| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | String becomes slack timing |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics question combining elastic strings with SHM. Part (i) is routine equation of motion setup using Hooke's law, (ii) requires identifying amplitude and period from the SHM equation, (iii) is a standard small-angle pendulum derivation, and (iv) applies SHM formulas with given parameters. While it requires multiple techniques and careful bookkeeping across four parts, each component follows well-established methods without requiring novel insight. The multi-part structure and need to link elastic and pendulum systems elevates it slightly above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(T = 4\text{mg}(4 + x - 3.2)/3.2\) \([\text{ma} = \text{mg} - 4\text{mg}(0.8 + x)/3.2]\) \(4\ddot{x} = -49x\) | B1 M1 A1 | For using Newton's second law; AG |
| (ii) Amplitude is 0.8m | B1 B1 M1 | (from \(4 + \mathcal{A} = 4.8\)) |
| Period is \(2\pi/\omega\) where \(\omega^2 = 49/4\) | String is instantaneously slack when shortest (\(4 - \mathcal{A} = 3.2 = L\)). Thus required interval length = period. | |
| Slack at intervals of 1.8s | A1 M1 | AG; For using Newton's second law tangentially |
| (iii) \([\text{ma} = -\text{mg}\sin\theta]\) | A1 A1 | |
| \(mL\ddot{\theta} = -\text{mg}\sin\theta\) | ||
| For using \(\sin\theta \approx \theta\) for small angles and obtaining \(\ddot{\theta} = -(\frac{g}{L})\theta\) | A1 | |
| (iv) \([\theta = 0.08\cos(3.5 \times 0.25)] = 0.05127...]\) | M1 | For using \(\theta = \omega\cos\) where \(\omega^2 = 12.25\) |
| \([\dot{\theta} = -3.5(0.08)\sin(3.5 \times 0.25), \dot{\theta}^2 = 12.25(0.08^2 - 0.05127...)^2]\) | M1 | For differentiating \(\omega\cos\theta\) and using \(\dot{\theta}\) or for using \(\dot{\theta}^2 = \omega^2(\theta_0^2 - \theta^2)\) where \(\omega^2 = 12.25\) |
| \(\dot{\theta} = \mp 0.215\) | A1 | May be implied by final answer |
| \([v = 0.215 \times 9.8/12.25]\) | M1 | For using \(v = L\dot{\theta}\) and \(L = g/\omega^2\) |
| Speed is 0.172 ms⁻¹ | A1 |
**(i)** $T = 4\text{mg}(4 + x - 3.2)/3.2$ $[\text{ma} = \text{mg} - 4\text{mg}(0.8 + x)/3.2]$ $4\ddot{x} = -49x$ | B1 M1 A1 | For using Newton's second law; AG
**(ii)** Amplitude is 0.8m | B1 B1 M1 | (from $4 + \mathcal{A} = 4.8$)
Period is $2\pi/\omega$ where $\omega^2 = 49/4$ | | String is instantaneously slack when shortest ($4 - \mathcal{A} = 3.2 = L$). Thus required interval length = period.
Slack at intervals of 1.8s | A1 M1 | AG; For using Newton's second law tangentially
**(iii)** $[\text{ma} = -\text{mg}\sin\theta]$ | A1 A1 |
$mL\ddot{\theta} = -\text{mg}\sin\theta$ | |
For using $\sin\theta \approx \theta$ for small angles and obtaining $\ddot{\theta} = -(\frac{g}{L})\theta$ | A1 |
**(iv)** $[\theta = 0.08\cos(3.5 \times 0.25)] = 0.05127...]$ | M1 | For using $\theta = \omega\cos$ where $\omega^2 = 12.25$
$[\dot{\theta} = -3.5(0.08)\sin(3.5 \times 0.25), \dot{\theta}^2 = 12.25(0.08^2 - 0.05127...)^2]$ | M1 | For differentiating $\omega\cos\theta$ and using $\dot{\theta}$ or for using $\dot{\theta}^2 = \omega^2(\theta_0^2 - \theta^2)$ where $\omega^2 = 12.25$
$\dot{\theta} = \mp 0.215$ | A1 | May be implied by final answer
$[v = 0.215 \times 9.8/12.25]$ | M1 | For using $v = L\dot{\theta}$ and $L = g/\omega^2$
Speed is 0.172 ms⁻¹ | A1 |
---7 A particle $P$, of mass $m \mathrm {~kg}$, is attached to one end of a light elastic string of natural length 3.2 m and modulus of elasticity $4 m g \mathrm {~N}$. The other end of the string is attached to a fixed point $A$. The particle is released from rest at a point 4.8 m vertically below $A$. At time $t \mathrm {~s}$ after $P$ 's release $P$ is ( $4 + x ) \mathrm { m }$ below $A$.\\
(i) Show that $4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 49 x$.\\
$P$ 's motion is simple harmonic.\\
(ii) Write down the amplitude of $P$ 's motion and show that the string becomes slack instantaneously at intervals of approximately 1.8 s .
A particle $Q$ is attached to one end of a light inextensible string of length $L \mathrm {~m}$. The other end of the string is attached to a fixed point $B$. The particle is released from rest with the string taut and inclined at a small angle with the downward vertical. At time $t \mathrm {~s}$ after $Q$ 's release $B Q$ makes an angle of $\theta$ radians with the downward vertical.\\
(iii) Show that $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } \approx - \frac { g } { L } \theta$.
The period of the simple harmonic motion to which $Q$ 's motion approximates is the same as the period of $P$ 's motion.\\
(iv) Given that $\theta = 0.08$ when $t = 0$, find the speed of $Q$ when $t = 0.25$.
\hfill \mbox{\textit{OCR M3 2008 Q7 [15]}}