definition (2 marks). Parts (ii)-(v) involve reading values from a graph and applying standard SHM formulas. While multi-part with several marks, all techniques are routine for M3 students with no novel problem-solving required. Slightly easier than average due to the mechanical nature of the tasks. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dot{x} = -A\omega\sin(\omega t - \phi)\) | B1 | |
| \(\ddot{x} = -A\omega^2\cos(\omega t - \phi)\) | M1 | Obtaining second derivative; allow one error |
| \(\ddot{x} = -\omega^2(x - c)\) | E1 [3] | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(c = 10\) | B1 | |
| \(A = 6\) | B1 | Accept \(A = -6\) |
| \(\frac{2\pi}{\omega} = 10\) | M1 | Using \(\frac{2\pi}{\omega}\); or other complete method for finding \(\omega\) |
| \(\omega = \frac{\pi}{5}\) | A1 | Accept \(\omega = -\frac{\pi}{5}\); allow \(\frac{2\pi}{10}\) etc |
| \(x = 16\) when \(t = 3 \Rightarrow 3\omega - \phi = 0\) | M1 | Obtaining simple relationship between \(\phi\) and \(\omega\); NB \(\phi = 3\) is M0; or \(x = 10 + 6\cos\{\frac{\pi}{5}(t-3)\}\) |
| \(\phi = \frac{3\pi}{5}\) | A1 [6] | NB other values possible; if exact values not seen give A0A1 for both \(\omega = 0.63\) and \(\phi = 1.9\); e.g. \(\phi = -\frac{7\pi}{5}\), \(\phi = \frac{13\pi}{5}\), \(x = 10 - 6\cos(\frac{\pi}{5}t - \frac{8\pi}{5})\) etc; max 5/6 if values not consistent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum speed is \(A\omega\) | M1 | Or e.g. evaluating \(\dot{x}\) when \(t = 5.5\) |
| Maximum speed is \(\frac{6\pi}{5}\) or \(3.77\ \text{ms}^{-1}\) (3 sf) | A1 [2] | FT is \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 0\), height is \(8.15\) m (3 sf) | B1 | FT is \(c + A\cos\phi\) (provided \(4 < x < 16\)); must use radians |
| \(v = -\frac{6\pi}{5}\sin(\frac{\pi t}{5} - \frac{3\pi}{5})\) | M1 | Or \(v^2 = \left(\frac{\pi}{5}\right)^2(6^2 - 1.854^2)\); allow one error in differentiation |
| When \(t = 0\), velocity is \(3.59\ \text{ms}^{-1}\) (3 sf) | A1 [3] | FT is \(A\omega\sin\phi\) (must be positive); \((\phi = 3\) gives \(x = 4.06\), \(v = 0.532)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 0\), \(x = 8.146\) | Correct FT value, or evidence of substitution, required \((\phi = 3\) gives \(x = 15.3)\) | |
| When \(t = 14\), \(x = 14.854\) | M1 | Finding \(x\) when \(t = 14\); requires \(4 < x(14) < 16\) |
| \((16 - 14.854)\) used | M1 | Also requires \(4 < x(0) < 16\) |
| \((16 - 8.146) + 12 + 12 + (16 - 14.854)\) | M1 | Fully correct strategy |
| Distance is \(33\) m | A1 [4] | CAO |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x} = -A\omega\sin(\omega t - \phi)$ | B1 | |
| $\ddot{x} = -A\omega^2\cos(\omega t - \phi)$ | M1 | Obtaining second derivative; allow one error |
| $\ddot{x} = -\omega^2(x - c)$ | E1 [3] | Correctly shown |
---
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $c = 10$ | B1 | |
| $A = 6$ | B1 | Accept $A = -6$ |
| $\frac{2\pi}{\omega} = 10$ | M1 | Using $\frac{2\pi}{\omega}$; or other complete method for finding $\omega$ |
| $\omega = \frac{\pi}{5}$ | A1 | Accept $\omega = -\frac{\pi}{5}$; allow $\frac{2\pi}{10}$ etc |
| $x = 16$ when $t = 3 \Rightarrow 3\omega - \phi = 0$ | M1 | Obtaining simple relationship between $\phi$ and $\omega$; NB $\phi = 3$ is M0; or $x = 10 + 6\cos\{\frac{\pi}{5}(t-3)\}$ |
| $\phi = \frac{3\pi}{5}$ | A1 [6] | NB other values possible; if exact values not seen give A0A1 for both $\omega = 0.63$ and $\phi = 1.9$; e.g. $\phi = -\frac{7\pi}{5}$, $\phi = \frac{13\pi}{5}$, $x = 10 - 6\cos(\frac{\pi}{5}t - \frac{8\pi}{5})$ etc; max 5/6 if values not consistent |
---
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum speed is $A\omega$ | M1 | Or e.g. evaluating $\dot{x}$ when $t = 5.5$ |
| Maximum speed is $\frac{6\pi}{5}$ or $3.77\ \text{ms}^{-1}$ (3 sf) | A1 [2] | FT is $|A\omega|$ (must be positive) |
---
## Question 3(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 0$, height is $8.15$ m (3 sf) | B1 | FT is $c + A\cos\phi$ (provided $4 < x < 16$); must use radians |
| $v = -\frac{6\pi}{5}\sin(\frac{\pi t}{5} - \frac{3\pi}{5})$ | M1 | Or $v^2 = \left(\frac{\pi}{5}\right)^2(6^2 - 1.854^2)$; allow one error in differentiation |
| When $t = 0$, velocity is $3.59\ \text{ms}^{-1}$ (3 sf) | A1 [3] | FT is $A\omega\sin\phi$ (must be positive); $(\phi = 3$ gives $x = 4.06$, $v = 0.532)$ |
---
## Question 3(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 0$, $x = 8.146$ | | Correct FT value, or evidence of substitution, required $(\phi = 3$ gives $x = 15.3)$ |
| When $t = 14$, $x = 14.854$ | M1 | Finding $x$ when $t = 14$; requires $4 < x(14) < 16$ |
| $(16 - 14.854)$ used | M1 | Also requires $4 < x(0) < 16$ |
| $(16 - 8.146) + 12 + 12 + (16 - 14.854)$ | M1 | Fully correct strategy |
| Distance is $33$ m | A1 [4] | CAO |
---3 A particle Q is performing simple harmonic motion in a vertical line. Its height, $x$ metres, above a fixed level at time $t$ seconds is given by
$$x = c + A \cos ( \omega t - \phi )$$
where $c , A , \omega$ and $\phi$ are constants.\\
(i) Show that $\ddot { x } = - \omega ^ { 2 } ( x - c )$.
Fig. 3 shows the displacement-time graph of Q for $0 \leqslant t \leqslant 14$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86dd0c01-970d-4b67-9a6c-5df276a4a2be-4_547_1079_703_495}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(ii) Find exact values for $c , A , \omega$ and $\phi$.\\
(iii) Find the maximum speed of Q .\\
(iv) Find the height and the velocity of Q when $t = 0$.\\
(v) Find the distance travelled by Q between $t = 0$ and $t = 14$.
\hfill \mbox{\textit{OCR MEI M3 2012 Q3 [18]}}