| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | SHM on inclined plane |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring: (i) equilibrium analysis with elastic strings on an inclined plane, (ii) derivation of SHM conditions and period calculation, (iii) application of SHM equations at a specific time. While it follows standard M3 methodology, the inclined plane context, elastic string mechanics, and multi-step reasoning across three parts make it moderately challenging—above average but not requiring exceptional insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg\times 0.2 = \frac{2.45m\times e}{0.3}\); \(e = 0.24\) | M1, A1 [2] | Allow sin/cos, wrong sign, missing \(g\); no errors; must show all numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(F = ma\) down slope: \(mg\sin\alpha - \frac{2.45m(x-0.3)}{0.3} = m\ddot{x}\) | M1 | 3 terms needed; allow sign error, sin/cos, missing \(g\) or \(m\); could use \(x\) in place of \(x-0.3\), leading to \(\ddot{x} = -\frac{49}{6}(x-0.24)\) (about \(x=0.24\)) |
| \(\ddot{x} = -\frac{49}{6}(x-0.54)\) | A1 | oe Accept \(2.45/0.3\) for \(\omega^2\); or \(x+0.24\) in place of \(x-0.3\) leading to \(\ddot{x} = -\frac{49}{6}x\) (about \(x=0\)) |
| SHM (about \(x = 0.54\)) | A1 | Dep M1A1; must be in correct form, and \(\omega^2\) in simplified form |
| \(\omega = 7/\sqrt{6}\ (2.8577)\) | B1 | soi; may see \(\omega^2 = 8\frac{1}{6}\) |
| \(T = 2.20\); \(a = 0.105\ \text{m}\ (0.1049795)\) | B1CAO, B1ft [6] | AG; need to see \(2\pi/\omega\) oe; ft their \(\omega\ \frac{3\sqrt{6}}{70}\); NB can find \(a\) by energy, leading to \(\omega\) and \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use SHM eqn for distance: \(x = -0.0956(227...)\) | M1, A1ft | \(x = a\sin\omega t\); their \(a\); allow M1 for \(x = a\cos\omega t\); or \(-0.9553\) or \(-0.09577\) |
| Dist from \(O\) is \(0.444(377...)\ \text{(m)}\) | A1CAO | |
| Use SHM equation for velocity: \(v = a\omega\cos\omega t\); \(v = -0.124\ (-0.123949...)\) | M1, A1 [5] | \(v = a\omega\cos\omega t\); must be clear velocity is towards \(O\); allow M1 for \(v = -a\omega\sin\omega t\) if consistent with \(x\) eqn for sin/cos, \(a\), \(\omega\); use of \(v^2 = \omega^2(a^2-x^2)\) will not gain A1 unless direction is established |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg\times 0.2 = \frac{2.45m\times e}{0.3}$; $e = 0.24$ | M1, A1 [2] | Allow sin/cos, wrong sign, missing $g$; no errors; must show all numbers |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $F = ma$ down slope: $mg\sin\alpha - \frac{2.45m(x-0.3)}{0.3} = m\ddot{x}$ | M1 | 3 terms needed; allow sign error, sin/cos, missing $g$ or $m$; could use $x$ in place of $x-0.3$, leading to $\ddot{x} = -\frac{49}{6}(x-0.24)$ (about $x=0.24$) |
| $\ddot{x} = -\frac{49}{6}(x-0.54)$ | A1 | oe Accept $2.45/0.3$ for $\omega^2$; or $x+0.24$ in place of $x-0.3$ leading to $\ddot{x} = -\frac{49}{6}x$ (about $x=0$) |
| SHM (about $x = 0.54$) | A1 | Dep M1A1; must be in correct form, and $\omega^2$ in simplified form |
| $\omega = 7/\sqrt{6}\ (2.8577)$ | B1 | soi; may see $\omega^2 = 8\frac{1}{6}$ |
| $T = 2.20$; $a = 0.105\ \text{m}\ (0.1049795)$ | B1CAO, B1ft [6] | AG; need to see $2\pi/\omega$ oe; ft their $\omega\ \frac{3\sqrt{6}}{70}$; NB can find $a$ by energy, leading to $\omega$ and $T$ |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use SHM eqn for distance: $x = -0.0956(227...)$ | M1, A1ft | $x = a\sin\omega t$; their $a$; allow M1 for $x = a\cos\omega t$; or $-0.9553$ or $-0.09577$ |
| Dist from $O$ is $0.444(377...)\ \text{(m)}$ | A1CAO | |
| Use SHM equation for velocity: $v = a\omega\cos\omega t$; $v = -0.124\ (-0.123949...)$ | M1, A1 [5] | $v = a\omega\cos\omega t$; must be clear velocity is towards $O$; allow M1 for $v = -a\omega\sin\omega t$ if consistent with $x$ eqn for sin/cos, $a$, $\omega$; use of $v^2 = \omega^2(a^2-x^2)$ will not gain A1 unless direction is established |7\\
\includegraphics[max width=\textwidth, alt={}, center]{3243c326-a51c-462f-a57c-a150d0044ea9-4_382_773_1567_648}
One end of a light elastic string, of natural length 0.3 m , is attached to a fixed point $O$ on a smooth plane that is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.2$. A particle $P$ of mass $m \mathrm {~kg}$ is attached to the other end of the string. The string lies along a line of greatest slope of the plane and has modulus of elasticity $2.45 m \mathrm {~N}$ (see diagram).\\
(i) Show that in the equilibrium position the extension of the string is 0.24 m .\\
$P$ is given a velocity of $0.3 \mathrm {~ms} ^ { - 1 }$ down the plane from the equilibrium position.\\
(ii) Show that $P$ performs simple harmonic motion with period 2.20 s (correct to 3 significant figures), and find the amplitude of the motion.\\
(iii) Find the distance of $P$ from $O$ and the velocity of $P$ at the instant 1.5 seconds after $P$ is set in motion.
\hfill \mbox{\textit{OCR M3 2014 Q7 [13]}}