| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.2 This is a standard M4 mechanics question involving elastic springs and differential equations. Part (a) requires basic kinematics reasoning about relative motion. Part (b) applies Hooke's law and Newton's second law—routine for M4 students. Parts (c) and (d) involve applying initial conditions to find constants and then finding maximum tension using calculus, all following established procedures. While it requires multiple techniques and careful algebra across 18 marks, it doesn't demand novel insight—it's a textbook application of M4 methods, making it moderately above average difficulty. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation4.10d 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 |
|---|---|---|
| - The end of string has moved \(\frac{1}{2}f t^2\), so \(a + \frac{1}{2}ft^2 = x + a + y\), giving \(x + y = \frac{1}{2}ft^2\) | B1 | Constraint relationship |
| Part (b): \(F = ma\) to particle: \(T = m\ddot{x}\) | M1 | Newton's second law on particle |
| \(\frac{man^2}{a}y = m\ddot{x}\) | M1 | Tension in string |
| \(n^2\left(\frac{1}{2}ft^2 - x\right) = \ddot{x}\) | M1 | Substitution using constraint |
| \(\ddot{x} + n^2x = \frac{1}{2}n^2ft^2\) | A1 | Differential equation |
| Part (c): At \(x = 0, t = 0\): \(0 = A - \frac{f}{n^2}, A = \frac{f}{n^2}\) | M1 | Particular integral constant |
| Differentiating: \(\dot{x} = -nA\sin nt + nB\cos nt + ft\) | M1 | First derivative |
| \(\dot{x} = 0, t = 0\): \(0 = nB, B = 0\) | A1 | Initial velocity condition |
| Part (d): Differentiating again: \(\ddot{x} = -n^2A\cos nt - n^2B\sin nt + f = -f\cos nt + f\) | M1 | Second derivative |
| \(T = m\ddot{x} = mf(1 - \cos nt)\) | A1 | Tension expression |
| Maximum value of \(2mf\) when \(\cos nt = -1\) | B1 | Maximum tension |
**Part (a):** At time $t$:
- The particle has moved distance $x$
- The string length is $(a + y)$
- The end of string has moved $\frac{1}{2}f t^2$, so $a + \frac{1}{2}ft^2 = x + a + y$, giving $x + y = \frac{1}{2}ft^2$ | B1 | Constraint relationship |
**Part (b):** $F = ma$ to particle: $T = m\ddot{x}$ | M1 | Newton's second law on particle |
$\frac{man^2}{a}y = m\ddot{x}$ | M1 | Tension in string |
$n^2\left(\frac{1}{2}ft^2 - x\right) = \ddot{x}$ | M1 | Substitution using constraint |
$\ddot{x} + n^2x = \frac{1}{2}n^2ft^2$ | A1 | Differential equation |
**Part (c):** At $x = 0, t = 0$: $0 = A - \frac{f}{n^2}, A = \frac{f}{n^2}$ | M1 | Particular integral constant |
Differentiating: $\dot{x} = -nA\sin nt + nB\cos nt + ft$ | M1 | First derivative |
$\dot{x} = 0, t = 0$: $0 = nB, B = 0$ | A1 | Initial velocity condition |
**Part (d):** Differentiating again: $\ddot{x} = -n^2A\cos nt - n^2B\sin nt + f = -f\cos nt + f$ | M1 | Second derivative |
$T = m\ddot{x} = mf(1 - \cos nt)$ | A1 | Tension expression |
Maximum value of $2mf$ when $\cos nt = -1$ | B1 | Maximum tension |A particle of mass $m$ is attached to one end $P$ of a light elastic spring $PQ$, of natural length $a$ and modulus of elasticity $man^2$. At time $t = 0$, the particle and the spring are at rest on a smooth horizontal table, with the spring straight but unstretched and uncompressed. The end $Q$ of the spring is then moved in a straight line, in the direction $PQ$, with constant acceleration $f$. At time $t$, the displacement of the particle in the direction $PQ$ from its initial position is $x$ and the length of the spring is $(a + y)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x + y = \frac{1}{2}ft^2$. [2]
\item Hence show that
$$\frac{d^2x}{dt^2} + n^2x = \frac{1}{2}n^2ft^2.$$ [6]
\end{enumerate}
You are given that the general solution of this differential equation is
$$x = A\cos nt + B\sin nt + \frac{1}{2}ft^2 - \frac{f}{n^2},$$
where $A$ and $B$ are constants.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the values of $A$ and $B$. [6]
\item Find the maximum tension in the spring. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q7 [18]}}