CAIE FP2 2011 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeFind period from given information
DifficultyStandard +0.3 This is a standard SHM problem requiring application of the energy equation v² = ω²(a² - x²) with two given conditions to form simultaneous equations. While it involves algebraic manipulation and understanding of SHM formulas, it follows a well-established method taught in Further Maths with no novel insight required, making it slightly easier than average.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x

1 A particle oscillates in simple harmonic motion with centre \(O\). When its distance from \(O\) is 3 m its speed is \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and when its distance from \(O\) is 4 m its speed is \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the period and amplitude of the motion.

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(16^2 = \omega^2(A^2 - 3^2)\), \(12^2 = \omega^2(A^2 - 4^2)\)M1 A1 Apply \(v^2 = \omega^2(A^2 - x^2)\) at both points
\(A = 5\) [m], \(\omega = 4\)B1 B1 Combine to find \(A\) and \(\omega\)
\(T = 2\pi/\omega = \pi/2\) or \(1.57\) [s]B1\(\sqrt{}\) Find period using \(\omega\)
Total: [5]
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $16^2 = \omega^2(A^2 - 3^2)$, $12^2 = \omega^2(A^2 - 4^2)$ | M1 A1 | Apply $v^2 = \omega^2(A^2 - x^2)$ at both points |
| $A = 5$ [m], $\omega = 4$ | B1 B1 | Combine to find $A$ and $\omega$ |
| $T = 2\pi/\omega = \pi/2$ or $1.57$ [s] | B1$\sqrt{}$ | Find period using $\omega$ |

**Total: [5]**

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1 A particle oscillates in simple harmonic motion with centre $O$. When its distance from $O$ is 3 m its speed is $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and when its distance from $O$ is 4 m its speed is $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the period and amplitude of the motion.

\hfill \mbox{\textit{CAIE FP2 2011 Q1 [5]}}