M4 June 2013 (R) Q7

EdexcelOld spec17 marksFurther Dynamics Variable Force & Kinematics

7. A particle \(P\) of mass 0.5 kg is attached to the end \(A\) of a light elastic spring \(AB\), of natural length 0.6 m and modulus of elasticity 2.7 N. At time \(t = 0\) the end \(B\) of the spring is held at rest and \(P\) hangs at rest at the point \(C\) which is vertically below \(B\). The end \(B\) is then moved along the line of the spring so that, at time \(t\) seconds, the downwards displacement of \(B\) from its initial position is \(4\sin 2t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of \(P\) below \(C\) is \(y\) metres.

(a) Show that \[y + \frac{49}{45} = x + 4\sin 2t\] (3)

(b) Hence show that \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 9y = 36\sin 2t\] (5)

Given that \(y = \dfrac{36}{5}\sin 2t\) is a particular integral of this differential equation,

(d) find the speed of \(P\) when \(t = \dfrac{1}{3}\pi\). (4)

Diagram: B displaced 4 sin 2t; P in equilibrium with T up and 0.5g down, then displaced y with acceleration y double dot — Mark scheme (a)

Scheme	Marks
In equilibrium \(\quad T = 0.5g = \dfrac{2.7e}{0.6}\)	M1
\(e = \dfrac{g}{9} = \dfrac{9.8}{9} = \dfrac{49}{45}\)	A1
\(0.6 + \dfrac{49}{45} - 4\sin 2t + y = 0.6 + x\)
\(y + \dfrac{49}{45} = x + 4\sin 2t\)	A1
	(3)

Mark scheme (b)

Scheme	Marks
\(0.5g - \dfrac{2.7x}{0.6} = 0.5\ddot{y}\)	M1A1
\(g - 9x = \ddot{y}\)
\(g - 9\left(y + \dfrac{g}{9} - 4\sin 2t\right) = \ddot{y}\)	DM1 A1
\(\ddot{y} + 9y = 36\sin 2t\)	A1
	(5)

Notes

M1A1 Equation of motion for \(P\)

DM1 Substitute for \(x\)

A1 Given Answer

Mark scheme (c)

Scheme	Marks
C.F. is \(y = A\cos 3t + B\sin 3t\)	M1
Gen. soln. is \(y = A\cos 3t + B\sin 3t + \dfrac{36}{5}\sin 2t\)	A1
\(t = 0\ \ y = 0\ \Rightarrow A = 0\)	B1
\(\dot{y} = 3B\cos 3t + \dfrac{72}{5}\cos 2t\)	M1
\(t = 0\ \ \dot{y} = 0\ \Rightarrow 3B = -\dfrac{72}{5} \qquad B = -\dfrac{24}{5}\)
\(\therefore\ y = -\dfrac{24}{5}\sin 3t + \dfrac{36}{5}\sin 2t\)	A1
	(5)

Notes

M1 Independent. Differentiate and use initial conditions to find \(B\)

Mark scheme (d)

Scheme	Marks
\(\dot{y} = -\dfrac{72}{5}\cos 3t + \dfrac{72}{5}\cos 2t\)	M1A1
\(\dot{y} = -\dfrac{72}{5}\cos\pi + \dfrac{72}{5}\cos\dfrac{2}{3}\pi\)	M1
\(\dot{y} = 7.2\)	A1
	(4)
	(17 marks)

Notes

M1 Substitute \(t = \dfrac{\pi}{3}\) in derivative to find \(\dot{y}\)

A1 Final answer