A2 June 2019 Paper 2 Q6

OCR ACurrent spec6 marksSecond Order Differentials

6 \(A\) is a fixed point on a smooth horizontal surface. A particle \(P\) is initially held at \(A\) and released from rest.

It subsequently performs simple harmonic motion in a straight line on the surface. After its release it is next at rest after 0.2 seconds at point \(B\) whose displacement is 0.2 m from \(A\). The point \(M\) is halfway between \(A\) and \(B\).

The displacement of \(P\) from \(M\) at time \(t\) seconds after release is denoted by \(x\) m.

(a) On the axes below, sketch a graph of \(x\) against \(t\) for \(0 \leqslant t \leqslant 0.4\).

Blank axes from the Printed Answer Booklet: vertical x-axis and horizontal t-axis meeting at O

[4]

(b) Find the displacement of \(P\) from \(M\) at 0.75 seconds after release. [2]

Graph of x against t for 0 to 0.4: an inverted cosine curve starting at x = -0.1 when t = 0, crossing the t-axis at 0.1, reaching 0.1 at t = 0.2, crossing again at 0.3 and returning to -0.1 at t = 0.4 — Mark scheme (a)

Scheme	Marks	AO
	B1 B1 B1 B1	1.2 3.4 3.4 3.1b
	[4]

Mark scheme (b)

Scheme	Marks	AO
So \(x = \pm 0.1\cos(5\pi t)\) or \(x = 0.1\sin(5\pi t \pm \frac{1}{2}\pi)\) when \(t = 0.75\) or 0.35	M1	3.1b
\(x = -\dfrac{\sqrt{2}}{20}\) \((= -0.0707\) to 3 sf\()\)	A1	3.4
	[2]

Notes

M1: or by argument from sketch. Condone amplitude of 0.2 for M1