A2 June 2019 Paper 1 Q14
14 In this question use \(g = 10\) m s−2
A light spring is attached to the base of a long tube and has a mass \(m\) attached to the other end, as shown in the diagram.
The tube is filled with oil.
When the compression of the spring is \(\varepsilon\) metres, the thrust in the spring is \(9m\varepsilon\) newtons.
The mass is held at rest in a position where the compression of the spring is \(\dfrac{20}{9}\) metres.
The mass is then released from rest. During the subsequent motion the oil causes a resistive force of \(6mv\) newtons to act on the mass, where \(v\) m s−1 is the speed of the mass.
At time \(t\) seconds after the mass is released, the displacement of the mass above its starting position is \(x\) metres.
| Scheme | Marks | AO |
|---|---|---|
| Forms general force equation (at least three terms) with at least two terms correct (allow equivalent notation for derivatives – condone \(a\) and \(v\)). | M1 | 3.1b |
| Obtains fully correct general force equation & cancels down into 2nd order DE form (allow equivalent notation for derivatives). | A1 | 1.1b |
| Obtains correct solution to their Auxiliary Equation | M1 | 1.1a |
| Obtains their correct RHS of Complementary Function | A1F | 1.1b |
| Obtains their correct (non-zero) Particular Integral | B1F | 1.1b |
| Obtains correct RHS of General Solution (ft their CF & non-zero PI, but must have two unknowns) | A1F | 2.2a |
| Uses \(x = 0\) when \(t = 0\) to obtain correct \(A\) | B1 | 1.1b |
| Sets their correct \(\dot{x} = 0\) when \(t = 0\) | M1 | 3.3 |
| Obtains correct \(B\) – can be unsimplified. | A1 | 1.1b |
| Obtains correct final equation – can be unsimplified. | R1 | 2.1 |
Typical solution
\[9m\left(\frac{20}{9} - x\right) - mg - 6m\dot{x} = m\ddot{x}\]\[\ddot{x} + 6\dot{x} + 9x = 10\]\[\lambda^2 + 6\lambda + 9 = 0\]\[\lambda = -3 \ (\text{twice})\]CF:
\[x = A\mathrm{e}^{-3t} + Bt\mathrm{e}^{-3t}\]PI:
\[x = \frac{10}{9}\]General Solution:
\[x = A\mathrm{e}^{-3t} + Bt\mathrm{e}^{-3t} + \frac{10}{9}\]\[x = 0,\ t = 0 \Rightarrow A = \frac{-10}{9}\]\[\dot{x} = -3A\mathrm{e}^{-3t} + B\mathrm{e}^{-3t} - 3Bt\mathrm{e}^{-3t}\]\[0 = -3A + B\]\[B = -\frac{30}{9}\]\[x = -\frac{10}{9}\mathrm{e}^{-3t} - \frac{10}{3}t\mathrm{e}^{-3t} + \frac{10}{9}\]| Scheme | Marks | AO |
|---|---|---|
| States critical damping because the Auxiliary Equation has equal roots (or equivalent) | B1 | 1.2 |
| (11 marks) |
Typical solution
Critical damping, because the Auxiliary Equation has equal roots