AQA Further Paper 1 2022 June — Question 1 1 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2022
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeFind period from given information
DifficultyModerate -0.8 This is a standard SHM recognition question requiring only recall of the formula relating ω² to the coefficient in d²x/dt² = -ω²x, then applying T = 2π/ω. It's a 1-mark multiple choice question testing direct formula application with no problem-solving or derivation required, making it easier than average even for Further Maths.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x

The displacement of a particle from its equilibrium position is \(x\) metres at time \(t\) seconds. The motion of the particle obeys the differential equation $$\frac{d^2x}{dt^2} = -9x$$ Calculate the period of its motion in seconds. Circle your answer. [1 mark] \(\frac{\pi}{9}\) \(\quad\) \(\frac{2\pi}{9}\) \(\quad\) \(\frac{\pi}{3}\) \(\quad\) \(\frac{2\pi}{3}\)

Question 1:
AnswerMarks Guidance
1Circles correct answer 2.2a
3
AnswerMarks Guidance
Total1
QMarking instructions AO
Question 1:
1 | Circles correct answer | 2.2a | B1 | 2π
3
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
The displacement of a particle from its equilibrium position is $x$ metres at time $t$ seconds.

The motion of the particle obeys the differential equation
$$\frac{d^2x}{dt^2} = -9x$$

Calculate the period of its motion in seconds.

Circle your answer.
[1 mark]

$\frac{\pi}{9}$ $\quad$ $\frac{2\pi}{9}$ $\quad$ $\frac{\pi}{3}$ $\quad$ $\frac{2\pi}{3}$

\hfill \mbox{\textit{AQA Further Paper 1 2022 Q1 [1]}}