| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring equilibrium analysis, proving SHM from Hooke's law, finding amplitude from energy/initial conditions, and evaluating position/velocity at a given time. All steps follow well-established procedures with no novel insight required, making it slightly easier than average for M3 level. |
| 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 | Mark | Guidance |
| (i) | M1 | For using \(T = mg\) and \(T = \lambda e/L\) |
| \(3.5e/0.7 = 0.2g\) \([e = 0.392]\) | A1 | |
| Position is \(1.092\text{m}\) below O | A1 | 3 AG |
| (ii) | M1 | For using Newton's second law |
| \(0.2g - 3.5(0.392+x)/0.7 = 0.2a\) | A1ft | ft incorrect e |
| \(a = -25x\) | A1ft | ft incorrect e |
| \([25A^2 = 1.6^2\) or \(\frac{1}{2}(0.2)1.6^2 + 3.5\times0.392^2/(2\times0.7) + 0.2gA = 3.5(0.392+A)^2/(2\times0.7)]\) | M1 | For using \(A^2n^2 = v_{max}^2\) or Energy at lowest point = energy at equilibrium point (4 terms needed including 2 EE terms) |
| Amplitude is \(0.32\text{m}\) | A1ft | 5 |
| (iii) \([x = 0.32\sin2^c]\) | M1 | For using \(x = A\sin nt\) or \(A\cos(\pi/2 - nt)\) |
| \(x = 0.291\) | A1 | |
| \([v = 0.32\times5\cos2^c\) or \(v^2 = 25(0.32^2 - 0.291^2)]\) or \(0.256 + 0.38416 + 0.2g(0.291) = \frac{1}{2}(0.2)v^2 + 2.5(0.683)^2\) | M1 | For using \(v = An\cos nt\) or \(v^2 = n^2(A^2-x^2)\) or Energy at equilibrium = energy at \(x=0.291\) |
| \(v^2 = 0.443\) | A1 | May be implied |
| \(v = -0.666\) (or \(0.666\) upwards) | A1 | 5 |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** | M1 | For using $T = mg$ and $T = \lambda e/L$ |
| $3.5e/0.7 = 0.2g$ $[e = 0.392]$ | A1 | |
| Position is $1.092\text{m}$ below O | A1 | **3** AG |
| **(ii)** | M1 | For using Newton's second law |
| $0.2g - 3.5(0.392+x)/0.7 = 0.2a$ | A1ft | ft incorrect e |
| $a = -25x$ | A1ft | ft incorrect e |
| $[25A^2 = 1.6^2$ or $\frac{1}{2}(0.2)1.6^2 + 3.5\times0.392^2/(2\times0.7) + 0.2gA = 3.5(0.392+A)^2/(2\times0.7)]$ | M1 | For using $A^2n^2 = v_{max}^2$ or Energy at lowest point = energy at equilibrium point (4 terms needed including 2 EE terms) |
| Amplitude is $0.32\text{m}$ | A1ft | **5** |
| **(iii)** $[x = 0.32\sin2^c]$ | M1 | For using $x = A\sin nt$ or $A\cos(\pi/2 - nt)$ |
| $x = 0.291$ | A1 | |
| $[v = 0.32\times5\cos2^c$ or $v^2 = 25(0.32^2 - 0.291^2)]$ or $0.256 + 0.38416 + 0.2g(0.291) = \frac{1}{2}(0.2)v^2 + 2.5(0.683)^2$ | M1 | For using $v = An\cos nt$ or $v^2 = n^2(A^2-x^2)$ or Energy at equilibrium = energy at $x=0.291$ |
| $v^2 = 0.443$ | A1 | May be implied |
| $v = -0.666$ (or $0.666$ upwards) | A1 | **5** |
---4 A particle $P$ of mass 0.2 kg is suspended from a fixed point $O$ by a light elastic string of natural length 0.7 m and modulus of elasticity $3.5 \mathrm {~N} . P$ is at the equilibrium position when it is projected vertically downwards with speed $1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$ after being set in motion $P$ is $x \mathrm {~m}$ below the equilibrium position and has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that the equilibrium position of $P$ is 1.092 m below $O$.\\
(ii) Prove that $P$ moves with simple harmonic motion, and calculate the amplitude.\\
(iii) Calculate $x$ and $v$ when $t = 0.4$.
\hfill \mbox{\textit{OCR M3 2007 Q4 [13]}}