| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: simple pendulum (particle on string) |
| Difficulty | Standard +0.3 This is a straightforward application of small-angle SHM for a simple pendulum. Part (i) requires showing T=2π√(l/g) with given values—routine verification. Parts (ii) and (iii) use standard SHM equations (ω²=ω₀²(A²-x²) and x=Acos(ωt)) with small amplitude approximation already established. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05q Trig in context: vectors, kinematics, forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For using Newton's 2nd Law perp. to string with \(a = L\ddot{\theta}\) |
| \(L(m)\ddot{\theta} = -(m)g\sin\theta\) or \((m)\ddot{s} = -(m)g\sin\theta\) | A1 | |
| \(\ddot{\theta} \approx -k\theta\) or \(\ddot{s} = -ks\) [and motion is therefore approx. simple harmonic] | B1 | |
| (implied method) | M1 | For using \(T = 2\pi/n\) and \(k = w^2\) or \(T = 2\pi\sqrt{L/g}\) for simple pendulum |
| Period is \(3.14\text{s}\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For using \(\dot{\theta}^2 = n^2(\theta_0^2 - \theta^2)\) or the principle of conservation of energy |
| \(\dot{\theta}^2 = 4(0.1^2 - 0.06^2)\) or \(\frac{1}{2}m(2.45\dot{\theta})^2 = 2.45mg(\cos 0.06 - \cos 0.1)\) | A1 | |
| Angular speed is \(0.16 \text{ rad s}^{-1}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\dot{\theta} = -0.2\sin 2t]\) | M1 | For using \(\dot{\theta} = d(A\cos nt)/dt\) |
| \(\dot{\theta} = -0.2\sin(2 \times 0.464)\) | A1ft | |
| Angular speed is \(0.16 \text{ rad s}^{-1}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For using \(\theta = A\cos nt\) or \(A\sin(\pi/2 - nt)\) or for using \(\theta = A\sin nt\) and \(T = t_{0.1} - t_{0.06}\) |
| \(0.06 = 0.1\cos 2t\) or \(0.1\sin(\pi/2 - 2t)\) or \(2T = \pi/2 - \sin^{-1}0.6\) | A1ft | ft angular displacement of \(0.04\) instead of \(0.06\) |
| Time taken is \(0.464\text{s}\) | A1 | 3 |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For using Newton's 2nd Law perp. to string with $a = L\ddot{\theta}$ |
| $L(m)\ddot{\theta} = -(m)g\sin\theta$ or $(m)\ddot{s} = -(m)g\sin\theta$ | A1 | |
| $\ddot{\theta} \approx -k\theta$ or $\ddot{s} = -ks$ [and motion is therefore approx. simple harmonic] | B1 | |
| (implied method) | M1 | For using $T = 2\pi/n$ and $k = w^2$ or $T = 2\pi\sqrt{L/g}$ for simple pendulum |
| Period is $3.14\text{s}$ | A1 | 5 | AG |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For using $\dot{\theta}^2 = n^2(\theta_0^2 - \theta^2)$ or the principle of conservation of energy |
| $\dot{\theta}^2 = 4(0.1^2 - 0.06^2)$ or $\frac{1}{2}m(2.45\dot{\theta})^2 = 2.45mg(\cos 0.06 - \cos 0.1)$ | A1 | |
| Angular speed is $0.16 \text{ rad s}^{-1}$ | A1 | 3 | ($0.1599\ldots$ from energy method) |
**OR (ii) [if (iii) attempted before (ii)]:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\dot{\theta} = -0.2\sin 2t]$ | M1 | For using $\dot{\theta} = d(A\cos nt)/dt$ |
| $\dot{\theta} = -0.2\sin(2 \times 0.464)$ | A1ft | |
| Angular speed is $0.16 \text{ rad s}^{-1}$ | A1 | 3 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For using $\theta = A\cos nt$ or $A\sin(\pi/2 - nt)$ or for using $\theta = A\sin nt$ and $T = t_{0.1} - t_{0.06}$ |
| $0.06 = 0.1\cos 2t$ or $0.1\sin(\pi/2 - 2t)$ or $2T = \pi/2 - \sin^{-1}0.6$ | A1ft | ft angular displacement of $0.04$ instead of $0.06$ |
| Time taken is $0.464\text{s}$ | A1 | 3 | |
---4 A particle is connected to a fixed point by a light inextensible string of length 2.45 m to make a simple pendulum. The particle is released from rest with the string taut and inclined at 0.1 radians to the downward vertical.\\
(i) Show that the motion of the particle is approximately simple harmonic with period 3.14 s , correct to 3 significant figures.
Calculate, in either order,\\
(ii) the angular speed of the pendulum when it has moved 0.04 radians from the initial position,\\
(iii) the time taken by the pendulum to move 0.04 radians from the initial position.
\hfill \mbox{\textit{OCR M3 2006 Q4 [11]}}