M4 January 2006 Q4

4. 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 \(x\).

(a) Show that \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0\). (5)

Given that the solution of this differential equation is \(x = \mathrm{e}^{-\omega t}(A\cos\omega t + B\sin\omega t)\), where \(A\) and \(B\) are constants,

(b) find \(A\) and \(B\). (4)

Spring of natural length a, equilibrium extension e, displacement x; tension T up, weight mg down, resistance 2a omega dx/dt up — Mark scheme (a)

Scheme	Marks
R\((\downarrow)\) \(\ m\dfrac{d^2x}{dt^2} = mg - T - 2m\omega\dfrac{dx}{dt}\qquad\) (4 terms)	M1 A1
\(m\dfrac{d^2x}{dt^2} = mg - \dfrac{2m\omega^2a}{a}(e + x) - 2m\omega\dfrac{dx}{dt}\)	M1
\(\rightarrow \dfrac{d^2x}{dt^2} + 2\omega\dfrac{dx}{dt} + 2\omega^2x = 0\qquad (*)\)	M1 A1
	(5)

Mark scheme (b)

Scheme	Marks
\(x = e^{-\omega t}(A\cos\omega t + b\sin\omega t)\)
\(t = 0,\ x = 0 \Rightarrow A = 0\)	B1
\(\dfrac{dx}{dt} = -\omega e^{-\omega t}.B\sin\omega t + e^{-\omega t}.B\omega\cos\omega t\qquad\) (use of product rule)	M1
\(t = 0,\ \dfrac{dx}{dt} = U\ \colon\ U = B\omega \Rightarrow B = \dfrac{U}{\omega}\)	M1 A1
	(4)

Notes

(Corrected from the printed mark scheme: the condition is printed as \(t - 0\).)

Mark scheme (c)

Scheme	Marks
\(\dfrac{dx}{dt} = -Ue^{-\omega t}\sin\omega t + Ue^{-\omega t}\cos\omega t = 0\)	M1
\(\Rightarrow \tan\omega t = 1\qquad\) (solve for \(\tan\omega t\))	M1
\(\Rightarrow t = \dfrac{\pi}{4\omega}\)	A1
	(3)
	(12 marks)