| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: given force or equation of motion directly |
| Difficulty | Standard +0.8 This is a standard M4 damped harmonic motion problem requiring force equation setup, application of given solution to find constants from initial conditions, and finding when velocity equals zero. While it involves multiple steps and careful algebraic manipulation (especially differentiating the exponential-trigonometric product in part c), the techniques are all standard for M4 with no novel insight required. The difficulty is elevated above average due to the algebraic complexity and the three-part structure requiring sustained accuracy. |
| 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 |
|---|---|---|
| \(\Rightarrow \frac{d^2x}{dt^2} + 2\omega\frac{dx}{dt} + 2\omega^2 x = 0\) (*) | M1 A1, M1, M1 A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 0, \frac{dx}{dt} = U : U = B\omega \Rightarrow B = \frac{U}{\omega}\) | B1, M1, M1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow t = \frac{\pi}{4\omega}\) | M1, M1, A1 | (3 marks) |
## Part (a)
R(↓): $m\frac{d^2x}{dt^2} = mg - T - 2m\omega\frac{dx}{dt}$ (4 terms)
$m\frac{d^2x}{dt^2} = mg - \frac{2m\omega^2 a}{a}(e+x) - 2m\omega\frac{dx}{dt}$
$\Rightarrow \frac{d^2x}{dt^2} + 2\omega\frac{dx}{dt} + 2\omega^2 x = 0$ (*) | M1 A1, M1, M1 A1 | (5 marks)
## Part (b)
$x = e^{-\alpha t}(A\cos\omega t + b\sin\omega t)$
$t = 0, x = 0 \Rightarrow A = 0$
$\frac{dx}{dt} = -\omega e^{-\alpha t}B\sin\omega t + e^{-\alpha t}B\omega\cos\omega t$ (use of product rule)
$t = 0, \frac{dx}{dt} = U : U = B\omega \Rightarrow B = \frac{U}{\omega}$ | B1, M1, M1 A1 | (4 marks)
## Part (c)
$\frac{dx}{dt} = -Ue^{-\alpha t}\sin\omega t + Ue^{-\alpha t}\cos\omega t = 0$
$\Rightarrow \tan\omega t = 1$ (solve for $\tan\omega t$)
$\Rightarrow t = \frac{\pi}{4\omega}$ | M1, M1, A1 | (3 marks)
**Total: 12 marks**
---A particle $P$ of mass $m$ is suspended from a fixed point by a light elastic spring. The spring has natural length $a$ and modulus of elasticity $2m\omega^2a$, where $\omega$ is a positive constant. At time $t = 0$ the particle is projected vertically downwards with speed $U$ from its equilibrium position. The motion of the particle is resisted by a force of magnitude $2m\omega v$, where $v$ is the speed of the particle. At time $t$, the displacement of $P$ downwards from its equilibrium position is $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega \frac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0$.
[5]
Given that the solution of this differential equation is $x = e^{-\omega t}(A \cos \omega t + B \sin \omega t)$, where $A$ and $B$ are constants,
\item find $A$ and $B$.
[4]
\item Find an expression for the time at which $P$ first comes to rest.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2006 Q4 [12]}}