| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Standard +0.8 This is a multi-part SHM question requiring equilibrium analysis with multiple elastic ropes, deriving the SHM equation from first principles using Hooke's law, and applying standard SHM formulas. The setup with five ropes adds complexity to parts (i)-(iii), but the SHM application in (iv)-(vi) is routine once the equation is established. The derivation and multi-step nature elevate it 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/l |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Tension is \(180(10-7)\) | M1 | Using \(T = k \times \text{extension}\) |
| \(= 540\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4\times540 = T + 200\times9.8\) | M1 | Resolving vertically |
| \(T = 200\) | A1 | |
| Extension is \(\frac{200}{80}\) \((=2.5)\) | M1 | |
| Natural length is 5.5 m | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(80(2.5-x) + 200\times9.8 - 4\times180(3+x) = 200\frac{d^2x}{dt^2}\) | B1 ft, M1, A1 | For \(180(3+x)\) or \(80(2.5-x)\); Equation of motion (condone one missing force) |
| \(200 - 80x + 1960 - 2160 - 720x = 200\frac{d^2x}{dt^2}\) | ||
| \(\frac{d^2x}{dt^2} = -4x\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Maximum acceleration is \(\omega^2 A\) | M1 | |
| \(= 4\times2.2 = 8.8 \text{ ms}^{-2}\) | A1 | Condone \(-8.8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x=-1.6\), \(v^2 = \omega^2(A^2-x^2)\) | Using \(v^2 = \omega^2(A^2-x^2)\) (or other complete method) | |
| \(= 4(2.2^2 - 1.6^2)\) | M1 | Allow M1 if \(\omega^2=2\) or 16 used but M0 if \(x=3.8\) is used |
| Speed is \(3.02 \text{ ms}^{-1}\) (3 sf) | A1 | Condone \(-3.02\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 2.2\cos2t\) | B1 | Condone \(x=-2.2\cos2t\); This B1 can be earned in (v) |
| When \(t=5\), \(x=-1.846\) | M1 | Obtaining \(x\) when \(t=5\) (from \(x=A\cos\omega t\) or \(x=A\sin\omega t\)) |
| Period is \(\frac{2\pi}{\omega}=\pi\), \(5\text{ s}\) is \(\frac{5}{\pi}\approx1.6\) periods | ||
| Distance travelled is \(6\times2.2 + (2.2-1.846)\) | M1 | Correct strategy for finding distance |
| \(= 13.6\) m (3 sf) | A1 |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Tension is $180(10-7)$ | M1 | Using $T = k \times \text{extension}$ |
| $= 540$ N | A1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $4\times540 = T + 200\times9.8$ | M1 | Resolving vertically |
| $T = 200$ | A1 | |
| Extension is $\frac{200}{80}$ $(=2.5)$ | M1 | |
| Natural length is 5.5 m | A1 cao | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $80(2.5-x) + 200\times9.8 - 4\times180(3+x) = 200\frac{d^2x}{dt^2}$ | B1 ft, M1, A1 | For $180(3+x)$ or $80(2.5-x)$; Equation of motion (condone one missing force) |
| $200 - 80x + 1960 - 2160 - 720x = 200\frac{d^2x}{dt^2}$ | | |
| $\frac{d^2x}{dt^2} = -4x$ | E1 | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Maximum acceleration is $\omega^2 A$ | M1 | |
| $= 4\times2.2 = 8.8 \text{ ms}^{-2}$ | A1 | Condone $-8.8$ |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x=-1.6$, $v^2 = \omega^2(A^2-x^2)$ | | Using $v^2 = \omega^2(A^2-x^2)$ (or other complete method) |
| $= 4(2.2^2 - 1.6^2)$ | M1 | Allow M1 if $\omega^2=2$ or 16 used but M0 if $x=3.8$ is used |
| Speed is $3.02 \text{ ms}^{-1}$ (3 sf) | A1 | Condone $-3.02$ |
## Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 2.2\cos2t$ | B1 | Condone $x=-2.2\cos2t$; This B1 can be earned in (v) |
| When $t=5$, $x=-1.846$ | M1 | Obtaining $x$ when $t=5$ (from $x=A\cos\omega t$ or $x=A\sin\omega t$) |
| Period is $\frac{2\pi}{\omega}=\pi$, $5\text{ s}$ is $\frac{5}{\pi}\approx1.6$ periods | | |
| Distance travelled is $6\times2.2 + (2.2-1.846)$ | M1 | Correct strategy for finding distance |
| $= 13.6$ m (3 sf) | A1 | |
---3 A block of mass 200 kg is connected to a horizontal ceiling by four identical light elastic ropes, each having natural length 7 m and stiffness $180 \mathrm {~N} \mathrm {~m} ^ { - 1 }$. It is also connected to the floor by a single light elastic rope having stiffness $80 \mathrm { Nm } ^ { - 1 }$. Throughout this question you may assume that all five ropes are stretched and vertical, and you may neglect air resistance.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2dd5719-bef3-45f2-afd2-c481e6a4b129-4_665_623_482_760}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Fig. 3 shows the block resting in equilibrium, with each of the top ropes having length 10 m and the bottom rope having length 8 m .\\
(i) Find the tension in one of the top ropes.\\
(ii) Find the natural length of the bottom rope.
The block now moves vertically up and down. At time $t$ seconds, the block is $x$ metres below its equilibrium position.\\
(iii) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 4 x$.
The motion is started by pulling the block down 2.2 m below its equilibrium position and releasing it from rest. The block then executes simple harmonic motion with amplitude 2.2 m .\\
(iv) Find the maximum magnitude of the acceleration of the block.\\
(v) Find the speed of the block when it has travelled 3.8 m from its starting point.\\
(vi) Find the distance travelled by the block in the first 5 s .
\hfill \mbox{\textit{OCR MEI M3 2011 Q3 [18]}}