| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Two springs/strings system equilibrium |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question with two springs. Part (i) is routine equilibrium (balancing tensions). Parts (ii)-(iv) involve standard SHM formulas (period from ω, amplitude from energy, time from SHM equation). While it requires careful bookkeeping of two springs and multiple steps, all techniques are textbook-standard with no novel insight required. Slightly above average difficulty due to the two-spring setup and multi-part nature, but well within typical M3 scope. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) For \(180e/1\) or \(360(0.8-e)/1.2\) | M1 | For using \(T = \lambda e/L\) once |
| \(T_A = 180\times0.5/1\) or \(T_B = 360\times0.3/1.2\) | A1 | |
| \(480e = 240\) or \(T_A = 90,\; T_B = 90\) | M1 | For using \(T_A(e) = T_B(e)\) or attempting to show \(T_A = T_B\) when \(BQ = 1.5\) |
| \(BQ = 1 + 0.5 = 1.5\) m or \(T_A = T_B\) | A1 | [4] AG |
| (ii) \(T_B = 360(0.3-x)/1.2\), \(T_A = 180(0.5+x)/1\) | B1, B1 | |
| \(1.2\,d^2x/dt^2 = 300(0.3-x) - 180(0.5+x)\) | M1 | For using Newton's 2nd law |
| \(d^2x/dt^2 = -400x\) | A1 | |
| Period is \(2\pi/\sqrt{400} = 0.314\) s | A1 | [5] AG |
| (iii) Max amplitude \(= 1.5 - 1.2 = 0.3\) m | M1, A1 | For using \(T_B = 0\) |
| amplitude \(= u/\sqrt{400}\) or \(180\times0.5^2/(2\times1) + 360\times0.3^2/(2\times1.2) + \frac{1}{2}\times1.2u_{\max}^2 = 180\times0.8^2/(2\times1)\) | M1 | For using Amp. \(= u/\omega\) or 'energy at equilibrium position \(=\) energy at max. displacement' |
| Maximum value of \(u\) is \(6\) | A1 | [4] AG |
| (iv) \(-0.2 = 0.3\sin20t\) | M1 | For relevant trig. equation |
| \(20t = 0.7297 + 3.142\) | M1 | For method of obtaining relevant solution |
| Time taken is \(0.194\) s | A1 | [3] |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** For $180e/1$ or $360(0.8-e)/1.2$ | M1 | For using $T = \lambda e/L$ once |
| $T_A = 180\times0.5/1$ or $T_B = 360\times0.3/1.2$ | A1 | |
| $480e = 240$ or $T_A = 90,\; T_B = 90$ | M1 | For using $T_A(e) = T_B(e)$ or attempting to show $T_A = T_B$ when $BQ = 1.5$ |
| $BQ = 1 + 0.5 = 1.5$ m or $T_A = T_B$ | A1 | [4] AG |
| **(ii)** $T_B = 360(0.3-x)/1.2$, $T_A = 180(0.5+x)/1$ | B1, B1 | |
| $1.2\,d^2x/dt^2 = 300(0.3-x) - 180(0.5+x)$ | M1 | For using Newton's 2nd law |
| $d^2x/dt^2 = -400x$ | A1 | |
| Period is $2\pi/\sqrt{400} = 0.314$ s | A1 | [5] AG |
| **(iii)** Max amplitude $= 1.5 - 1.2 = 0.3$ m | M1, A1 | For using $T_B = 0$ |
| amplitude $= u/\sqrt{400}$ or $180\times0.5^2/(2\times1) + 360\times0.3^2/(2\times1.2) + \frac{1}{2}\times1.2u_{\max}^2 = 180\times0.8^2/(2\times1)$ | M1 | For using Amp. $= u/\omega$ or 'energy at equilibrium position $=$ energy at max. displacement' |
| Maximum value of $u$ is $6$ | A1 | [4] AG |
| **(iv)** $-0.2 = 0.3\sin20t$ | M1 | For relevant trig. equation |
| $20t = 0.7297 + 3.142$ | M1 | For method of obtaining relevant solution |
| Time taken is $0.194$ s | A1 | [3] |7\\
\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-4_122_1009_265_571}
As shown in the diagram, $A$ and $B$ are fixed points on a smooth horizontal table, where $A B = 3 \mathrm {~m}$. A particle $Q$ of mass 1.2 kg is attached to $A$ by a light elastic string of natural length 1 m and modulus of elasticity $180 \mathrm {~N} . Q$ is attached to $B$ by a light elastic string of natural length 1.2 m and modulus of elasticity 360 N .\\
(i) Verify that when $Q$ is in equilibrium $B Q = 1.5 \mathrm {~m}$.\\
$Q$ is projected towards $B$ from the equilibrium position with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Subsequently $Q$ oscillates with simple harmonic motion.\\
(ii) Show that the period of the motion is 0.314 s approximately.\\
(iii) Show that $u \leqslant 6$.\\
(iv) Given that $u = 6$, find the time taken for $Q$ to move from the equilibrium position to a position 1.3 m from $A$ for the first time.
\hfill \mbox{\textit{OCR M3 2006 Q7 [16]}}