| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a mechanics-based second order DE problem requiring careful setup of geometry/forces, deriving the DE from Hooke's law and Newton's second law, then solving using complementary function + particular integral. The geometric relationship in part (a) requires clear spatial reasoning, and the multi-step nature with physical interpretation elevates it above routine DE questions, though the mathematical techniques themselves are standard for M4. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Natural length = 0.6 m, so equilibrium extension: \(\frac{2.7 \times e_0}{0.6} = 0.5 \times 9.8\), giving \(e_0 = \frac{49}{45}\) (or equivalent) | M1 | Finding equilibrium extension using Hooke's Law and weight |
| Displacement of B downward = \(4\sin 2t\), extension \(x = y - 4\sin 2t + \frac{49}{45}\)... rearranged to give \(y + \frac{49}{45} = x + 4\sin 2t\) | A1 A1 | Correct geometric/kinematic relationship; correct equation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Newton's second law: \(0.5\ddot{y} = 0.5g - \frac{2.7x}{0.6}\) | M1 | Applying N2L with weight and tension |
| \(= 0.5g - \frac{9x}{2}\) | A1 | Correct tension term |
| From part (a): \(x = y + \frac{49}{45} - 4\sin 2t\) | M1 | Substituting for \(x\) |
| \(0.5\ddot{y} = 0.5g - \frac{9}{2}(y + \frac{49}{45} - 4\sin 2t)\) | M1 | Substituting and expanding |
| Using \(0.5g = \frac{9}{2} \times \frac{49}{45}\), terms cancel to give \(\frac{d^2y}{dt^2} + 9y = 36\sin 2t\) | A1 | Correct cancellation and final result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Auxiliary equation: \(m^2 + 9 = 0 \Rightarrow m = \pm 3i\) | M1 | Correct auxiliary equation solved |
| Complementary function: \(y = A\cos 3t + B\sin 3t\) | A1 | Correct CF |
| General solution: \(y = A\cos 3t + B\sin 3t + \frac{36}{5}\sin 2t\) | A1 | Adding given PI |
| At \(t=0\): \(y=0\) (P starts at C, displacement = 0) \(\Rightarrow A = 0\) | M1 | Applying initial condition \(y(0)=0\) |
| \(\dot{y} = -3A\sin 3t + 3B\cos 3t + \frac{72}{5}\cos 2t\); at \(t=0\), \(\dot{y}=0 \Rightarrow 3B + \frac{72}{5} = 0 \Rightarrow B = -\frac{24}{5}\) | A1 | Applying \(\dot{y}(0)=0\) and correct \(B\) |
| \(y = -\frac{24}{5}\sin 3t + \frac{36}{5}\sin 2t\) | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\dot{y} = -\frac{72}{5}\cos 3t + \frac{72}{5}\cos 2t\) | M1 | Differentiating \(y\) |
| At \(t = \frac{\pi}{3}\): \(\cos\pi = -1\), \(\cos\frac{2\pi}{3} = -\frac{1}{2}\) | M1 | Substituting \(t = \frac{\pi}{3}\) |
| \(\dot{y} = -\frac{72}{5}(-1) + \frac{72}{5}(-\frac{1}{2}) = \frac{72}{5} - \frac{36}{5} = \frac{36}{5}\) | A1 | Correct value |
| Speed \(= \frac{36}{5} = 7.2\) m/s | A1 | Final answer |
# Question 7:
## Part (a): Show that $y + \frac{49}{45} = x + 4\sin 2t$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Natural length = 0.6 m, so equilibrium extension: $\frac{2.7 \times e_0}{0.6} = 0.5 \times 9.8$, giving $e_0 = \frac{49}{45}$ (or equivalent) | M1 | Finding equilibrium extension using Hooke's Law and weight |
| Displacement of B downward = $4\sin 2t$, extension $x = y - 4\sin 2t + \frac{49}{45}$... rearranged to give $y + \frac{49}{45} = x + 4\sin 2t$ | A1 A1 | Correct geometric/kinematic relationship; correct equation shown |
## Part (b): Show that $\frac{d^2y}{dt^2} + 9y = 36\sin 2t$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Newton's second law: $0.5\ddot{y} = 0.5g - \frac{2.7x}{0.6}$ | M1 | Applying N2L with weight and tension |
| $= 0.5g - \frac{9x}{2}$ | A1 | Correct tension term |
| From part (a): $x = y + \frac{49}{45} - 4\sin 2t$ | M1 | Substituting for $x$ |
| $0.5\ddot{y} = 0.5g - \frac{9}{2}(y + \frac{49}{45} - 4\sin 2t)$ | M1 | Substituting and expanding |
| Using $0.5g = \frac{9}{2} \times \frac{49}{45}$, terms cancel to give $\frac{d^2y}{dt^2} + 9y = 36\sin 2t$ | A1 | Correct cancellation and final result shown |
## Part (c): Find $y$ in terms of $t$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Auxiliary equation: $m^2 + 9 = 0 \Rightarrow m = \pm 3i$ | M1 | Correct auxiliary equation solved |
| Complementary function: $y = A\cos 3t + B\sin 3t$ | A1 | Correct CF |
| General solution: $y = A\cos 3t + B\sin 3t + \frac{36}{5}\sin 2t$ | A1 | Adding given PI |
| At $t=0$: $y=0$ (P starts at C, displacement = 0) $\Rightarrow A = 0$ | M1 | Applying initial condition $y(0)=0$ |
| $\dot{y} = -3A\sin 3t + 3B\cos 3t + \frac{72}{5}\cos 2t$; at $t=0$, $\dot{y}=0 \Rightarrow 3B + \frac{72}{5} = 0 \Rightarrow B = -\frac{24}{5}$ | A1 | Applying $\dot{y}(0)=0$ and correct $B$ |
| $y = -\frac{24}{5}\sin 3t + \frac{36}{5}\sin 2t$ | | Final answer |
## Part (d): Find speed of $P$ when $t = \frac{1}{3}\pi$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dot{y} = -\frac{72}{5}\cos 3t + \frac{72}{5}\cos 2t$ | M1 | Differentiating $y$ |
| At $t = \frac{\pi}{3}$: $\cos\pi = -1$, $\cos\frac{2\pi}{3} = -\frac{1}{2}$ | M1 | Substituting $t = \frac{\pi}{3}$ |
| $\dot{y} = -\frac{72}{5}(-1) + \frac{72}{5}(-\frac{1}{2}) = \frac{72}{5} - \frac{36}{5} = \frac{36}{5}$ | A1 | Correct value |
| Speed $= \frac{36}{5} = 7.2$ m/s | A1 | Final answer |7. A particle $P$ of mass 0.5 kg is attached to the end $A$ of a light elastic spring $A B$, of natural length 0.6 m and modulus of elasticity 2.7 N . At time $t = 0$ the end $B$ of the spring is held at rest and $P$ hangs at rest at the point $C$ which is vertically below $B$. The end $B$ is then moved along the line of the spring so that, at time $t$ seconds, the downwards displacement of $B$ from its initial position is $4 \sin 2 t$ metres. At time $t$ seconds, the extension of the spring is $x$ metres and the displacement of $P$ below $C$ is $y$ metres.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y + \frac { 49 } { 45 } = x + 4 \sin 2 t$$
\item Hence show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 36 \sin 2 t$$
Given that $y = \frac { 36 } { 5 } \sin 2 t$ is a particular integral of this differential equation,
\item find $y$ in terms of $t$,
\item find the speed of $P$ when $t = \frac { 1 } { 3 } \pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2013 Q7 [17]}}