| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Challenging +1.2 This is a standard M4/FM mechanics question on damped SHM with two springs. Part (a) is routine equilibrium with Hooke's law, part (b) requires setting up forces and applying F=ma (standard technique), and part (c) solves a second-order ODE with given auxiliary equation roots. While it requires multiple steps and careful algebra, all techniques are standard M4 material with no novel insight required—slightly above average due to the two-spring setup and damped motion. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_1 = \dfrac{2mge}{a}\); \(T_2 = \dfrac{mg(2a-e)}{a}\) | B1 | Either tension |
| \(T_1 = T_2 \Rightarrow 2e = (2a-e)\) | M1 A1 | |
| \(e = \dfrac{2a}{3}\) | ||
| \(AP = a + \dfrac{2a}{3} = \dfrac{5a}{3}\) ** | A1 | Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_2 - T_1 - 4m\omega\dot{x} = m\ddot{x}\) | ||
| \(\dfrac{mg}{a}\!\left(\dfrac{4a}{3}-x\right) - \dfrac{2mg}{a}\!\left(\dfrac{2a}{3}+x\right) - 4m\omega\dot{x} = m\ddot{x}\) | M1 A3 | |
| \(\ddot{x} + 4\omega\dot{x} + \dfrac{3g}{a}x = 0\) | ||
| \(\ddot{x} + 4\omega\dot{x} + 3\omega^2 x = 0\) ** | A1 | Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda^2 + 4\omega\lambda + 3\omega^2 = 0\) | ||
| \((\lambda+3\omega)(\lambda+\omega)=0\) | M1 | |
| \(\lambda = -3\omega\) or \(\lambda = -\omega\) | ||
| \(x = Ae^{-\omega t} + Be^{-3\omega t}\) | A1 | |
| \(\dot{x} = -\omega Ae^{-\omega t} - 3\omega Be^{-3\omega t}\) | M1 A1 | |
| \(t=0,\ x=\tfrac{1}{2}a,\ \dot{x}=0\) | M1 | Apply initial conditions |
| \(\tfrac{1}{2}a = A+B\) | ||
| \(0 = -\omega A - 3\omega B\) | A1 | |
| \(A = \tfrac{3}{4}a,\ B = -\tfrac{1}{4}a\) | A1 | |
| \(\dot{x} = v = \tfrac{3}{4}a\omega(e^{-3\omega t} - e^{-\omega t})\) | A1 | Total: 8 marks |
## Question 6:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_1 = \dfrac{2mge}{a}$; $T_2 = \dfrac{mg(2a-e)}{a}$ | B1 | Either tension |
| $T_1 = T_2 \Rightarrow 2e = (2a-e)$ | M1 A1 | |
| $e = \dfrac{2a}{3}$ | | |
| $AP = a + \dfrac{2a}{3} = \dfrac{5a}{3}$ ** | A1 | Total: 4 marks |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_2 - T_1 - 4m\omega\dot{x} = m\ddot{x}$ | | |
| $\dfrac{mg}{a}\!\left(\dfrac{4a}{3}-x\right) - \dfrac{2mg}{a}\!\left(\dfrac{2a}{3}+x\right) - 4m\omega\dot{x} = m\ddot{x}$ | M1 A3 | |
| $\ddot{x} + 4\omega\dot{x} + \dfrac{3g}{a}x = 0$ | | |
| $\ddot{x} + 4\omega\dot{x} + 3\omega^2 x = 0$ ** | A1 | Total: 5 marks |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda^2 + 4\omega\lambda + 3\omega^2 = 0$ | | |
| $(\lambda+3\omega)(\lambda+\omega)=0$ | M1 | |
| $\lambda = -3\omega$ or $\lambda = -\omega$ | | |
| $x = Ae^{-\omega t} + Be^{-3\omega t}$ | A1 | |
| $\dot{x} = -\omega Ae^{-\omega t} - 3\omega Be^{-3\omega t}$ | M1 A1 | |
| $t=0,\ x=\tfrac{1}{2}a,\ \dot{x}=0$ | M1 | Apply initial conditions |
| $\tfrac{1}{2}a = A+B$ | | |
| $0 = -\omega A - 3\omega B$ | A1 | |
| $A = \tfrac{3}{4}a,\ B = -\tfrac{1}{4}a$ | A1 | |
| $\dot{x} = v = \tfrac{3}{4}a\omega(e^{-3\omega t} - e^{-\omega t})$ | A1 | Total: 8 marks |
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\item Two points $A$ and $B$ lie on a smooth horizontal table with $A B = 4 a$. One end of a light elastic spring, of natural length $a$ and modulus of elasticity $2 m g$, is attached to $A$. The other end of the spring is attached to a particle $P$ of mass $m$. Another light elastic spring, of natural length $a$ and modulus of elasticity $m g$, has one end attached to $B$ and the other end attached to $P$. The particle $P$ is on the table at rest and in equilibrium.\\
(a) Show that $A P = \frac { 5 a } { 3 }$.
\end{enumerate}
The particle $P$ is now moved along the table from its equilibrium position through a distance $0.5 a$ towards $B$ and released from rest at time $t = 0$. At time $t , P$ is moving with speed $v$ and has displacement $x$ from its equilibrium position. There is a resistance to motion of magnitude $4 m \omega v$ where $\omega = \sqrt { } \left( \frac { g } { a } \right)$.\\
(b) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + 3 \omega ^ { 2 } x = 0$.\\
(c) Find the velocity, $\frac { \mathrm { d } x } { \mathrm {~d} t }$, of $P$ in terms of $a , \omega$ and $t$.
\hfill \mbox{\textit{Edexcel M4 2010 Q6 [17]}}