| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: given force or equation of motion directly |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring the substitution to identify the centre of oscillation, application of standard SHM formulas for period and amplitude, and careful analysis of velocity conditions with timing calculations. While the techniques are standard for M3, the question requires multiple steps, careful bookkeeping of positions, and understanding of the symmetry of SHM motion across several parts. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([0.36 - 0.144x = 0.1a]\) | M1 | For applying Newton's second law |
| \(\ddot{x} = 3.6 - 1.44x\) | A1 | |
| \(\ddot{y} = -1.44y \Rightarrow \text{SHM}\) or \(d^2(x-2.5)/dt^2 = -1.44(x-2.5) \Rightarrow \text{SHM}\) | B1 | |
| M1 | For using \(T = 2\pi/n\) | |
| Of period \(5.24\text{s}\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Amplitude is \(0.5\text{m}\) | B1 | |
| \(0.48^2 = 1.2^2(0.5^2 - y^2)\) | M1 | For using \(v^2 = n^2(a^2 - y^2)\) |
| A1ft | ||
| Possible values are \(2.2\) and \(2.8\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([t_0 = (\sin^{-1}0.6)/1.2;\; t_1 = (\cos^{-1}0.6)/1.2]\) | M1 | For using \(y = 0.5\sin1.2t\) to find \(t_0\) or \(y = 0.5\cos1.2t\) to find \(t_1\) |
| \(t_0 = 0.53625\ldots\) or \(t_1 = 0.7727\ldots\) | A1 | Principal value may be implied |
| (a) \([2(\sin^{-1}0.6)/1.2\) or \((\pi - 2\cos^{-1}0.6)/1.2]\) | M1 | For using \(\Delta t = 2t_0\) or \(\Delta t = T/2 - 2t_1\) |
| Time interval is \(1.07\text{s}\) | A1ft | ft incorrect \(t_0\) or \(t_1\) |
| (b) | From \(\Delta t = T/2 - 2t_0\) or \(\Delta t = 2t_1\); ft \(2.62 - \text{ans(a)}\) or | |
| Time interval is \(1.55\text{s}\) | B1ft | 5 |
# Question 6:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[0.36 - 0.144x = 0.1a]$ | M1 | For applying Newton's second law |
| $\ddot{x} = 3.6 - 1.44x$ | A1 | |
| $\ddot{y} = -1.44y \Rightarrow \text{SHM}$ or $d^2(x-2.5)/dt^2 = -1.44(x-2.5) \Rightarrow \text{SHM}$ | B1 | |
| | M1 | For using $T = 2\pi/n$ |
| Of period $5.24\text{s}$ | A1 | 5 | AG |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Amplitude is $0.5\text{m}$ | B1 | |
| $0.48^2 = 1.2^2(0.5^2 - y^2)$ | M1 | For using $v^2 = n^2(a^2 - y^2)$ |
| | A1ft | |
| Possible values are $2.2$ and $2.8$ | A1 | 4 | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[t_0 = (\sin^{-1}0.6)/1.2;\; t_1 = (\cos^{-1}0.6)/1.2]$ | M1 | For using $y = 0.5\sin1.2t$ to find $t_0$ or $y = 0.5\cos1.2t$ to find $t_1$ |
| $t_0 = 0.53625\ldots$ or $t_1 = 0.7727\ldots$ | A1 | Principal value may be implied |
| **(a)** $[2(\sin^{-1}0.6)/1.2$ or $(\pi - 2\cos^{-1}0.6)/1.2]$ | M1 | For using $\Delta t = 2t_0$ or $\Delta t = T/2 - 2t_1$ |
| Time interval is $1.07\text{s}$ | A1ft | ft incorrect $t_0$ or $t_1$ |
| **(b)** | | From $\Delta t = T/2 - 2t_0$ or $\Delta t = 2t_1$; ft $2.62 - \text{ans(a)}$ or |
| Time interval is $1.55\text{s}$ | B1ft | 5 | incorrect $t_0$ or $t_1$ |
---6 A particle $P$ of mass 0.1 kg moves in a straight line on a smooth horizontal surface. A force of $( 0.36 - 0.144 x ) \mathrm { N }$ acts on $P$ in the direction from $O$ to $P$, where $x \mathrm {~m}$ is the displacement of $P$ from a point $O$ on the surface at time $t \mathrm {~s}$.\\
(i) By using the substitution $x = y + 2.5$, or otherwise, show that $P$ moves with simple harmonic motion of period 5.24 s , correct to 3 significant figures.
The maximum value of $x$ during the motion is 3 .\\
(ii) Write down the amplitude of $P$ 's motion and find the two possible values of $x$ for which $P$ 's speed is $0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) On each of the first two occasions when $P$ has speed $0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 } , P$ is moving towards $O$. Find the time interval between
\begin{enumerate}[label=(\alph*)]
\item these first two occasions,
\item the second and third occasions when $P$ has speed $0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2008 Q6 [14]}}