| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Challenging +1.2 This is a standard SHM question requiring Hooke's law application, proving the SHM equation, and using standard SHM formulas (amplitude, period, energy). While it has multiple parts and requires careful setup of the elastic force equation, all techniques are routine for M3 students with no novel insights needed. The multi-part structure and modelling assumptions push it slightly above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02d Mechanical energy: KE and PE concepts6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{980}{25}y = 5\times9.8\) | M1 | Using \(\frac{\lambda y}{l_0}\) (Allow M1 for \(980y = mg\)) |
| Extension is 1.25 m | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = \frac{980}{25}(1.25 + x)\) | B1 (ft) | *(ft) indicates ft from previous parts as for A marks* |
| \(5\times9.8 - 39.2(1.25+x) = 5\frac{d^2x}{dt^2}\) | M1 | Equation of motion with three terms |
| \(-39.2x = 5\frac{d^2x}{dt^2}\) | F1 | Must have \(\ddot{x}\) in terms of \(x\) only |
| \(\frac{d^2x}{dt^2} = -7.84x\) | E1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(8.4^2 = 7.84(A^2 - 1.25^2)\) | M2 | Using \(v^2 = \omega^2(A^2 - x^2)\) |
| Amplitude is 3.25 m | A1, A1 | |
| OR | ||
| \(\frac{980}{2\times25}y^2 = 5\times9.8y + \frac{1}{2}\times5\times8.4^2\) | M2 | Equation involving EE, PE and KE |
| \(y = 4.5\); Amplitude is \(4.5 - 1.25 = 3.25\) m | A1, A1 | |
| OR \(x = A\sin2.8t + B\cos2.8t\), \(x=-1.25\), \(v=8.4\) when \(t=0\) | M2 | Obtaining \(A\) and \(B\), both correct |
| \(\Rightarrow A=3\), \(B=-1.25\); Amplitude is \(\sqrt{A^2+B^2} = 3.25\) | A1, A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum speed is \(A\omega = 3.25\times2.8 = 9.1\ \text{ms}^{-1}\) | M1 | or equation involving EE, PE and KE |
| A1 | ft only if answer is greater than 8.4 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 3.25\cos2.8t\) | B1 (ft) | or \(x=3.25\sin2.8t\); or \(v=9.1\cos2.8t\); or \(v=9.1\sin2.8t\); or \(x=3.25\sin(2.8t+\varepsilon)\) etc; or \(x=\pm3\sin2.8t\pm1.25\cos2.8t\) |
| \(-1.25 = 3.25\cos2.8t\) | M1 | Obtaining equation for \(t\) or \(\varepsilon\) by setting \(x=(\pm)1.25\) or \(v=(\pm)8.4\) or solving \(\pm3\sin2.8t\pm1.25\cos2.8t = 3.25\) |
| M1 | Strategy for finding the required time, e.g. \(\frac{1}{2.8}\sin^{-1}\frac{1.25}{3.25}+\frac{1}{4}\times\frac{2\pi}{2.8}\) | |
| Time is 0.702 s | A1 cao | \(2.8t - 0.3948 = \frac{1}{2}\pi\) or \(2.8t - 1.966 = 0\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. Rope is light; Rock is a particle; No air resistance/friction/external forces; Rope obeys Hooke's Law/Perfectly elastic/Within elastic limit/No energy loss in rope | B1B1B1 | Three modelling assumptions |
| Total: 3 |
# Question 3:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{980}{25}y = 5\times9.8$ | M1 | Using $\frac{\lambda y}{l_0}$ (Allow M1 for $980y = mg$) |
| Extension is 1.25 m | A1 | |
| **Total: 2** | | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = \frac{980}{25}(1.25 + x)$ | B1 (ft) | *(ft) indicates ft from previous parts as for A marks* |
| $5\times9.8 - 39.2(1.25+x) = 5\frac{d^2x}{dt^2}$ | M1 | Equation of motion with three terms |
| $-39.2x = 5\frac{d^2x}{dt^2}$ | F1 | Must have $\ddot{x}$ in terms of $x$ only |
| $\frac{d^2x}{dt^2} = -7.84x$ | E1 | |
| **Total: 4** | | |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8.4^2 = 7.84(A^2 - 1.25^2)$ | M2 | Using $v^2 = \omega^2(A^2 - x^2)$ |
| Amplitude is 3.25 m | A1, A1 | |
| **OR** | | |
| $\frac{980}{2\times25}y^2 = 5\times9.8y + \frac{1}{2}\times5\times8.4^2$ | M2 | Equation involving EE, PE and KE |
| $y = 4.5$; Amplitude is $4.5 - 1.25 = 3.25$ m | A1, A1 | |
| **OR** $x = A\sin2.8t + B\cos2.8t$, $x=-1.25$, $v=8.4$ when $t=0$ | M2 | Obtaining $A$ and $B$, both correct |
| $\Rightarrow A=3$, $B=-1.25$; Amplitude is $\sqrt{A^2+B^2} = 3.25$ | A1, A1 | |
| **Total: 4** | | |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum speed is $A\omega = 3.25\times2.8 = 9.1\ \text{ms}^{-1}$ | M1 | or equation involving EE, PE and KE |
| | A1 | ft only if answer is greater than 8.4 |
| **Total: 2** | | |
## Part (v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 3.25\cos2.8t$ | B1 (ft) | or $x=3.25\sin2.8t$; or $v=9.1\cos2.8t$; or $v=9.1\sin2.8t$; or $x=3.25\sin(2.8t+\varepsilon)$ etc; or $x=\pm3\sin2.8t\pm1.25\cos2.8t$ |
| $-1.25 = 3.25\cos2.8t$ | M1 | Obtaining equation for $t$ or $\varepsilon$ by setting $x=(\pm)1.25$ or $v=(\pm)8.4$ or solving $\pm3\sin2.8t\pm1.25\cos2.8t = 3.25$ |
| | M1 | Strategy for finding the required time, e.g. $\frac{1}{2.8}\sin^{-1}\frac{1.25}{3.25}+\frac{1}{4}\times\frac{2\pi}{2.8}$ |
| Time is 0.702 s | A1 cao | $2.8t - 0.3948 = \frac{1}{2}\pi$ or $2.8t - 1.966 = 0$ |
| **Total: 4** | | |
## Part (vi):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. Rope is light; Rock is a particle; No air resistance/friction/external forces; Rope obeys Hooke's Law/Perfectly elastic/Within elastic limit/No energy loss in rope | B1B1B1 | Three modelling assumptions |
| **Total: 3** | | |
---3 An elastic rope has natural length 25 m and modulus of elasticity 980 N . One end of the rope is attached to a fixed point O , and a rock of mass 5 kg is attached to the other end; the rock is always vertically below O.\\
(i) Find the extension of the rope when the rock is hanging in equilibrium.
When the rock is moving with the rope stretched, its displacement is $x$ metres below the equilibrium position at time $t$ seconds.\\
(ii) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 7.84 x$.
The rock is released from a position where the rope is slack, and when the rope just becomes taut the speed of the rock is $8.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Find the distance below the equilibrium position at which the rock first comes instantaneously to rest.\\
(iv) Find the maximum speed of the rock.\\
(v) Find the time between the rope becoming taut and the rock first coming to rest.\\
(vi) State three modelling assumptions you have made in answering this question.
\hfill \mbox{\textit{OCR MEI M3 2009 Q3 [19]}}