| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2007 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.8 This M4 question combines forced harmonic motion with moving support points, requiring students to set up geometric relationships, derive a second-order differential equation from Hooke's law and Newton's second law, solve it using given particular integral, and apply initial conditions. While the steps are systematic, the moving boundary condition adds conceptual complexity beyond standard SHM, and the multi-stage solution with careful bookkeeping of equilibrium position, extensions, and displacements requires strong technical facility typical of Further Maths mechanics. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10e Second order non-homogeneous: complementary + particular integral6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks |
|---|---|
| At E, \(\frac{2.45e}{0.5} = 0.1g\) | M1 |
| \(\Rightarrow e = 0.2\) | A1 |
| \(\Rightarrow 0.5(orl) + 0.2 + y = 2\sin 2t + 0.5(orl) + x\) | |
| \(\Rightarrow 0.2 + y = 2\sin 2t + x\) * | B1 |
| Total: 3 marks |
| Answer | Marks |
|---|---|
| \(0.1g - T = 0.1\ddot{y}\) | M*1 |
| R(\(\downarrow\)) \(0.1g - \frac{2.45x}{0.5} = 0.1\ddot{y}\) | M1 |
| \(0.98 - 4.9(0.2 + y - 2\sin 2t) = 0.1\ddot{y}\) | DM*1A1 |
| \((-4.9y + 9.8\sin 2t = 0.1\ddot{y})\) | |
| \(\Rightarrow \frac{d^2y}{dt^2} + 49y = 98\sin 2t\) * | A1 cso |
| Total: 5 marks |
| Answer | Marks |
|---|---|
| CF is \(y = A\cos 7t + B\sin 7t\) | M1 |
| Hence GS is \(y = A\cos 7t + B\sin 7t + \frac{98}{45}\sin 2t\) | A1 |
| \(t = 0, y = 0:\) \(0 = A\) so, \(y = B\sin 7t + \frac{98}{45}\sin 2t\) | B1 |
| \(\dot{y} = 7B\cos 7t + \frac{196}{45}\cos 2t\) | M1 |
| \(t = 0, \dot{y} = 0: 0 = 7B + \frac{196}{45}\) \(\Rightarrow B = -\frac{28}{45}\) | A1 |
| \(\Rightarrow y = \frac{14}{45}(7\sin 2t - 2\sin 7t)\) | A1 |
| Total: 5 marks |
| Answer | Marks |
|---|---|
| \(y = \frac{14}{45}(14\cos 2t - 14\cos 7t)\) | B1 |
| \(y = 0 \Rightarrow \cos 2t = \cos 7t\) | M1 |
| \(\Rightarrow 7t = 2k\pi \pm 2t\) | M1 |
| \(k = 1 \Rightarrow 9t = 2\pi\) (or \(5t = 2\pi\)) | A1 |
| \(t = \frac{2\pi}{9}\), accept \(0.698s, 0.70s.\) | A1 |
| Total: 4 marks |
## Part (a)
| At E, $\frac{2.45e}{0.5} = 0.1g$ | M1 |
| $\Rightarrow e = 0.2$ | A1 |
| $\Rightarrow 0.5(orl) + 0.2 + y = 2\sin 2t + 0.5(orl) + x$ | |
| $\Rightarrow 0.2 + y = 2\sin 2t + x$ * | B1 |
| **Total: 3 marks** | |
## Part (b)
| $0.1g - T = 0.1\ddot{y}$ | M*1 |
| R($\downarrow$) $0.1g - \frac{2.45x}{0.5} = 0.1\ddot{y}$ | M1 |
| $0.98 - 4.9(0.2 + y - 2\sin 2t) = 0.1\ddot{y}$ | DM*1A1 |
| $(-4.9y + 9.8\sin 2t = 0.1\ddot{y})$ | |
| $\Rightarrow \frac{d^2y}{dt^2} + 49y = 98\sin 2t$ * | A1 cso |
| **Total: 5 marks** | |
## Part (c)
| CF is $y = A\cos 7t + B\sin 7t$ | M1 |
| Hence GS is $y = A\cos 7t + B\sin 7t + \frac{98}{45}\sin 2t$ | A1 |
| $t = 0, y = 0:$ $0 = A$ so, $y = B\sin 7t + \frac{98}{45}\sin 2t$ | B1 |
| $\dot{y} = 7B\cos 7t + \frac{196}{45}\cos 2t$ | M1 |
| $t = 0, \dot{y} = 0: 0 = 7B + \frac{196}{45}$ $\Rightarrow B = -\frac{28}{45}$ | A1 |
| $\Rightarrow y = \frac{14}{45}(7\sin 2t - 2\sin 7t)$ | A1 |
| **Total: 5 marks** | |
## Part (d)
| $y = \frac{14}{45}(14\cos 2t - 14\cos 7t)$ | B1 |
| $y = 0 \Rightarrow \cos 2t = \cos 7t$ | M1 |
| $\Rightarrow 7t = 2k\pi \pm 2t$ | M1 |
| $k = 1 \Rightarrow 9t = 2\pi$ (or $5t = 2\pi$) | A1 |
| $t = \frac{2\pi}{9}$, accept $0.698s, 0.70s.$ | A1 |
| **Total: 4 marks** | |
### Guidance:
**a)** M1: Hooke's law to find extension at equilibrium. A1: cao. B1: Q specifies reference to a diagram. Correct reasoning leading to given answer.
**b)** M1: Use of F=ma. Weight, tension and acceleration. Condone sign errors. M1: Substitute for tension in terms of $x$. M1: Use given result to substitute for $x$ in terms of $y$. A1: Correct unsimplified equation. A1: Rearrange to given form cso.
**c)** M1: Correct form for CF. A1: GS for $y$ correct. B1: Deduce coefficient of $\cos\theta = 0$. M1: Differentiate their $y$ and substitute $t = 0, \dot{y} = 0$. A1: $y$ in terms of $t$. Any exact equivalent.
**d)** B1: $\dot{y}$ correct. M1: set $\dot{y} = 0$. M1: solve for general solution for $r: 7t = 2k\pi \pm 2t$. Or: $\sin\frac{9t}{2} \times\sin\frac{5t}{2} = 0 \Rightarrow \sin\frac{9t}{2} = 0$ or $\sin\frac{5t}{2} = 0$. A1: Select smallest value.
---A small ball is attached to one end of a spring. The ball is modelled as a particle of mass 0.1 kg and the spring is modelled as a light elastic spring $AB$, of natural length 0.5 m and modulus of elasticity 2.45 N. The particle is attached to the end $B$ of the spring. Initially, at time $t = 0$, $A$ is held at rest and the particle hangs at rest in equilibrium below $A$ at the point $E$. The end $A$ then begins to move along the line of the spring in such a way that, at time $t$ seconds, $t \leq 1$, the downward displacement of $A$ from its initial position is $2 \sin 2t$ metres. At time $t$ seconds, the extension of the spring is $x$ metres and the displacement of the particle below $E$ is $y$ metres.
\begin{enumerate}[label=(\alph*)]
\item Show, by referring to a simple diagram, that $y + 0.2 = x + 2 \sin 2t$. [3]
\item Hence show that $\frac{d^2y}{dt^2} + 49y = 98 \sin 2t$. [5]
\end{enumerate}
Given that $y = \frac{98}{45} \sin 2t$ is a particular integral of this differential equation,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find $y$ in terms of $t$. [5]
\item Find the time at which the particle first comes to instantaneous rest. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2007 Q6 [17]}}