M4 June 2015 Q5

Figure 2: spring AB on the floor with P at its midpoint O, AP = 3 m and PB = 3 m — Figure 2

A particle \(P\) of mass 1.5 kg is attached to the midpoint of a light elastic spring \(AB\), of natural length 2 m and modulus of elasticity 12 N. The end \(A\) of the spring is attached to a fixed point on a smooth horizontal floor. The end \(B\) is held at a point on the floor where \(AB = 6\) m.

At time \(t = 0\), \(P\) is at rest on the floor at the point \(O\), where \(AO = 3\) m, as shown in Figure 2. The end \(B\) is now moved along the floor in such a way that \(AOB\) remains a straight line and at time \(t\) seconds, \(t \geqslant 0\), \[AB = \left(6 + \frac{1}{4}\sin 2t\right)\text{ m}\]

At time \(t\) seconds, \(AP = (3 + x)\) m.

(a) Show that, for \(t \geqslant 0\), \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 16x = 2\sin 2t\] (5)

The general solution of this differential equation is \[x = C\cos 4t + D\sin 4t + \frac{1}{6}\sin 2t\] where \(C\) and \(D\) are constants.

(b) Find the time at which \(P\) first comes to instantaneous rest. (5)

Spring with P at O, tensions T1 towards A and T2 towards B; AP = 3 + x and PB = 6 + (1/4) sin 2t - 3 - x — Mark scheme (a)

Scheme	Marks
\(T_1 = \dfrac{12(2 + x)}{1}\)	B1
\(T_2 = 12\left(2 + \dfrac{1}{4}\sin 2t - x\right)\)	B1
\(1.5\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = T_2 - T_1 = 3\sin 2t - 24x\)	M1 A1
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 16x = 2\sin 2t\)	A1
	(5)

Mark scheme (b)

Scheme	Marks
\(t = 0,\ x = 0 \quad \Rightarrow C = 0\)	B1
\(t = 0,\ \dot{x} = 0 = 4D\cos 4t + \dfrac{1}{3}\cos 2t\)	M1
\(D = -\dfrac{1}{12}\)	A1
\(\dot{x} = 0 \Rightarrow \cos 4t = \cos 2t\)	M1
\(2\cos^2 2t - 1 = \cos 2t\)
\(\cos 2t = 1, -\dfrac{1}{2} \quad 2t = \dfrac{2\pi}{3},\quad t = \dfrac{\pi}{3}\ \ (1.05)\)	A1
	(5)
	(10 marks)

Notes

M1 At rest: set \(\dot{x} = 0\)

A1 Not \(\dfrac{1}{2}\cos^{-1}\left(\dfrac{-1}{2}\right)\)?