| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a mechanics-based second order differential equation requiring force analysis to derive the equation, then solving with damping and initial conditions. While the derivation is structured and the DE solution follows standard methods (complementary function + particular integral), it requires multiple sophisticated steps: setting up forces correctly, handling the damped oscillator, and applying two initial conditions (x=a and v=0 at t=0). This is significantly harder than routine DE questions but not exceptionally difficult for Further Maths students familiar with mechanics applications. |
| Spec | 4.10g Damped oscillations: model and interpret6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg - T - 2mn\dot{x} = m\ddot{x}\) | M1 A1 A1 | |
| \(mg - \dfrac{2man^2x}{a} - 2mn\dot{x} = m\ddot{x}\) | M1 | |
| \(\ddot{x} + 2n\dot{x} + 2n^2x = g\) (*) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| AE: \(u^2 + 2nu + 2n^2 = 0\) | M1 | |
| \((u+n)^2 = -n^2\) | ||
| \(u = -n \pm ni\) | A1 | |
| CF: \(x = e^{-nt}(A\cos nt + B\sin nt)\), PI: \(x = \dfrac{g}{2n^2}\) | M1 | |
| GS: \(x = e^{-nt}(A\cos nt + B\sin nt) + \dfrac{g}{2n^2}\) | A1 | |
| \(t=0, x=a, \dot{x}=0\): \(A = a - \dfrac{g}{2n^2}\) | M1 | |
| \(\dot{x} = e^{-nt}(-An\sin nt + Bn\cos nt) - ne^{-nt}(A\cos nt + B\sin nt)\) | M1 | |
| \(x = e^{-nt}\!\left(a - \dfrac{g}{2n^2}\right)(\cos nt + \sin nt) + \dfrac{g}{2n^2}\) | A1 | (7), (12) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg - T - 2mn\dot{x} = m\ddot{x}$ | M1 A1 A1 | |
| $mg - \dfrac{2man^2x}{a} - 2mn\dot{x} = m\ddot{x}$ | M1 | |
| $\ddot{x} + 2n\dot{x} + 2n^2x = g$ (*) | A1 | **(5)** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| AE: $u^2 + 2nu + 2n^2 = 0$ | M1 | |
| $(u+n)^2 = -n^2$ | | |
| $u = -n \pm ni$ | A1 | |
| CF: $x = e^{-nt}(A\cos nt + B\sin nt)$, PI: $x = \dfrac{g}{2n^2}$ | M1 | |
| GS: $x = e^{-nt}(A\cos nt + B\sin nt) + \dfrac{g}{2n^2}$ | A1 | |
| $t=0, x=a, \dot{x}=0$: $A = a - \dfrac{g}{2n^2}$ | M1 | |
| $\dot{x} = e^{-nt}(-An\sin nt + Bn\cos nt) - ne^{-nt}(A\cos nt + B\sin nt)$ | M1 | |
| $x = e^{-nt}\!\left(a - \dfrac{g}{2n^2}\right)(\cos nt + \sin nt) + \dfrac{g}{2n^2}$ | A1 | **(7), (12)** |
---5. A particle $P$ of mass $m$ is fixed to one end of a light elastic string, of natural length $a$ and modulus of elasticity $2 m a n ^ { 2 }$. The other end of the string is attached to a fixed point $O$. The particle $P$ is released from rest at a point which is a distance $2 a$ vertically below $O$. The air resistance is modelled as having magnitude $2 m n v$, where $v$ is the speed of $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that, when the extension of the string is $x$,
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 n \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 n ^ { 2 } x = g$$
\item Find $x$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q5 [12]}}