| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.8 This M4 question requires setting up forces from elastic string tension (using Hooke's law with geometry), air resistance, and deriving a second-order differential equation with damping. Part (b) involves solving the characteristic equation and applying initial conditions. While systematic, it demands careful geometric analysis of string extensions, vector resolution, and fluency with damped harmonic motion—significantly above average difficulty but follows established M4 patterns. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| HL \(T_1 = \frac{2mk^2L(0.5L + x)}{L}\) | M1 (either) | |
| HL \(T_2 = \frac{2mk^2L(0.5L - x)}{L}\) | A1 (both) | |
| N2L \(T_2 - T_1 - 2mk\frac{dx}{dt} = m\frac{d^2x}{dt^2}\) | M1, A1, A1 | |
| \(4mk^2x - 2mk\frac{dx}{dt} = m\frac{d^2x}{dt^2}\) | ||
| \(\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 4k^2x = 0\) (with *) | A1 (cao) | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m^2 + 2km + 4m^2 = 0\) | M1 (ae) | |
| \(m = -k \pm k\sqrt{3}i\) | M1 | |
| \(x = e^{-kt}(A\cos\sqrt{3}kt + B\sin\sqrt{3}kt)\) | A1 (oe) | |
| \(t = 0, x = \frac{L}{2} \Rightarrow A = \frac{L}{2}\) | B1 | |
| \(\dot{x} = -ke^{-kt}(A\cos\sqrt{3}kt + B\sin\sqrt{3}kt) + \sqrt{3}ke^{-kt}(-A\sin\sqrt{3}kt + B\cos\sqrt{3}kt)\) | M1 | |
| \(t = 0, \dot{x} = 0 \Rightarrow 0 = -kA + \sqrt{3}kB\) | M1 | |
| \(B = \frac{1}{\sqrt{3}}A = \frac{L}{2\sqrt{3}}\) | A1 | |
| \(AP = 1.5L + \frac{L}{2\sqrt{3}}e^{-kt}(\sqrt{3}\cos\sqrt{3}kt + \sin\sqrt{3}kt)\) | A1 (oe) | (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = Ae^{-kt}\cos(\sqrt{3}kt - \varepsilon)\) | M1, M1, A1 | |
| \(t = 0, x = \frac{L}{2} \Rightarrow \frac{L}{2} = A\cos(-\varepsilon) = A\cos\varepsilon\) | B1 | |
| \(\dot{x} = -kAe^{-kt}\cos(\sqrt{3}kt - \varepsilon) - \sqrt{3}kAe^{-kt}\sin(\sqrt{3}kt - \varepsilon)\) | M1 | |
| \(t = 0, \dot{x} = 0 \Rightarrow 0 = -kA\cos\varepsilon - \sqrt{3}kA\sin(-\varepsilon)\) | M1 | |
| Leading to \(\tan\varepsilon = \frac{1}{\sqrt{3}} \Rightarrow \varepsilon = \frac{\pi}{6}\) and \(A = \frac{L}{\sqrt{3}}\) (both) | A1 | |
| \(AP = 1.5L + \frac{L}{\sqrt{3}}e^{-kt}\cos(\sqrt{3}kt - \frac{\pi}{6})\) | A1 | (8 marks) |
## Part (a)
**HL** $T_1 = \frac{2mk^2L(0.5L + x)}{L}$ | M1 (either) |
**HL** $T_2 = \frac{2mk^2L(0.5L - x)}{L}$ | A1 (both) |
**N2L** $T_2 - T_1 - 2mk\frac{dx}{dt} = m\frac{d^2x}{dt^2}$ | M1, A1, A1 |
$4mk^2x - 2mk\frac{dx}{dt} = m\frac{d^2x}{dt^2}$ | |
$\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 4k^2x = 0$ (with *) | A1 (cao) | (6 marks)
## Part (b)
$m^2 + 2km + 4m^2 = 0$ | M1 (ae) |
$m = -k \pm k\sqrt{3}i$ | M1 |
$x = e^{-kt}(A\cos\sqrt{3}kt + B\sin\sqrt{3}kt)$ | A1 (oe) |
$t = 0, x = \frac{L}{2} \Rightarrow A = \frac{L}{2}$ | B1 |
$\dot{x} = -ke^{-kt}(A\cos\sqrt{3}kt + B\sin\sqrt{3}kt) + \sqrt{3}ke^{-kt}(-A\sin\sqrt{3}kt + B\cos\sqrt{3}kt)$ | M1 |
$t = 0, \dot{x} = 0 \Rightarrow 0 = -kA + \sqrt{3}kB$ | M1 |
$B = \frac{1}{\sqrt{3}}A = \frac{L}{2\sqrt{3}}$ | A1 |
$AP = 1.5L + \frac{L}{2\sqrt{3}}e^{-kt}(\sqrt{3}\cos\sqrt{3}kt + \sin\sqrt{3}kt)$ | A1 (oe) | (8 marks)
**Total: 14 marks**
---
# Question 4(b) - Alternative form of General Solution:
$x = Ae^{-kt}\cos(\sqrt{3}kt - \varepsilon)$ | M1, M1, A1 |
$t = 0, x = \frac{L}{2} \Rightarrow \frac{L}{2} = A\cos(-\varepsilon) = A\cos\varepsilon$ | B1 |
$\dot{x} = -kAe^{-kt}\cos(\sqrt{3}kt - \varepsilon) - \sqrt{3}kAe^{-kt}\sin(\sqrt{3}kt - \varepsilon)$ | M1 |
$t = 0, \dot{x} = 0 \Rightarrow 0 = -kA\cos\varepsilon - \sqrt{3}kA\sin(-\varepsilon)$ | M1 |
Leading to $\tan\varepsilon = \frac{1}{\sqrt{3}} \Rightarrow \varepsilon = \frac{\pi}{6}$ and $A = \frac{L}{\sqrt{3}}$ (both) | A1 |
$AP = 1.5L + \frac{L}{\sqrt{3}}e^{-kt}\cos(\sqrt{3}kt - \frac{\pi}{6})$ | A1 | (8 marks)
**Note:** Another possible trig form is $\sin(\sqrt{3}kt + \frac{\pi}{3})$
---A particle $P$ of mass $m$ is attached to the mid-point of a light elastic string, of natural length $2L$ and modulus of elasticity $2mk^2L$, where $k$ is a positive constant. The ends of the string are attached to points $A$ and $B$ on a smooth horizontal surface, where $AB = 3L$. The particle is released from rest at the point $C$, where $AC = 2L$ and $ACB$ is a straight line. During the subsequent motion $P$ experiences air resistance of magnitude $2mkv$, where $v$ is the speed of $P$. At time $t$, $AP = 1.5L + x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 4k^2x = 0$.
[6]
\item Find an expression, in terms of $t$, $k$ and $L$, for the distance $AP$ at time $t$.
[8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q4 [14]}}