M4 June 2012 Q6

6. Two points \(A\) and \(B\) are in a vertical line, with \(A\) above \(B\) and \(AB = 4a\). One end of a light elastic spring, of natural length \(a\) and modulus of elasticity \(3mg\), is attached to \(A\). The other end of the spring is attached to a particle \(P\) of mass \(m\). Another light elastic spring, of natural length \(a\) and modulus of elasticity \(mg\), has one end attached to \(B\) and the other end attached to \(P\). The particle \(P\) hangs at rest in equilibrium.

(a) Show that \(AP = \dfrac{7a}{4}\) (3)

The particle \(P\) is now pulled down vertically from its equilibrium position towards \(B\) and at time \(t = 0\) it is released from rest. At time \(t\), the particle \(P\) is moving with speed \(v\) and has displacement \(x\) from its equilibrium position. The particle \(P\) is subject to air resistance of magnitude \(mkv\), where \(k\) is a positive constant.

(b) Show that\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + k\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{4g}{a}x = 0\] (5)

Mark scheme (a)

Scheme	Marks
\(T_1 = mg + T_2\)	M1
\(\dfrac{3mge}{a} = mg + \dfrac{mg(2a - e)}{a}\)	A1
\(e = \dfrac{3a}{4} \Rightarrow AP = \dfrac{7a}{4}\ *\)	A1
	(3)

Notes

M1 No resultant force and use of Hooke’s law

A1 Correct equation in one unknown. \(\dfrac{3mg(AP - a)}{a} = mg + \dfrac{mg(3a - AP)}{a}\), \(3AP - 3a = a + 3a - AP\)

A1 Derive given result correctly. Condone verification for 3/3

Mark scheme (b)

Scheme	Marks
\(mg + T_2 - T_1 - mkv = m\ddot{x}\)	M1 A1
\(mg + \dfrac{mg\left(\frac{5}{4}a - x\right)}{a} - \dfrac{3mg\left(\frac{3}{4}a + x\right)}{a} - mkv = m\ddot{x}\)	DM1 A1
\(\ddot{x} + k\dot{x} + \dfrac{4g}{a}x = 0\ *\)	A1
	(5)

Notes

M1 Equation of motion – requires all terms but condone sign errors.

A1 o.e. Correct equation in \(T_1\) & \(T_2\).

DM1 Use Hooke’s law with extensions of the form \(ka \pm x\)

A1 o.e. Correct unsimplified

A1 Given answer derived correctly

Mark scheme (c)

Scheme	Marks
For a damped oscillation, \(k^2 \lt \dfrac{16g}{a}\)	M1 A1
i.e. \(k \lt 4\sqrt{\dfrac{g}{a}}\)	A1
	(3)
	(11 marks)

Notes

M1 AE will have complex roots

A1 Correctly substituted inequality

A1 Only (Q gives k>0) \(-4\sqrt{\dfrac{g}{a}} \lt k \lt 4\sqrt{\dfrac{g}{a}}\) is A0.