A2 June 2019 Paper 1 Q11
11 A particle is suspended in a resistive medium from one end of a light spring. The other end of the spring is attached to a point which is made to oscillate in a vertical line.
The displacement of the particle may be modelled by the differential equation
\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\frac{\mathrm{d}x}{\mathrm{d}t} + 5x = 10\sin t\]where \(x\) is the displacement of the particle below the equilibrium position at time \(t\).
When \(t = 0\) the particle is stationary and its displacement is 2.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 5x = 10\sin t\) A.E. \(n^2 + 2n + 5 = 0 \Rightarrow n = -1 \pm 2\mathrm{i}\) | M1 | 1.1a |
| \(\Rightarrow x = \mathrm{e}^{-t}(A\cos 2t + B\sin 2t)\) | A1 | 1.1 |
| P.I. \(x = a\sin t + b\cos t\) \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = a\cos t - b\sin t\) \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -a\sin t - b\cos t\) | M1 | 3.1a |
| \(\Rightarrow -a\sin t - b\cos t + 2(a\cos t - b\sin t) + 5(a\sin t + b\cos t) = 10\sin t\) \(\Rightarrow -b + 2a + 5b = 0,\ -a - 2b + 5a = 10\) | M1 | 1.1 |
| i.e. \(a = -2b,\ 2a - b = 5 \Rightarrow b = -1,\ a = 2\) | A1 | 1.1 |
| \(\Rightarrow\) G.S. \(x = \mathrm{e}^{-t}(A\cos 2t + B\sin 2t) + 2\sin t - \cos t\) | A1 | 1.1 |
| When \(t = 0\) we have \(x = 2\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\) | B1 | 3.3 |
| \(\Rightarrow A - 1 = 2 \Rightarrow A = 3\) | B1 | 3.3 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -\mathrm{e}^{-t}(A\cos 2t + B\sin 2t) + \mathrm{e}^{-t}(-2A\sin 2t + 2B\cos 2t) + 2\cos t + \sin t\) | M1 | 3.4 |
| \(t = 0 \Rightarrow -A + 2B + 2 = 0 \Rightarrow B = \dfrac{1}{2}\) | A1 | 3.3 |
| \(\Rightarrow x = \mathrm{e}^{-t}\left(3\cos 2t + \dfrac{1}{2}\sin 2t\right) + 2\sin t - \cos t\) | A1 | 3.4 |
| [11] |
Notes
M1: For obtaining the AE
A1: Solving AE and interpreting. Or equivalent forms (i.e. exponential form)
M1: Correct form of PI and differentiate twice
M1: Dep on previous M. Substitute their PI with two constants into DE
A1: Values of \(a\) and \(b\)
A1: Correct GS
B1: For substituting into 2 eqns
B1: For \(A\)
M1: Dep on previous M. For diffn their GS. Their value for A could be substituted
A1: For B
| Scheme | Marks | AO |
|---|---|---|
| \(x = \mathrm{e}^{-t}(A\cos 2t + B\sin 2t) + 2\sin t - \cos t\) As \(t \to \infty,\ \mathrm{e}^{-t} \to 0\) | M1 | 3.4 |
| \(\Rightarrow x \approx 2\sin t - \cos t\) | A1 | 3.4 |
| [2] |
Notes
M1: Consider behaviour of \(\mathrm{e}^{-t}\)
A1: Accept =
ft their equation from (a)
Accept this final line for 2 marks