M4 June 2006 Q7

Figure 4: plane inclined at alpha; spring along a line of greatest slope from A down to particle P at B, AB = l — Figure 4

A light elastic spring has natural length \(l\) and modulus of elasticity \(4mg\). One end of the spring is attached to a point \(A\) on a plane that is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \tfrac{3}{4}\). The other end of the spring is attached to a particle \(P\) of mass \(m\). The plane is rough and the coefficient of friction between \(P\) and the plane is \(\tfrac{1}{2}\). The particle \(P\) is held at a point \(B\) on the plane where \(B\) is below \(A\) and \(AB = l\), with the spring lying along a line of greatest slope of the plane, as shown in Figure 4. At time \(t = 0\), the particle is projected up the plane towards \(A\) with speed \(\tfrac{1}{2}\sqrt{(gl)}\). At time \(t\), the compression of the spring is \(x\).

(a) Show that \(\quad\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\omega^2x = -g,\ \) where \(\omega = \sqrt{\left(\dfrac{g}{l}\right)}\). (6)

(b) Find \(x\) in terms of \(l\), \(\omega\) and \(t\). (7)

Particle on the inclined plane with acceleration x double-dot up the plane, normal reaction R, thrust T of the spring and friction half R down the plane, weight — Mark scheme (a)

Scheme	Marks
\(F = \dfrac{1}{2}R\)	M1
\(R = mg\cos\alpha\)	B1
\(T = \dfrac{4mgx}{l}\)	B1
\((\nearrow)\colon\ -F - mg\sin\alpha - T = m\ddot{x}\)	M1 A1
\(-\dfrac{1}{2}.\dfrac{4}{5}mg - \dfrac{3}{5}mg - \dfrac{4mgx}{l} = m\ddot{x}\)
\(\Rightarrow \dfrac{d^2x}{dt^2} + 4\omega^2x = -g\ \ *\qquad \left(\omega = \sqrt{g/l}\right)\)	A1
	(6)

Mark scheme (b)

Scheme	Marks
\(m^2 + 4\omega^2 = 0 \Rightarrow m = \pm 2\omega\mathrm{i}\)
C.F. is \(\ x = A\sin 2\omega t + B\cos 2\omega t\)	M1
P.I. is \(\ x = \dfrac{-g}{4\omega^2} = -l/4\)	B1
G.S. is \(\ x = A\sin 2\omega t + B\cos 2\omega t - \dfrac{l}{4}\)	B1
\(t = 0,\ x = 0\colon\ \ B = l/4\)
\(\dot{x} = 2\omega A\cos 2\omega t - 2\omega B\sin 2\omega t\)	M1 A1
\(t = 0,\ \dot{x} = \tfrac{1}{2}\sqrt{gl}\colon\ \ \dfrac{\sqrt{gl}}{2} = 2\sqrt{\dfrac{g}{l}}A \Rightarrow A = l/4\)	M1
\(\Rightarrow x = \dfrac{l}{4}(\sin 2\omega t + \cos 2\omega t - 1)\)	A1
	(7)

Mark scheme (c)

Scheme	Marks
\(\dot{x} = 0 \Rightarrow 2\omega A\cos 2\omega t - 2\omega B\sin 2\omega t = 0\)	M1
\(\Rightarrow \tan 2\omega t = \dfrac{A}{B} = 1\)
\(\Rightarrow 2\omega t = \pi/4\quad\) (first value)	A1
\(\Rightarrow x = \dfrac{l}{4}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} - 1\right)\)	M1
\(= \dfrac{l}{4}\left(\sqrt{2} - 1\right)\)	A1
	(4)
	(17 marks)

Notes