| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a standard M4 damped harmonic motion question requiring force equation setup, solving a second-order linear ODE with given initial conditions, and interpretation. While it involves multiple steps (deriving the differential equation, finding auxiliary equation roots, applying initial conditions), these are well-practiced techniques at this level. The critical damping case (λ = -1, -2) makes it slightly more involved than simple harmonic motion but remains a textbook application of standard methods. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 2mL \times \frac{x}{L}\) | B1 | |
| Equation of motion: \(-3m\dot{x} - T = m\ddot{x}\) | M1A1 | |
| \(\Rightarrow \ddot{x} + 3\dot{x} + 2x = 0\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| A.E. \(m^2 + 3m + 2 = 0\) \(\Rightarrow\) \(m = -1\) or \(-2\) | M1A1 | |
| G.S. \(x = Ae^{-t} + Be^{-2t}\) | A1√ | |
| \(t = 0, x = 2\): \(\Rightarrow\) \(A + B = 2\) | B1 | |
| Differentiating: \(\dot{x} = -Ae^{-t} - 2Be^{-2t}\) | M1 | |
| \(t = 0, \dot{x} = -4\): \(\Rightarrow\) \(A + 2B = 4\) (any equivalent form) | A1 | |
| Correctly solving simultaneous equations (\(A = 0\), \(B = 2\)) | M1 | |
| \(x = 2e^{-2t}\) | A1 | (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Shape: (0,2), \(x = 0\) asymptote | B1√, B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| No, with reason, e.g. \(P\) always moving | B1 | (1 mark) |
## Part (a)
$T = 2mL \times \frac{x}{L}$ | B1
Equation of motion: $-3m\dot{x} - T = m\ddot{x}$ | M1A1
$\Rightarrow \ddot{x} + 3\dot{x} + 2x = 0$ | A1 | (4 marks)
## Part (b)
A.E. $m^2 + 3m + 2 = 0$ $\Rightarrow$ $m = -1$ or $-2$ | M1A1
G.S. $x = Ae^{-t} + Be^{-2t}$ | A1√
$t = 0, x = 2$: $\Rightarrow$ $A + B = 2$ | B1
Differentiating: $\dot{x} = -Ae^{-t} - 2Be^{-2t}$ | M1
$t = 0, \dot{x} = -4$: $\Rightarrow$ $A + 2B = 4$ (any equivalent form) | A1
Correctly solving simultaneous equations ($A = 0$, $B = 2$) | M1
$x = 2e^{-2t}$ | A1 | (8 marks)
## Part (c)
Shape: (0,2), $x = 0$ asymptote | B1√, B1 | (2 marks)
## Part (d)
No, with reason, e.g. $P$ always moving | B1 | (1 mark)
---\includegraphics{figure_2}
In a simple model of a shock absorber, a particle $P$ of mass $m$ kg is attached to one end of a light elastic horizontal spring. The other end of the spring is fixed at $A$ and the motion of $P$ takes place along a fixed horizontal line through $A$. The spring has natural length $L$ metres and modulus of elasticity $2mL$ newtons. The whole system is immersed in a fluid which exerts a resistance on $P$ of magnitude $3mv$ newtons, where $v$ m s$^{-1}$ is the speed of $P$ at time $t$ seconds. The compression of the spring at time $t$ seconds is $x$ metres, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{\text{d}^2 x}{\text{d}t^2} + 3\frac{\text{d}x}{\text{d}t} + 2x = 0.$$ [4]
\end{enumerate}
Given that when $t = 0$, $x = 2$ and $\frac{\text{d}x}{\text{d}t} = -4$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find $x$ in terms of $t$. [8]
\item Sketch the graph of $x$ against $t$. [2]
\item State, with a reason, whether the model is realistic. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q6 [15]}}