| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.3 This is a mechanics question requiring setup of a differential equation from Hooke's law with a moving boundary condition, then solving a second-order ODE with particular integral. While it involves multiple steps (finding tensions in both spring sections, applying F=ma, solving the ODE, applying initial conditions), each step follows standard M4 procedures without requiring novel insight. The algebra is moderately involved but the problem structure is typical for this module. |
| Spec | 4.10e 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 |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T_1 = 4(1+x);\; T_2 = 4(1+\tfrac{1}{2}\sin 4t - x)\) | B1; B1 | |
| \(T_2 - T_1 = 2\ddot{x}\) | M1 | |
| \(2\ddot{x} = 4(1+\tfrac{1}{2}\sin 4t - x)-4(1+x)\) | A1 | |
| \(\Rightarrow \ddot{x}+4x = \sin 4t\) (*) | A1 | (5) Given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| CF: \(x = A\sin 2t + B\cos 2t\) | B1 | |
| PI: \(x = P\sin 4t\) | M1 | |
| \(-16P\sin 4t + 4P\sin 4t = \sin 4t\) | M1 | |
| \(\Rightarrow P = -\tfrac{1}{12}\) | A1 | |
| \(x = A\sin 2t + B\cos 2t - \tfrac{1}{12}\sin 4t\) | M1 | |
| \(t=0, x=0 \Rightarrow B=0\) | A1 | |
| \(\dot{x} = 2A\cos 2t - \tfrac{1}{3}\cos 4t\) | M1 | |
| \(t=0, \dot{x}=0 \Rightarrow A=\tfrac{1}{6}\) | A1 ft | (8) |
| \(\dot{x}=0 \Rightarrow \tfrac{1}{3}\cos 2t - \tfrac{1}{3}\cos 4t = 0\) | M1 | |
| \(\Rightarrow \cos 4t = \cos 2t\) | ||
| \(\Rightarrow 4t = 2t+2\pi \text{ or } 2\pi-2t\) | M1 | |
| \(\Rightarrow t=\pi \text{ or } \tfrac{\pi}{3}\) | A1 | |
| First at rest after \(t=0\) when \(t=\dfrac{\pi}{3}\) | A1 cso | (4) |
# Question 6:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T_1 = 4(1+x);\; T_2 = 4(1+\tfrac{1}{2}\sin 4t - x)$ | B1; B1 | |
| $T_2 - T_1 = 2\ddot{x}$ | M1 | |
| $2\ddot{x} = 4(1+\tfrac{1}{2}\sin 4t - x)-4(1+x)$ | A1 | |
| $\Rightarrow \ddot{x}+4x = \sin 4t$ (*) | A1 | (5) Given answer |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| CF: $x = A\sin 2t + B\cos 2t$ | B1 | |
| PI: $x = P\sin 4t$ | M1 | |
| $-16P\sin 4t + 4P\sin 4t = \sin 4t$ | M1 | |
| $\Rightarrow P = -\tfrac{1}{12}$ | A1 | |
| $x = A\sin 2t + B\cos 2t - \tfrac{1}{12}\sin 4t$ | M1 | |
| $t=0, x=0 \Rightarrow B=0$ | A1 | |
| $\dot{x} = 2A\cos 2t - \tfrac{1}{3}\cos 4t$ | M1 | |
| $t=0, \dot{x}=0 \Rightarrow A=\tfrac{1}{6}$ | A1 ft | (8) |
| $\dot{x}=0 \Rightarrow \tfrac{1}{3}\cos 2t - \tfrac{1}{3}\cos 4t = 0$ | M1 | |
| $\Rightarrow \cos 4t = \cos 2t$ | | |
| $\Rightarrow 4t = 2t+2\pi \text{ or } 2\pi-2t$ | M1 | |
| $\Rightarrow t=\pi \text{ or } \tfrac{\pi}{3}$ | A1 | |
| First at rest after $t=0$ when $t=\dfrac{\pi}{3}$ | A1 cso | (4) |6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{c68c85a1-9d80-4ced-bfb6-c7b5347e9bb8-4_244_1264_1314_382}
\end{center}
\end{figure}
A particle $P$ of mass 2 kg is attached to the mid-point of a light elastic spring of natural length 2 m and modulus of elasticity 4 N . One end $A$ of the elastic spring is attached to a fixed point on a smooth horizontal table. The spring is then stretched until its length is 4 m and its other end $B$ is held at a point on the table where $A B = 4 \mathrm {~m}$. At time $t = 0 , P$ is at rest on the table at the point $O$ where $A O = 2 \mathrm {~m}$, as shown in Fig. 3. The end $B$ is now moved on the table in such a way that $A O B$ remains a straight line. At time $t$ seconds, $A B = \left( 4 + \frac { 1 } { 2 } \sin 4 t \right) \mathrm { m }$ and $A P = ( 2 + x ) \mathrm { m }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 x = \sin 4 t$$
\item Hence find the time when $P$ first comes to instantaneous rest.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q6 [17]}}