M4 June 2017 Q6
6. A particle \(P\) of mass 0.2 kg is suspended from a fixed point by a light elastic spring. The spring has natural length 0.8 m and modulus of elasticity 7 N. At time \(t = 0\) the particle is released from rest from a point 0.2 metres vertically below its equilibrium position. The motion of \(P\) is resisted by a force of magnitude \(2v\) newtons, where \(v\) m s\(^{-1}\) is the speed of \(P\). At time \(t\) seconds, \(P\) is \(x\) metres below its equilibrium position.
| Scheme | Marks |
|---|---|
| \(\dfrac{7e}{0.8} = 0.2g\) | B1 |
| \(0.2\ddot{x} = 0.2g - \dfrac{7(x + e)}{0.8} - 2v\) | M1 |
| A1 | |
| \(\left(0.2\ddot{x} = -\dfrac{7x}{0.8} - 2\dot{x}\right)\) | A1 |
| \(\ddot{x} + 10\dot{x} + 43.75x = 0\) | A1 |
| (5) |
Notes
B1 \(T = mg\) at equilibrium \((e = 0.224)\)
M1 Equation of motion. No missing/additional terms. Condone sign error(s). Allow with \(m\) and \(l\)
A1 At most one error. \(m\) and \(l\) substituted
A1 Correct unsimplified equation. \(m\) and \(l\) substituted
A1 *given answer*
| Scheme | Marks |
|---|---|
| AE: \(m^2 + 10m + 43.75 = 0,\) \(m = -5 \pm \mathrm{i}\dfrac{5\sqrt{3}}{2}\) | M1 |
| \(x = \mathrm{e}^{-5t}\left(A\cos\dfrac{5\sqrt{3}}{2}t + B\sin\dfrac{5\sqrt{3}}{2}t\right)\) | A1 |
| \(t = 0, x = 0.2\quad \Rightarrow A = 0.2\) | B1 |
| \(\dot{x} = -5\mathrm{e}^{-5t}(A\cos\omega t + B\sin\omega t)\) \(+\ \mathrm{e}^{-5t}(-A\omega\sin\omega t + B\omega\cos\omega t)\) | M1 |
| \(0 = -5A + B\omega = -1 + B\omega,\quad B = \dfrac{2}{5\sqrt{3}}\) | M1 |
| \(x = \dfrac{\mathrm{e}^{-5t}}{5}\left(\cos\dfrac{5\sqrt{3}}{2}t + \dfrac{2}{\sqrt{3}}\sin\dfrac{5\sqrt{3}}{2}t\right)\) | A1 |
| (6) |
Notes
M1 Form and solve AE
M1 Use of \(t = 0, \dot{x} = 0\) to find \(B\)
| Scheme | Marks |
|---|---|
| instantaneous rest when \(t = \dfrac{\pi}{\omega}\) (s) | M1 |
| \(= \dfrac{2\pi}{5\sqrt{3}} = 0.73\) | A1 |
| (2) | |
| (13 marks) |
Notes
M1 Use of periodic time \(= \dfrac{2\pi}{\omega}\)
6c alt
| \(\dot{x} = -5\mathrm{e}^{-5t}\left(\dfrac{1}{5}\cos\omega t + \dfrac{2}{5\sqrt{3}}\sin\omega t\right)\) \(+\ \mathrm{e}^{-5t}\left(-\dfrac{1}{5}\omega\sin\omega t + \dfrac{2}{5\sqrt{3}}\omega\cos\omega t\right)\) | |
| \(= -\dfrac{7\sqrt{3}}{6}\mathrm{e}^{-5t}\sin\dfrac{5\sqrt{3}}{2}t = 0\) | M1 |
| \(\Rightarrow \dfrac{5\sqrt{3}}{2}t = \pi,\quad t = \dfrac{2\pi}{5\sqrt{3}} = 0.73\) s | A1 |
| (2) |
M1 Solve \(\dot{x} = 0\)
(Corrected from the printed mark scheme: the coefficient in the second line is printed as \(\sqrt{3}\); it is \(-\dfrac{7\sqrt{3}}{6}\).)