A2 June 2022 Paper 1 Q15

OCR MEICurrent spec23 marksSecond Order Differentials

15 In an oscillating system, a particle of mass \(m\) kg moves in a horizontal line. Its displacement from its equilibrium position O at time \(t\) seconds is \(x\) metres, its velocity is \(v\) m s⁻¹, and it is acted on by a force \(2mx\) newtons acting towards O as shown in the diagram.

A particle on a horizontal dashed line, displaced x m to the right of O, moving with velocity v m/s to the right, with a force 2mx N acting on it towards O

Initially, the particle is projected away from O with speed 1 m s⁻¹ from a point 2 m from O in the positive direction.

(a)

(i) Show that the motion is modelled by the differential equation \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2x = 0\). [1]

(ii) State the type of motion. [1]

(iii) Write down the period of the motion. [1]

(iv) Find \(x\) in terms of \(t\). [4]

(v) Find the amplitude of the motion. [2]

(b) The motion is now damped by a force \(2mv\) newtons.

(i) Show that \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x = 0\). [1]

(ii) State, giving a reason, whether the system is under-damped, critically damped or over-damped. [1]

(iii) Determine the general solution of this differential equation. [3]

(c) Finally, a variable force \(2m\cos 2t\) newtons is added, so that the motion is now modelled by the differential equation
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x = 2\cos 2t\).

(i) Find \(x\) in terms of \(t\). [7]

In the long term, the particle is seen to perform simple harmonic motion with a period of just over 3 seconds.

(ii) Verify that this behaviour is consistent with the answer to part (c)(i). [2]

Mark scheme (a)

Scheme	Marks	AO
(i) \(m\dfrac{d^2x}{dt^2} = -2mx \Rightarrow \dfrac{d^2x}{dt^2} + 2x = 0\)	B1	3.1b
	[1]

Notes

B1: AG

Scheme	Marks	AO
(ii) Simple harmonic motion	B1	3.3
	[1]

Scheme	Marks	AO
(iii) \(\frac{2\pi}{\sqrt{2}} = \sqrt{2}\pi\) (s)	B1	1.2
	[1]

Notes

B1: Accept anything wrt 4.4

Scheme	Marks	AO
(iv) General solution: \(x = A\cos\sqrt{2}t + B\sin\sqrt{2}t\)	B1	1.1
\(t = 0,\ x = 2 \Rightarrow A = 2\)	B1	3.3
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -\sqrt{2}A\sin\sqrt{2}t + \sqrt{2}B\cos\sqrt{2}t\) \(t = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 1 \Rightarrow B = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)	M1	1.1
so \(x = 2\cos\sqrt{2}t + \frac{\sqrt{2}}{2}\sin\sqrt{2}t\)	A1	3.3
	[4]

Notes

B1: Must come from correct GS

M1: Must be differentiating a function in terms of cos and sin

Scheme	Marks	AO
(v) Amplitude \(= \sqrt{2^2 + \left(\dfrac{1}{2}\sqrt{2}\right)^2}\)	M1	3.4
\(= \frac{3\sqrt{2}}{2}\) (m)	A1	1.1
	[2]

Notes

A1: Accept anything wrt 2.1

Mark scheme (b)

Scheme	Marks	AO
(i) \(m\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -2mx - 2m\frac{\mathrm{d}x}{\mathrm{d}t}\) \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x = 0\)	B1	3.1b
	[1]

Notes

B1: AG

Scheme	Marks	AO
(ii) Underdamped as \(2^2 - 4 \times 1 \times 2 \lt 0\)	B1	3.5b
	[1]

Notes

B1: Stating complex roots of auxillary equation is acceptable \((-1 \pm i)\)

Scheme	Marks	AO
(iii) Auxiliary equation: \(\lambda^2 + 2\lambda + 2 = 0\)	M1	1.2
\(\lambda = \dfrac{-2 \pm \sqrt{-4}}{2} = -1 \pm i\)	A1	1.1
General solution: \(x = e^{-t}(A\cos t + B\sin t)\)	A1	1.1
	[3]

Notes

A1: soi

Mark scheme (c)

Scheme	Marks	AO
(i) Particular integral: \(x = C\cos 2t + D\sin 2t\) \(-4C + 4D + 2C = 2,\ -4D - 4C + 2D = 0\) \(\Rightarrow C = -0.2,\ D = 0.4\)	M1 M1 A1	2.1 3.3 2.2a
\(x = e^{-t}(A\cos t + B\sin t) - 0.2\cos 2t + 0.4\sin 2t\) \(t = 0,\ x = 2 \Rightarrow A = 2.2\)	B1	3.3
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -e^{-t}(A\cos t + B\sin t) + e^{-t}(-A\sin t + B\cos t)\) \(+0.4\sin 2t + 0.8\cos 2t\)	M1*	2.1
\(t = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 1 \Rightarrow 1 = -A + B + 0.8\) \(\Rightarrow B = 2.4\)	M1dep*	3.3
\(x = \mathrm{e}^{-t}(2.2\cos t + 2.4\sin t) - 0.2\cos 2t + 0.4\sin 2t\)	A1cao	2.2a
	[7]

Notes

B1: Has to come from a correct CF

M1*: Must include use of product rule from x = CF + PI

M1dep*: For substitution to lead to an equation in A and B

Scheme	Marks	AO
(ii) As \(t \to \infty,\ x \to -0.2\cos 2t + 0.4\sin 2t\) [this is SHM] with period \(\frac{2\pi}{2} = \pi \approx 3.14\) \(s\)	M1 A1	3.5a 3.5a
	[2]

Notes

M1: Dependent on a function containing \(e^{-kt}\) and \(\cos pt\) and \(\sin pt\)