| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.2 This is a structured mechanics problem requiring equilibrium analysis, Newton's second law with damping, and solving a second-order linear ODE with constant coefficients. While it involves multiple steps and careful bookkeeping of forces in an elastic string system, each part is clearly signposted and uses standard A-level techniques (Hooke's law, F=ma, auxiliary equation method). The critical damping case makes part (c) straightforward once the differential equation is established. |
| Spec | 4.10d 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_2 = T_1 + mg\) | M1 | |
| \(\frac{mge}{a} = \frac{mg}{a}(2a-e) + mg\) | M1 A1 | |
| \(e = \frac{3a}{2} \Rightarrow AE = \frac{5a}{2}\) | A1 A1 cso | (5) (*) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg + \frac{mg}{a}\left(\frac{1}{2}a - x\right) - \frac{mg}{a}\left(\frac{3}{2}a + x\right) - 2m\sqrt{\frac{g}{a}}\frac{dx}{dt} = m\frac{d^2x}{dt^2}\) | M1 A3 | (−1 each error or omission) |
| \(\Rightarrow \frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 2k^2x = 0\) | A1 | (5) (*) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| AE: \(m^2 + 2km + 2k^2 = 0\) | M1 | |
| \(m = -k \pm ki\) | A1 | |
| GS: \(x = e^{-kt}(A\cos kt + B\sin kt)\) | A1 ft | |
| \(t=0,\ x = \frac{1}{2}a \Rightarrow A = \frac{1}{2}a\) | B1 | |
| \(\frac{dx}{dt} = ke^{-kt}(A\cos kt + B\sin kt) + e^{-kt}(-kA\sin kt + kB\cos kt)\) | M1 | |
| \(t=0,\ \frac{dx}{dt} = 0 \Rightarrow -kA + kB = 0 \Rightarrow B = A = \frac{1}{2}a\) | M1 | |
| \(x = \frac{1}{2}a\,e^{-kt}(\cos kt + \sin kt)\) | A1 | (7) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_2 = T_1 + mg$ | M1 | |
| $\frac{mge}{a} = \frac{mg}{a}(2a-e) + mg$ | M1 A1 | |
| $e = \frac{3a}{2} \Rightarrow AE = \frac{5a}{2}$ | A1 A1 cso | (5) (*) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg + \frac{mg}{a}\left(\frac{1}{2}a - x\right) - \frac{mg}{a}\left(\frac{3}{2}a + x\right) - 2m\sqrt{\frac{g}{a}}\frac{dx}{dt} = m\frac{d^2x}{dt^2}$ | M1 A3 | (−1 each error or omission) |
| $\Rightarrow \frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 2k^2x = 0$ | A1 | (5) (*) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| AE: $m^2 + 2km + 2k^2 = 0$ | M1 | |
| $m = -k \pm ki$ | A1 | |
| GS: $x = e^{-kt}(A\cos kt + B\sin kt)$ | A1 ft | |
| $t=0,\ x = \frac{1}{2}a \Rightarrow A = \frac{1}{2}a$ | B1 | |
| $\frac{dx}{dt} = ke^{-kt}(A\cos kt + B\sin kt) + e^{-kt}(-kA\sin kt + kB\cos kt)$ | M1 | |
| $t=0,\ \frac{dx}{dt} = 0 \Rightarrow -kA + kB = 0 \Rightarrow B = A = \frac{1}{2}a$ | M1 | |
| $x = \frac{1}{2}a\,e^{-kt}(\cos kt + \sin kt)$ | A1 | (7) |
---5. A light elastic string, of natural length $2 a$ and modulus of elasticity $m g$, has a particle $P$ of mass $m$ attached to its mid-point. One end of the string is attached to a fixed point $A$ and the other end is attached to a fixed point $B$ which is at a distance $4 a$ vertically below $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that $P$ hangs in equilibrium at the point $E$ where $A E = \frac { 5 } { 2 } a$.
The particle $P$ is held at a distance $3 a$ vertically below $A$ and is released from rest at time $t = 0$. When the speed of the particle is $v$, there is a resistance to motion of magnitude $2 m k v$, where $k = \sqrt { } \left( \frac { g } { a } \right)$. At time $t$ the particle is at a distance $\left( \frac { 5 } { 2 } a + x \right)$ from $A$.
\item Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = 0$$
\item Hence find $x$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2003 Q5 [17]}}