CAIE FP2 2010 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeMaximum acceleration in SHM
DifficultyStandard +0.3 This is a straightforward SHM problem requiring standard formulas (ω = 2π/T, amplitude = 0.3m, F_max = mω²a) with direct substitution. While it's Further Maths content, it involves only routine application of well-known SHM relationships with no conceptual challenges or multi-step reasoning, making it slightly easier than average overall.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x

A particle \(P\), of mass \(0.2\) kg, moves in simple harmonic motion along a straight line under the action of a resultant force of magnitude \(F\) N. The distance between the end-points of the motion is \(0.6\) m, and the period of the motion is \(0.5\) s. Find the greatest value of \(F\) during the motion. [5]

Question 1:
AnswerMarks
1Find ω or ω2 from 2π/T: ω = 2π / 0⋅5 [= 4π = 12⋅57] B1
Relate F to acceleration: F = 0⋅2 d 2 x/dt 2 M1
Relate acceleration to ω and x: d 2 x/dt 2 = [–]ω 2 x M1
State or use value of x giving max of F [or d 2 x/dt 2 ]: Maximum when x = [±] 0⋅3 M1
Evaluate maximum F of F: 0⋅2 (4π) 2 0⋅3 = 0⋅96π2 or 9⋅47 A1
AnswerMarks Guidance
max5 [5]
Question 1:
1 | Find ω or ω2 from 2π/T: ω = 2π / 0⋅5 [= 4π = 12⋅57] B1
Relate F to acceleration: F = 0⋅2 d 2 x/dt 2 M1
Relate acceleration to ω and x: d 2 x/dt 2 = [–]ω 2 x M1
State or use value of x giving max of F [or d 2 x/dt 2 ]: Maximum when x = [±] 0⋅3 M1
Evaluate maximum F of F: 0⋅2 (4π) 2 0⋅3 = 0⋅96π2 or 9⋅47 A1
max | 5 | [5]
A particle $P$, of mass $0.2$ kg, moves in simple harmonic motion along a straight line under the action of a resultant force of magnitude $F$ N. The distance between the end-points of the motion is $0.6$ m, and the period of the motion is $0.5$ s. Find the greatest value of $F$ during the motion. [5]

\hfill \mbox{\textit{CAIE FP2 2010 Q1 [5]}}